diff git a/books/bookvol10.1.pamphlet b/books/bookvol10.1.pamphlet
index 97562f7..ad4fc73 100644
 a/books/bookvol10.1.pamphlet
+++ b/books/bookvol10.1.pamphlet
@@ 31,6 +31,12 @@
\newcommand{\pagefrom}[2]{% e.g. \pagefrom{name}{abb}
\ \\${\bf\Leftarrow{}}${``#1''} (#2) \ref{#1} on page~\pageref{#1}}
%%
+%% We need to use a dagger to mark footnotes because numbers can be
+%% mistaken as superscripts
+\renewcommand{\thefootnote}{\fnsymbol{footnote}}
+%% Before each footnote we explicitly set the counter to 1 so that
+%% when it increments it prints a dagger
+\setcounter{footnote}{1}
% special meanings for math characters
\providecommand{\N}{\mbox{\bbold N}}
\providecommand{\Natural}{\mbox{\bbold N}}
@@ 2408,6 +2414,5168 @@ Analysis or LSA. Although the domain is different, the concepts
are the same. We are trying to predict patterns of how words occur
in documents instead of trying to predict patterns of how players
score on holes.
+\chapter{Quaternions}
+\begin{center}
+\bigskip
+\large Peter Guthrie Tait, Robert S. Sutor, Timothy Daly
+
+\end{center}
+\section*{Preface}
+\addcontentsline{toc}{section}{Preface}
+The Theory of Quaternions is due to Sir William Rowan Hamilton,
+Royal Astronomer of Ireland, who presented his first paper on the
+subject to the Royal Irish Academy in 1843. His Lectures on
+Quaternions were published in 1853, and his Elements, in 1866,
+shortly after his death. The Elements of Quaternions by Tait\cite{33} is
+the accepted textbook for advanced students.
+
+Large portions of this file are derived from a public domain version
+of Tait's book combined with the algebra available in Axiom.
+The purpose is to develop a tutorial introduction to the Axiom
+domain and its uses.
+\newpage
+
+\section{Quaternions}
+
+\section{Vectors, and their Composition}
+
+{\bf 1}. For at least two centuries the geometrical representation
+of the negative and imaginary algebraic quantities, $1$ and $\sqrt{1}$
+has been a favourite subject of speculation with mathematicians.
+The essence of almost all of the proposed processes consists in
+employing such expressions to indicate the DIRECTION, not the
+{\sl length}, of lines.
+
+{\bf 2}. Thus it was long ago seen that if positive quantities were
+measured off in one direction along a fixed line, a useful and lawful
+convention enabled us to express negative quantities of the same
+kind by simply laying them off on the same line in the opposite
+direction. This convention is an essential part of the Cartesian
+method, and is constantly employed in Analytical Geometry and
+Applied Mathematics.
+
+{\bf 3}. Wallis, towards the end of the seventeenth century, proposed
+to represent the impossible roots of a quadratic equation by going
+{\sl out} of the line on which, if real, they would have been laid off.
+This construction is equivalent to the consideration of $\sqrt{1}$ as a
+directed unitline perpendicular to that on which real quantities
+are measured.
+
+{\bf 4}. In the usual notation of Analytical Geometry of two
+dimensions, when rectangular axes are employed, this amounts
+to reckoning each unit of length along $Oy$ as $+\sqrt{1}$, and on
+$Oy^{'}$ as $\sqrt{1}$ ; while on $Ox$ each unit is $+1$, and on
+$Ox$ it is $1$.
+
+If we look at these four lines in circular order, i.e. in the order of
+positive rotation (that of the northern hemisphere of the earth
+about its axis, or {\sl opposite} to that of the hands of a watch), they
+give
+$$ 1, \sqrt{1}, 1, \sqrt{1}$$
+
+\boxed{4.6in}{
+\vskip 0.1cm
+In Axiom the same elements would be written as complex numbers
+which are constructed using the function {\bf complex}:
+\spadcommand{complex(1,0)}
+$$1$$
+\returnType{Type: Complex Integer}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{complex(0,1)}
+$$\%i$$
+\returnType{Type: Complex Integer}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{complex(1,0)}
+$$1$$
+\returnType{Type: Complex Integer}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{complex(0,1)}
+$$i$$
+\returnType{Type: Complex Integer}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+Note that \%i is of type Complex(Integer), that is, the imaginary
+part of a complex number. The apparently equivalent expression
+\spadcommand{sqrt(1)}
+$$\sqrt{1}$$
+\returnType{Type: AlgebraicNumber}
+has the type AlgebraicNumber which means that it is the root of
+a polynomial with rational coefficients.\\
+}
+
+In this series each expression is derived from that which precedes
+it by multiplication by the factor $\sqrt{1}$. Hence we may consider
+$\sqrt{1}$ as an operator, analogous to a handle perpendicular to the
+plane of $xy$, whose effect on any line in that plane is to make it
+rotate (positively) about the origin through an angle of $90^{\circ}$.
+
+\boxed{4.6in}{
+\vskip 0.1cm
+In Axiom
+\spadcommand{\%i*\%i}
+$$1$$
+\returnType{Type: Complex Integer}
+}
+
+{\bf 5}. In such a system, (which seems to have been first developed,
+in 1805, by Bu\'ee) a point in the plane of reference is defined by a
+single imaginary expression. Thus $a + b\sqrt{1}$ may be considered
+as a single quantity, denoting the point, $P$, whose coordinates are
+$a$ and $b$. Or, it may be used as an expression for the line $OP$
+joining that point with the origin. In the latter sense, the expression
+$a + b\sqrt{1}$ implicitly contains the {\sl direction}, as well as the
+{\sl length}, of this line ; since, as we see at once, the direction is
+inclined at an angle $\tan^{1}(b/a)$ to the axis of $x$,
+and the length is $\sqrt{a^2+b^2}$. Thus, say we have
+$$OP = a + b\sqrt{1}$$
+the line $OP$ considered as that by which we pass from one
+extremity, $O$, to the other, $P$. In this sense it is called a VECTOR.
+Considering, in the plane, any other vector,
+$$OQ = a^{'}+b^{'}\sqrt{1}$$
+
+\boxed{4.6in}{
+\vskip 0.1cm
+In order to created superscripted variables we use the superscript
+function from the SYMBOL domain. So we can create $a^{'}$ as ``ap''
+(that is, ``aprime'') and $b^{'}$ as ``bp'' (``bprime'') thus
+(also note that the underscore character is Axiom's escape character
+which removes any special meaning of the next character, in this case,
+the quote character):
+\spadcommand{ap:=superscript(a,[\_'])}
+$$a^{'}$$
+\returnType{Type: Symbol}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{bp:=superscript(b,[\_'])}
+$$b^{'}$$
+\returnType{Type: Symbol}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+at this point we can type
+\spadcommand{ap+bp*\%i}
+$$a^{'}+b^{'}\ \%i$$
+\returnType{Type: Complex Polynomial Integer}
+}
+
+the addition of these two lines obviously gives
+$$OR = a + a^{'} + (b + b^{'})\sqrt{1}$$
+
+\boxed{4.6in}{
+\vskip 0.1cm
+In Axiom the computation looks like:
+\spadcommand{op:=complex(a,b)}
+$$a + b\ \%i$$
+\returnType{Type: Complex Polynomial Integer}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{oq:=complex(ap,bp)}
+$$a^{'} + b^{'}\ \%i$$
+\returnType{Type: Complex Polynomial Integer}
+}
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{op + oq}
+$$a + a^{'} + (b + b^{'})\%i$$
+\returnType{Type: Complex Polynomial Integer}
+}
+
+and we see that the sum is the diagonal of the parallelogram on
+$OP$, $OQ$. This is the law of the composition of simultaneous
+velocities; and it contains, of course, the law of subtraction of one
+directed line from another.
+
+{\bf 6}. Operating on the first of these symbols by the factor $\sqrt{1}$,
+it becomes $ b + a\sqrt{1}$; and now, of course, denotes the point
+whose $x$ and $y$ coordinates are $ b$ and $a$; or the line joining this
+point with the origin. The length is still $\sqrt{a^2+b^2}$, but the angle
+the line makes with the axis of $x$ is $\tan^{1}( a/b)$; which is
+evidently greater by $\pi/2$ than before the operation.
+
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{op*complex(0,1)}
+$$b+a\ i$$
+\returnType{Type: Complex Polynomial Integer}
+}
+
+{\bf 7}. De Moivre's Theorem tends to lead us still further in the
+same direction. In fact, it is easy to see that if we use, instead
+of $\sqrt{1}$, the more general factor $\cos \alpha + \sqrt{1} \sin \alpha$,
+its effect on
+any line is to turn it through the (positive) angle $\alpha$. in the plane
+of $x$, $y$. [Of course the former factor, $\sqrt{1}$, is merely the
+particular case of this, when $\alpha=\frac{\pi}{2}$].
+
+Thus
+$$
+\begin{array}{ll}
+ &(\cos \alpha + \sqrt{1} \sin \alpha) (a + b \sqrt{1})\\
+= & a \cos \alpha  b \sin \alpha + \sqrt{1} (a \sin \alpha + b \cos \alpha)
+\end{array}
+$$
+
+by direct multiplication. The reader will at once see that the new
+form indicates that a rotation through an angle $\alpha$ has taken place,
+if he compares it with the common formulae for turning the coordinate
+axes through a given angle. Or, in a less simple manner, thus
+
+$$
+\begin{array}{rcl}
+Length & = & \sqrt{(a \cos \alpha  b \sin \alpha)^2 +
+ (a \sin \alpha + b \cos \alpha)^2} \\
+ & = & \sqrt{a^2 + b^2}
+\end{array}
+$$
+as before.
+
+Inclination to axis of $x$
+$$
+\begin{array}{cl}
+= & \tan^{1}{\frac{a \sin \alpha + b \cos \alpha}
+ {a \cos \alpha  b \sin \alpha}}\\
+= & \tan^{1}{\frac{\tan \alpha + \frac{b}{a}}
+ {1  \frac{b}{a} \tan \alpha}}\\
+= & \alpha + \tan^{1}{\frac{b}{a}}
+\end{array}
+$$
+
+{\bf 8}. We see now, as it were, why it happens that
+
+$$(\cos \alpha + \sqrt{1} \sin \alpha)^m =
+\cos m\alpha + \sqrt{1} \sin m\alpha
+$$
+
+In fact, the first operator produces $m$ successive rotations in the
+same direction, each through the angle $\alpha$ ; the second, a single
+rotation through the angle $m\alpha$.
+
+{\bf 9}. It may be interesting, at this stage, to anticipate so far as to
+remark that in the theory of Quaternions the analogue of
+
+$$
+\begin{array}{lclr}
+ & \textrm{ } & \cos \theta + \sqrt{1} \sin \theta &\\
+\textrm{is} & \textrm{ } & \cos \theta + \omega \sin \theta &\\
+\textrm{where} & \textrm{ } & \omega^2 = 1 & \\
+\end{array}
+$$
+
+Here, however, $\omega$ is not the algebraic $\sqrt{1}$, but is
+{\sl any directed unitline} whatever in space.
+
+{\bf 10}. In the present century Argand, Warren, Mourey, and
+others, extended the results of Wallis and Bu\'ee. They attempted
+to express as a line the product of two lines each represented by a
+symbol such $a+b\sqrt{1}$. To a certain extent they succeeded,
+but all their results remained confined to two dimensions.
+
+The product, $\prod$, of two such lines was defined as the fourth
+proportional to unity and the two lines, thus
+
+$$
+\begin{array}{lclr}
+ & \textrm{ } &
+1 : a+b\sqrt{1} :: a^{'}+b^{'}\sqrt{1} : \prod\\
+\textrm{or} & \textrm{ } &
+\prod = (aa^{'}  bb^{'})+(a^{'}b+b^{'}a)\sqrt{1}
+\end{array}
+$$
+
+The length of $\prod$ is obviously the product of the lengths of the
+factor lines; and its direction makes an angle with the axis of $x$
+which is the sum of those made by the factor lines. From this
+result the quotient of two such lines follows immediately.
+
+{\bf 11}. A very curious speculation, due to Servois and published
+in 1813 in Gergonne's {\sl Annales}, is one of the very few, so far as has
+been discovered, in which a wellfounded guess at a possible mode
+of extension to three dimensions is contained. Endeavouring to
+extend to {\sl space} the form $a+b\sqrt{1}$ for the plane, he is guided by
+analogy to write for a directed unitline in space the form
+
+$$p \cos \alpha + q \cos \beta + r \cos \gamma$$
+
+where $\alpha$, $\beta$, $\gamma$
+are its inclinations to the three axes. He perceives
+easily that $p$, $q$, $r$ must be {\sl nonreals}:
+but, he asks, ``seraientelles
+{\sl imaginaires} r\'eductibles \`a la forme g\'en\'erale $A+B\sqrt{1}$?''
+The $i$,$j$, $k$ of the Quaternion Calculus furnish an answer to this
+question. (See Chap. II.) But it may be remarked that, in applying the
+idea to lines in a plane, a vector $OP$ will no longer be represented
+(as in \S 5) by
+$$
+\begin{array}{lclr}
+ & \textrm{ OP} & = & a + b\sqrt{1}\\
+\textrm{but by} & \textrm{ OP} & = & pa + qb\\
+\textrm{And if, similarly,} & \textrm{ OQ} & = & pa^{'} + qb^{'}\\
+\end{array}
+$$
+
+the addition of these two lines gives for $OR$ (which retains its
+previous signification)
+$$OR = p(a+a^{'} + q(b+b^{'})$$
+
+{\bf 12}. Beyond this, few attempts were made, or at least recorded,
+in earlier times, to extend the principle to space of three dimensions;
+and, though many such had been made before 1843, none,
+with the single exception of Hamilton's, have resulted in simple,
+practical methods; all, however ingenious, seeming to lead almost
+at once to processes and results of fearful complexity.
+
+For a lucid, complete, and most impartial statement of the
+claims of his predecessors in this field we refer to the Preface to
+Hamilton's {\sl Lectures on Quaternions}. He there shows how his long
+protracted investigations of Sets culminated in this unique system
+of tridimensionalspace geometry.
+
+{\bf 13}. It was reserved for Hamilton to discover the use and
+properties of a class of symbols which, though all in a certain sense
+square roots of 1, may be considered as {\sl real} unit lines, tied down
+to no particular direction in space ; the expression for a vector is,
+or may be taken to be,
+$$ \rho = ix + jy + kz$$
+
+but such vector is considered in connection with an {\sl extraspatial}
+magnitude $w$, and we have thus the notion of a QUATERNION
+
+$$w + \rho$$
+
+This is the fundamental notion in the singularly elegant, and
+enormously powerful, Calculus of Quaternions.
+
+While the schemes for using the algebraic $\sqrt{1}$ to indicate
+direction make one direction in space expressible by real numbers,
+the remainder being imaginaries of some kind, and thus lead to
+expressions which are heterogeneous ; Hamilton s system makes all
+directions in space equally imaginary, or rather equally real, thereby
+ensuring to his Calculus the power of dealing with space
+indifferently in all directions.
+
+In fact, as we shall see, the Quaternion method is independent
+of axes or any supposed directions in space, and takes its reference
+lines solely from the problem it is applied to.
+
+{\bf 14}. But, for the purpose of elementary exposition, it is best
+to begin by assimilating it as closely as we can to the ordinary
+Cartesian methods of Geometry of Three Dimensions, with which
+the student is supposed to be, to some extent at least, acquainted.
+Such assistance, it will be found, can (as a rule) soon be dispensed
+with; and Hamilton regarded any apparent necessity for an oc
+casional recurrence to it, in higher applications, as an indication
+of imperfect development in the proper methods of the new
+Calculus.
+
+We commence, therefore, with some very elementary geometrical
+ideas, relating to the theory of vectors in space. It will subsequently
+appear how we are thus led to the notion of a Quaternion.
+
+{\bf 15}. Suppose we have two points $A$ and $B$ in {\sl space}, and suppose
+$A$ given, on how many numbers does $B$'s relative position depend ?
+
+If we refer to Cartesian coordinates (rectangular or not) we find
+that the data required are the excesses of $B$'s three coordinates
+over those of $A$. Hence three numbers are required.
+
+Or we may take polar coordinates. To define the moon's
+position with respect to the earth we must have its Geocentric
+Latitude and Longitude, or its Right Ascension and Declination,
+and, in addition, its distance or radiusvector. {\sl Three} again.
+
+{\bf 16}. Here it is to be carefully noticed that nothing has been
+said of the {\sl actual} coordinates of either A or B, or of the earth
+and moon, in space; it is only the {\sl relative} coordinates that are
+contemplated.
+
+Hence any expression, as $\overline{AB}$, denoting a line considered with
+reference to direction and currency as well as length, (whatever
+may be its actual position in space) contains implicitly {\sl three}
+numbers, and all lines parallel and equal to $AB$, and concurrent
+with it, depend in the same way upon the same three. Hence, {\sl all
+lines which are equal, parallel, and concurrent, may be represented
+by a common symbol, and that symbol contains three distinct numbers}.
+In this sense a line is called a VECTOR, since by it we pass from
+the one extremity, $A$, to the other, $B$, and it may thus be
+considered as an instrument which {\sl carries} $A$ to $B$: so that a
+vector may be employed to indicate a definite {\sl translation} in space.
+
+[The term " currency " has been suggested by Cayley for use
+instead of the somewhat vague suggestion sometimes taken to
+be involved in the word "direction." Thus parallel lines have
+the same direction, though they may have similar or opposite
+currencies. The definition of a vector essentially includes its
+currency.]
+
+{\bf 17}. We may here remark, once for all, that in establishing a
+new Calculus, we are at liberty to give any definitions whatever
+of our symbols, provided that no two of these interfere with, or
+contradict, each other, and in doing so in Quaternions {sl simplicity}
+and (so to speak) {\sl naturalness} were the inventor's aim.
+
+{\bf 18}. Let $\overline{AB}$ be represented by $\alpha$, we know that
+$\alpha$ involves
+{\sl three} separate numbers, and that these depend solely upon the
+position of $B$ {\sl relatively} to $A$.
+Now if $CD$ be equal in length to $AB$
+and if these lines be parallel, and have the same currency, we may
+evidently write
+$$\overline{CD} = \overline{AB} = \alpha$$
+where it will be seen that the sign of equality between vectors
+contains implicitly {\sl equality in length}, {\sl parallelism in direction},
+and {\sl concurrency}. So far we have {\sl extended} the meaning of an
+algebraical symbol. And it is to be noticed that an equation
+between vectors, as
+$$\alpha = \beta$$
+contains {\sl three} distinct equations between mere numbers.
+
+{\bf 19}. We must now define $+$ (and the meaning of $$ will follow)
+in the new Calculus. Let $A$, $B$, $C$ be any three points, and (with
+the above meaning of $=$ ) let
+$$\overline{AB} = \alpha, \overline{BC} = \beta, \overline{AC} = \gamma$$
+If we define $+$ (in accordance with the idea (\S 16) that a vector
+represents a {\sl translation}) by the equation
+$$
+\begin{array}{lcl}
+ & \textrm{ } & \alpha + \beta = \gamma\\
+ & & \\
+\textrm{or} & \textrm{ } &
+\overline{AB} + \overline{BC} = \overline{AC}
+\end{array}
+$$
+we contradict nothing that precedes, but we at once introduce the
+idea that {\sl vectors are to be compounded},
+{\sl in direction and magnitude},
+{\sl like simultaneous velocities}. A reason for this may be seen in
+another way if we remember that by {\sl adding} the (algebraic) differences
+of the Cartesian coordinates of $B$ and $A$, to those of the
+coordinates of $C$ and $B$, we get those of the coordinates of $C$ and
+$A$. Hence these coordinates enter {\sl linearly} into the expression for
+a vector. (See, again, \S 5.)
+
+{\bf 20}. But we also see that if $C$ and $A$ coincide (and $C$ may be
+{\sl any} point)
+$$\overline{AC} = 0$$
+for no vector is then required to carry $A$ to $C$. Hence the above
+relation may be written, in this case,
+$$\overline{AB}+\overline{BA} = 0$$
+or, introducing, and by the same act defining, the symbol $$,
+$$\overline{AB} = \overline{BA}$$
+
+Hence, {\sl the symbol $$, applied to a vector, simply shows that its
+currency is to be reversed}.
+And this is consistent with all that precedes; for instance,
+$$
+\begin{array}{lcrcl}
+ & \textrm{ } & \overline{AB} + \overline{BC} &=& \overline{AC}\\
+\textrm{and} & \textrm{ } & \overline{AB} = \overline{AC} && \overline{BC} \\
+\textrm{or} & \textrm{ } & = \overline{AC} &+& \overline{CB} \\
+\end{array}
+$$
+are evidently but different expressions of the same truth.
+
+{\bf 21}. In any triangle, $ABC$, we have, of course,
+$$\overline{AB} + \overline{BC} + \overline{CA} = 0$$
+and, in any closed polygon, whether plane or gauche,
+$$\overline{AB}+\overline{BC}+\ldots+\overline{YZ}+\overline{ZA} = 0$$
+
+In the case of the polygon we have also
+$$\overline{AB}+\overline{BC}+\ldots+\overline{YZ} = \overline{AZ}$$
+
+These are the wellknown propositions regarding composition
+of velocities, which, by Newton's second law of motion, give us
+the geometrical laws of composition of forces acting at one point.
+
+{\bf 22}. If we compound any number of {\bf parallel} vectors, the result
+is obviously a numerical multiple of any one of them.
+Thus, if $A$, $B$, $C$ are in one straight line,
+$$\overline{BC} = x\overline{AB}$$
+where $x$ is a number, positive when $B$ lies between $A$ and $C$,
+otherwise negative; but such that its numerical value, independent
+of sign, is the ratio of the length of $BC$ to that of $AB$. This is
+at once evident if $AB$ and $BC$ be commensurable; and is easily
+extended to incommensurables by the usual {\sl reductio ad absurdum}.
+
+{\bf 23}. An important, but almost obvious, proposition is that {\sl any
+vector may be resolved, and in one way only, into three components
+parallel respectively to any three given vectors, no two of which are
+parallel, and which are not parallel to one plane}.
+
+\includegraphics{ps/quat1.ps}
+
+Let $OA$, $OB$, $OC$ be the three fixed
+vectors, $OP$ any other vector. From $P$ draw
+$PQ$ parallel to $CO$, meeting the plane $BOA$
+in $Q$. [There must be a definite point $Q$,
+else $PQ$, and therefore $CO$, would be parallel
+to $BOA$, a case specially excepted.] From $Q$
+draw $QR$ parallel to $BO$, meeting $OA$ in $R$.
+
+Then we have $\overline{OP}=\overline{OR} + \overline{RQ} + \overline{QP}$
+(\S 21),
+and these components are respectively parallel to the three given
+vectors. By \S 22 we may express $\overline{OR}$ as a numerical multiple
+of $\overline{OA}$, $\overline{RQ}$ of $\overline{OB}$, and
+$\overline{QP}$ of $\overline{OC}$. Hence we have, generally, for
+any vector in terms of three fixed noncoplanar vectors, $\alpha$,
+$\beta$, $\gamma$
+$$\overline{OP} = \rho = x\alpha + y\beta + z\gamma$$
+which exhibits, in one form, the {\sl three} numbers on which a vector
+depends (\S 16). Here $x$, $y$, $z$ are perfectly definite, and can have
+but single values.
+
+{\bf 24}. Similarly any vector, as $\overline{OQ}$, in the same plane with
+$\overline{OA}$ and $\overline{OB}$,
+can be resolved (in one way only) into components $\overline{OR}$,
+$\overline{RQ}$,
+parallel respectively to $\overline{OA}$ and
+$\overline{OB}$; so long, at least, as these
+two vectors are not parallel to each other.
+
+{\bf 25}. There is particular advantage, in certain cases, in employing
+a series of {\sl three mutually perpendicular unitvectors} as
+lines of reference. This system Hamilton denotes by $i$,$j$, $k$.
+
+Any other vector is then expressible as
+$$\rho= xi + yj + zk$$
+Since $i$, $j$, $k$ are unitvectors, $x$, $y$, $z$ are here the lengths of
+conterminous edges of a rectangular parallelepiped of which $\rho$
+is the vectordiagonal; so that the length of $\rho$ is, in this case,
+$$\sqrt{x^2+y^2+z^2}$$
+Let \hbox{\hskip 4cm}$\omega = \xi i + \eta j + \zeta k$\\
+be any other vector, then (by the proposition of \S 23) the vector
+$$
+\begin{array}{lcr}
+\textrm{equation} & \textrm{ } & \rho = \omega
+\end{array}
+$$
+obviously involves the following three equations among numbers,
+$$x=\xi, y=\eta, z=\zeta$$
+Suppose $i$ to be drawn eastwards, $j$ northwards, and $k$ upwards,
+this is equivalent merely to saying that {\sl if two points coincide, they
+are equally to the east (or west) of any third point, equally to the
+north (or south) of it, and equally elevated above (or depressed below)
+its level.}
+
+{\bf 26}. It is to be carefully noticed that it is only when
+$\alpha$, $\beta$, $\gamma$
+are not coplanar that a vector equation such as
+$$\rho = \omega$$
+or\hbox{\hskip 3cm}
+$x\alpha + y\beta + z\gamma = \xi \alpha + \eta \beta + \zeta \gamma$\\
+necessitates the three numerical equations
+$$x = \xi, y = \eta, z = \zeta$$
+For, if $\alpha$, $\beta$, $\gamma$ be coplanar (\S 24),
+a condition of the following form must hold
+$$\gamma = a\alpha + b\beta$$
+Hence,\hbox{\hskip 3cm}$\rho=(x+za)\alpha+(y+zb)\beta$\\
+\hbox{\hskip 4cm}$\omega=(\xi+\zeta a)\alpha + (\eta+\zeta b)\beta$\\
+and the equation\hbox{\hskip 3cm}$\rho=\omega$\\
+now requires only the two numerical conditions
+$$x+za=\xi+\zeta a\hbox{\hskip 1cm}y+zb = \eta+\zeta b$$
+
+{\bf 27}. {\sl The Commutative and Associative Laws hold in the combination
+of vectors by the signs $+$ and $$}. It is obvious that, if we
+prove this for the sign $+$, it will be equally proved for $$, because
+$$ before a vector (\S 20) merely indicates that it is to be reversed
+before being considered positive.
+
+Let $A$, $B$, $C$, $D$ be, in order, the corners of a parallelogram ; we
+have, obviously,
+$$\overline{AB} = \overline{DC}\hbox{\hskip 1cm}\overline{AD}=\overline{BC}$$
+And \hbox{\hskip 2cm}$\overline{AB}+\overline{BC}=
+\overline{AC}=
+\overline{AD}+\overline{DC}=
+\overline{BC}+\overline{AB}$
+
+Hence the commutative law is true for the addition of any two
+vectors, and is therefore generally true.
+
+Again, whatever four points are represented by $A$, $B$, $C$, $D$, we
+$$\overline{AD}=\overline{AB}+\overline{BD}=\overline{AC}+\overline{CD}$$
+or substituting their values for $\overline{AD}$, $\overline{BD}$,
+$\overline{AC}$ respectively, in these three expressions,
+$$\overline{AB}+\overline{BC}+\overline{CD}=
+\overline{AB}+(\overline{BC}+\overline{CD})=
+(\overline{AB}+\overline{BC})+\overline{CD}$$
+And thus the truth of the associative law is evident.
+
+{\bf 28}. The equation
+$$\rho = x\beta$$
+where $\rho$ is the vector connecting a variable point with the origin,
+$\beta$ a definite vector, and $x$ an indefinite number, represents the
+straight line drawn from the origin parallel to $\beta$ (\S 22).
+
+The straight line drawn from $A$, where $\overline{OA} = \alpha$,
+and parallel to $\beta$, has the equation
+$$\rho = \alpha + x\beta\eqno(1)$$
+In words, we may pass directly from $O$ to $P$ by the vector $\overline{OP}$
+or $\rho$; or we may pass first to $A$, by means of $\overline{OA}$ or
+$\alpha$, and then to $P$ along a vector parallel to $\beta$ (\S 16).
+
+Equation (1) is one of the many useful forms into which
+Quaternions enable us to throw the general equation of a straight
+line in space. As we have seen (\S 25) it is equivalent to three
+numerical equations; but, as these involve the indefinite quantity
+$x$, they are virtually equivalent to but {\sl two}, as in ordinary Geometry
+of Three Dimensions.
+
+{\bf 29}. A good illustration of this remark is furnished by the fact
+that the equation
+$$\rho = y\alpha + x\beta$$
+which contains two indefinite quantities, is virtually equivalent to
+only one numerical equation. And it is easy to see that it represents
+the plane in which the lines $\alpha$ and $\beta$ lie; or the surface
+which is formed by drawing, through every point of $OA$, a line
+parallel to $OB$. In fact, the equation, as written, is simply \S 24
+in symbols.
+
+And it is evident that the equation
+$$\rho = \gamma + y\alpha + x\beta$$
+is the equation of the plane passing through the extremity of $\gamma$,
+and parallel to $\alpha$ and $\beta$.
+
+It will now be obvious to the reader that the equation
+$$\rho = p_1\alpha_1+p_2\alpha_2+\ldots=\sum{p\alpha}$$
+where $\alpha_1$, $\alpha_2$ , \&c. are given vectors,
+and $p_1$, $p_2$, \&c. numerical quantities,
+{\sl represents a straight line}
+if $p_1$, $p_2$, \&c. be linear functions of
+{\sl one} indeterminate number; and a {\sl plane}, if they be linear
+expressions containing two indeterminate numbers. Later (\S 31 (l)),
+this theorem will be much extended.
+
+Again, the equation
+$$\rho = x\alpha + y\beta +z\gamma$$
+refers to {\sl any} point whatever in space, provided
+$\alpha$, $\beta$, $\gamma$ are not coplanar. (Ante, \S 23)
+
+{\bf 30}. The equation of the line joining any two points $A$ and $B$,
+where $\overline{OA} = \alpha$ and $\overline{OB} = \beta$, is obviously
+$$\rho = \alpha + x(\beta\alpha)$$
+or \hbox{\hskip 4.2cm}$\rho=\beta+y(\alpha\beta)$\\
+These equations are of course identical, as may be seen by putting
+$1y$ for $x$.
+
+The first may be written
+$$\rho+(x1)\alphax\beta = 0$$
+or\hbox{\hskip 4cm}$p\rho+q\alpha+r\beta=0$\\
+subject to the condition $p + q + r = 0$ identically. That is 
+A homogeneous linear function of three vectors, equated to zero,
+expresses that the extremities of these vectors are in one straight
+line, {\sl if the sum of the coefficients be identically zero}.
+
+Similarly, the equation of the plane containing the extremities
+$A$, $B$, $C$ of the three noncoplanar vectors
+$\alpha$, $\beta$, $\gamma$ is
+$$\rho=\alpha+x(\beta\alpha)+y(\gamma\beta)$$
+where $x$ and $y$ are each indeterminate.
+
+This may be written
+$$p\rho + q\alpha + r\beta +s\gamma = 0$$
+with the identical relation
+$$p+q+r+x=0$$
+which is one form of the condition that four points may lie in one plane.
+
+{\bf 31}. We have already the means of proving, in a very simple
+manner, numerous classes of propositions in plane and solid
+geometry. A very few examples, however, must suffice at this
+stage; since we have hardly, as yet, crossed the threshold of the
+subject, and are dealing with mere linear equations connecting two
+or more vectors, and even with them {\sl we are restricted as yet to
+operations of mere addition}. We will give these examples with a
+painful minuteness of detail, which the reader will soon find to be
+necessary only for a short time, if at all.
+
+(a) {\sl The diagonals of a parallelogram bisect each other}.
+
+Let $ABCD$ be the parallelogram, $O$ the point of intersection of
+its diagonals. Then
+$$\overline{AO}+\overline{OB}=\overline{AB}=\overline{DC}=
+\overline{DO}+\overline{OC}$$
+which gives\hbox{\hskip 2cm}$\overline{AO}\overline{OC}=
+\overline{DO}\overline{OB}$\\
+The two vectors here equated are parallel to the diagonals respectively.
+Such an equation is, of course, absurd unless
+\begin{enumerate}
+\item The diagonals are parallel, in which case the figure
+is not a parallelogram;
+\item $\overline{AO} = \overline{OC}$, and
+$\overline{DO} = \overline{OB}$, the proposition.
+\end{enumerate}
+
+(b) {\sl To shew that a triangle can be constructed, whose sides
+are parallel, and equal, to the bisectors of the sides of any
+triangle}.
+
+Let $ABC$ be any triangle, $Aa$, $Bb$, $Cc$ the bisectors of the
+sides.
+
+Then
+$$
+\begin{array}{ccc}
+\overline{Aa} & =\overline{AB}+\overline{Ba}
+ & =\overline{AB}+\frac{1}{2}\overline{BC}\\
+\overline{Bb} & \ldots & = \overline{BC} + \frac{1}{2}\overline{CA}\\
+\overline{Cc} & \ldots & = \overline{CA} + \frac{1}{2}\overline{AB}
+\end{array}
+$$
+Hence \hbox{\hskip 2cm}$\overline{Aa}+\overline{Bb}+\overline{Cc}=
+\frac{3}{2}(\overline{AB}+\overline{BC}+\overline{CA})=0$\\
+which (\S 21) proves the proposition.
+
+Also
+$$
+\begin{array}{rcl}
+\overline{Aa} & = & \overline{AB}+\frac{1}{2}\overline{BC}\\
+ & = & \overline{AB}\frac{1}{2}(\overline{CA}+\overline{AB})\\
+ & = & \frac{1}{2}(\overline{AB}\overline{CA})\\
+ & = & \frac{1}{2}(\overline{AB}+\overline{AC})
+\end{array}
+$$
+results which are sometimes useful. They may be easily verified
+by producing $\overline{Aa}$ to twice its length and joining the extremity
+with $B$.
+
+($b^{'}$) {\sl The bisectors of the sides of a triangle meet in a point,
+which trisects each of them}.
+
+Taking $A$ as origin, and putting $\alpha$, $\beta$, $\gamma$
+for vectors parallel, and
+equal, to the sides taken in order $BC$, $CA$, $AB$; the equation of
+$Bb$ is (\S 28 (1))
+$$\rho=\gamma+x(\gamma+\frac{\beta}{2})=(1+x)\gamma+\frac{x}{2}\beta$$
+That of $Cc$ is, in the same way,
+$$\rho=(1+y)\beta\frac{y}{2}\gamma$$
+At the point $O$, where $Bb$ and $Cc$ intersect,
+$$\rho=(1+x)\gamma+\frac{x}{2}\beta=(1+y)\beta\frac{y}{2}\gamma$$
+Since $\gamma$ and $\beta$ are not parallel, this equation gives
+$$1+x=\frac{y}{2}\textrm{\ \ and\ \ }\frac{x}{2}=(1+y)$$
+From these\hbox{\hskip 3cm}$x=y=\frac{2}{3}$
+
+Hence\hbox{\hskip 1cm}$\overline{AO}=\frac{1}{3}(\gamma\beta)=
+\frac{2}{3}\overline{Aa}$ (See Ex. (b))\\
+
+This equation shows, being a vector one, that $\overline{Aa}$ passes
+through $O$, and that $AO$ : $Oa$ :: 2:1.
+
+(c) If
+$$\overline{OA}=\alpha$$
+$$\overline{OB}=\beta$$
+$$\overline{OC}=l\alpha+m\beta$$
+{\sl be three given coplanar vectors, $c$ the intersection of $AB$, $OC$, and
+if the lines indicated in the figure be drawn, the points
+$a_1$,$b_1$,$c_1$ lie in a straight line. }
+\includegraphics{ps/quat2.ps}
+
+We see at once, by the process indicated in \S 30, that
+$$\overline{Oc}=\frac{l\alpha+m\beta}{l+m},\hbox{\hskip 1cm}
+\overline{Ob}=\frac{l\alpha}{1m},\hbox{\hskip 1cm}
+\overline{Oa}=\frac{m\beta}{1l}$$
+Hence we easily find
+$$\overline{Oa_1}=\frac{m\beta}{1l2m},\hbox{\hskip 0.5cm}
+\overline{Ob_1}=\frac{l\alpha}{12lm},\hbox{\hskip 0.5cm}
+\overline{Oc_1}=\frac{l\alpha+m\beta}{ml}$$
+These give
+$$(1l2m)\overline{Oa_1}+(12lm)\overline{Ob_1}(ml)\overline{Oc_1}=0$$
+But\hbox{\hskip 1cm}$(1l2m)+(12lm)(m  l)=0$ identically.
+
+This, by \S 30, proves the proposition.
+
+(d) Let $\overline{OA} = \alpha$,
+$\overline{OB} = \beta$, be any two vectors. If $MP$ be a
+given line parallel to $OB$; and $OQ$, $BQ$, be drawn parallel to $AP$,
+$OP$ respectively ; the locus of $Q$ is a straight line parallel to $OA$.
+
+\noindent
+\includegraphics{ps/quat3.ps}
+
+\noindent
+Let \hbox{\hskip 4cm}$\overline{OM}=e\alpha$\\
+Then \hbox{\hskip 3cm}$\overline{AP}=\overline{e1}\alpha+x\beta$
+
+Hence the equation of $OQ$ is
+$$\rho=y(\overline{e1}\alpha+x\beta)$$
+and that of $BQ$ is\hbox{\hskip 1cm}$\rho=\beta+z(e\alpha+x\beta)$\\
+At Q we have, therefore,
+$$
+\left.
+\begin{array}{c}
+xy=1+zx\\
+y(e1)=ze
+\end{array}
+\right\}
+$$
+These give $xy = e$, and the equation of the locus of $Q$ is
+$$\rho = e\beta+y^{'}\alpha$$
+i.e. a straight line parallel to $OA$, drawn through $N$ in $OB$
+produced, so that
+$$ON : OB :: OM : OA$$
+
+COR. If $BQ$ meet $MP$ in $q$, $\overline{Pq} = \beta$;
+and if $AP$ meet $NQ$ in $p$, $\overline{Qp} = \alpha$.
+
+Also, for the point $R$ we have $\overline{pR} = \overline{AP}$,
+$\overline{QR} = \overline{Bq}$.
+
+Further, the locus of $R$ is a hyperbola, of which $MP$ and $NQ$
+are the asymptotes. See, in this connection, \S 31 (k) below.
+
+Hence, {\sl if from any two points, $A$ and $B$, lines be drawn intercepting
+a given length $Pq$ on a given line $Mq$ ; and if, from $R$ their
+point of intersection, $Rp$ be laid off $= PA$, and $RQ = qB$ ; $Q$ and $p$
+lie on a fixed straight line, and the length of $Qp$ is constant}.
+
+(e) {\sl To find the centre of inertia of any system of masses.}
+
+If $\overline{OA} = \alpha$, $\overline{OB} = \alpha_1$,
+be the vector sides of any triangle, the
+vector from the vertex dividing the base $AB$ in $C$ so that
+$$BC : CA :: m : m_1$$
+is \hbox{\hskip 4cm}$\frac{m\alpha+m_1\alpha_1}{m+m_1}$\\
+
+For $AB$ is $\alpha_1\alpha$, and therefore $\overline{AC}$ is
+$$\frac{m_1}{m+m_1}(\alpha_1\alpha)$$
+
+Hence\hbox{\hskip 3cm}$\overline{OC}=\overline{OA}+\overline{AC}$
+$$=\alpha+\frac{m_1}{m+m_1}(\alpha_1\alpha)$$
+$$=\frac{m\alpha +m_1\alpha_1}{m+m_1}$$
+This expression shows how to find the centre of inertia of two
+masses ; $m$ at the extremity of $\alpha$, $m_1$ at that of $\alpha_1$.
+Introduce $m_2$ at the extremity of $a_2$,
+then the vector of the centre of inertia of the
+three is, by a second application of the formula,
+$$\frac{(m+m_1)(\frac{m\alpha+m_1\alpha_1}{m+m_1})+m_2\alpha_2}
+{(m+m_1)+m_2}=\frac{m\alpha+m_1\alpha_1+m_2\alpha_2}{m+m_1+m_2}$$
+From this it is clear that, for any number of masses, expressed
+generally by $m$ at the extremity of the vector $\alpha$, the vector of the
+centre of inertia is
+$$\beta=\frac{\sum(m\alpha)}{\sum(m)}$$
+This may be written\hbox{\hskip 1cm}$\sum m(\alpha\beta)=0$\\
+Now a $\alpha_1\beta$
+is the vector of $m_1$ with respect to the centre of inertia.
+Hence the theorem, {\sl If the vector of each element of a mass, drawn
+from the centre of inertia, be increased in length in proportion to the
+mass of the element, the sum of all these vectors is zero.}
+
+(f) We see at once that the equation
+
+\noindent
+\includegraphics{ps/quat4.ps}
+
+$$\rho=\alpha t +\frac{\beta t^2}{2}$$
+where $t$ is an indeterminate
+number, and $\alpha$, $\beta$ given vectors,
+represents a parabola.
+The origin, $O$, is a point on
+the curve, $\beta$ is parallel to
+the axis, i.e. is the diameter
+$OB$ drawn from the origin,
+and $\alpha$ is $OA$ the tangent at the origin. In the figure
+$$\overline{QP}=\alpha t,\hbox{\hskip 1cm}\overline{OQ}=\frac{\beta t^2}{2}$$
+
+The secant joining the points where $t$ has the values $t$ and $t^{'}$ is
+represented by the equation
+$$
+\begin{array}{rcl}
+\rho&=&\alpha t +\frac{\beta t^2}{2}+x\left(\alpha t^{'}+\frac{\beta t^{'2}}{2}
+\alpha t\frac{\beta t^2}{2}\right)\hbox{\hskip 1cm}(\S 30)\\
+&=&\alpha t+\frac{\beta t^2}{2}+x(t^{'}t)\left\{\alpha+\beta\frac{t^{'}t}{2}
+\right\}
+\end{array}
+$$
+Write $x$ for $x(t^{'}t)$ [which may have any value], then put
+$t^{'}=t$, and the equation of the tangent at the point ($t$) is
+$$\rho=\alpha t + \frac{\beta t^2}{2}+x(\alpha+\beta t)$$
+In this put $x = t$, and we have
+$$\rho=\frac{\beta t^2}{2}$$
+or the intercept of the tangent on the diameter is equal in length
+to the abscissa of the point of contact, but has the opposite
+currency.
+
+Otherwise: the tangent is parallel to the vector $\alpha+\beta t$ or
+$\alpha t + \beta t^2$ or $\frac{\beta t^2}{2}+\alpha t+\frac{\beta t^2}{2}$
+or $\overline{OQ}+\overline{OP}$.
+But $\overline{TP}=\overline{TO}+\overline{OP}$,
+hence $\overline{TO} = \overline{OQ}$.
+
+(g) Since the equation of any tangent to the parabola is
+$$\rho=\alpha t + \frac{\beta t^2}{2} + x(\alpha+\beta t)$$
+let us find the tangents which can be drawn from a given point.
+Let the vector of the point be
+$$\rho=p\alpha + q\beta\hbox{\hskip 0.5cm}(\S 24)$$
+Since the tangent is to pass through this point, we have, as con
+ditions to determine $t$ and $x$,
+$$t+x=p$$
+$$\frac{t^2}{2}+xt=q$$
+by equating respectively the coefficients of $\alpha$ and $\beta$.
+
+Hence\hbox{\hskip 3.5cm}$t=p \pm \sqrt{p^22q}$
+
+Thus, in general, {\sl two} tangents can be drawn from a given point.
+These coincide if $$p^2=2q$$
+that is, if the vector of the point from which they are to be drawn
+is $$\rho=p\alpha+q\beta=p\alpha+\frac{p^2}{2}\beta$$
+i.e. if the point lies on the parabola. They are imaginary if
+$2q > p^2$, that is, if the point be
+$$\rho=p\alpha+\left(\frac{p^2}{2}+r\right)\beta$$
+$r$ being {\sl positive}. Such a point is evidently {\sl within} the curve,
+as at $R$, where $\overline{OQ}=\frac{p^2}{2}\beta$,
+$\overline{QP}=p\alpha$, $\overline{PR} = r\beta$.
+
+(h) Calling the values of $t$ for the two tangents found in (g)
+$t_1$ and $t_2$ respectively, it is obvious that the vector joining the
+points of contact is
+$$\alpha t_1+\frac{\beta t_1^2}{2}\alpha t_2  \frac{\beta t_2^2}{2}$$
+which is parallel to\hbox{\hskip 2cm}
+$\alpha+\beta\frac{t_1+t_2}{2}$
+or, by the values of $t_1$ and $t_2$ in (g),
+$$\alpha+p\beta$$
+Its direction, therefore, does not depend on $q$. In words, {\sl If pairs of
+tangents be drawn to a parabola from points of a diameter produced,
+the chords of contact are parallel to the tangent at the vertex of the
+diameter.} This is also proved by a former result, for we must have
+$\overline{OT}$ for each tangent equal to $\overline{QO}$.
+
+(i) The equation of the chord of contact, for the point whose vector is
+$$\rho=p\alpha+q\beta$$
+is thus\hbox{\hskip 3cm}
+$\rho=\alpha t_1+\frac{\beta t_1^2}{2}+y(\alpha+p\beta)$
+
+Suppose this to pass always through the point whose vector is
+$$\rho=a\alpha+b\beta$$
+Then we must have
+$$
+\left.
+\begin{array}{rcl}
+t_1+y & = & a\\
+\frac{t_1^2}{2}+py & = & b
+\end{array}
+\right\}
+$$
+or\hbox{\hskip 4cm}$t_1=p\pm\sqrt{p^22p\alpha+2\beta}$
+
+Comparing this with the expression in (g), we have
+$$
+q = pa  b
+$$
+that is, the point from which the tangents are drawn has the vector
+a straight line (\S 28 (1)).
+
+The mere form of this expression contains the proof of the usual
+properties of the pole and polar in the parabola ; but, for the sake
+of the beginner, we adopt a simpler, though equally general, process.
+
+Suppose $\alpha = 0$. This merely restricts the pole to the particular
+diameter to which we have referred the parabola. Then the pole
+is $Q$, where $$\rho = b\beta$$
+and the polar is the line $TU$, for which
+$$\rho=b\beta+p\alpha$$
+{\sl Hence the polar of any point is parallel to the tangent at the
+extremity of the diameter on which the point lies, and its intersection
+with that diameter is as far beyond the vertex as the pole
+is within, and vice versa. }
+
+(j) As another example let us prove the following theorem.
+{\sl If a triangle be inscribed in a parabola, the three points in which
+the sides are met by tangents at the angles lie in a straight line. }
+
+Since $O$ is any point of the curve, we may take it as one corner
+of the triangle. Let $t$ and $t_1$ determine the others. Then, if
+$\omega_1$,$\omega_2$,$\omega_3$
+represent the vectors of the points of intersection of the
+tangents with the sides, we easily find
+$$
+\begin{array}{rcl}
+\omega_1 & = & \frac{t_1^2}{2t_1t}
+\left(\alpha+\frac{t}{2}\beta\right)\\
+&&\\
+\omega_2 & = & \frac{t^2}{2tt_1}
+\left(\alpha+\frac{t_1}{2}\beta\right)\\
+&&\\
+\omega_3 & = & \frac{tt_1}{t_1+t}\alpha
+\end{array}
+$$
+These values give
+$$\frac{2t_1t}{t_1}\omega_1 
+\frac{2tt_1}{t}\omega_2 
+\frac{t_1^2t^2}{tt_1}\omega_3 = 0$$
+Also
+$$\frac{2t_1t}{t_1} 
+\frac{2tt_1}{t} 
+\frac{t_1^2t^2}{tt_1} = 0$$
+identically.
+
+Hence, by \S 30, the proposition is proved.
+
+(k) Other interesting examples of this method of treating
+curves will, of course, suggest themselves to the student. Thus
+$$\rho = \alpha\cos t + \beta\sin t$$
+or
+$$\rho=\alpha x + \beta\sqrt{1x^2}$$
+represents an ellipse, of which the given vectors $\alpha$ and $\beta$
+are semiconjugate diameters. If $t$ represent time, the radiusvector of this
+ellipse traces out equal areas in equal times. [We may anticipate
+so far as to write the following :
+$$2 \textrm{Area} = T\int V \rho d\rho = TV\alpha\beta.\int dt$$
+which will be easily understood later.]
+
+Again,
+$$\rho=\alpha t+\frac{\beta}{t}\textrm{ or }
+\rho=\alpha\tan x + \beta\cot x$$
+evidently represents a hyperbola referred to its asymptotes. [If
+$t$ represent time, the sectorial area traced out is proportional to
+$\log t$, taken between proper limits.]
+Thus, also, the equation
+$$\rho = \alpha(t + \sin t)+\beta\cos t$$
+in which $\alpha$ and $\beta$ are of equal lengths, and at right angles to one
+another, represents a cycloid. The origin is at the middle point of
+the axis ($2\beta$) of the curve. [It may be added that, if t represent
+{\sl time}, this equation shows the motion of the tracing point, provided
+the generating circle rolls uniformly, revolving at the rate of a
+radian per second.]
+
+When the lengths of $\alpha$, $\beta$ are not equal, this equation gives the
+cycloid distorted by elongation of its ordinates or abscissae : {\sl not} a
+trochoid. The equation of a trochoid may be written
+$$\rho = \alpha(et + \sin t)+\beta\cos t$$
+$e$ being greater or less than 1 as the curve is prolate or curtate.
+The lengths of $\alpha$ and $\beta$ are still taken as equal.
+
+But, so far as we have yet gone with the explanation of the
+calculus, as we are not prepared to determine the lengths or
+inclinations of vectors, we can investigate only a very limited class of
+the properties of curves, represented by such equations as those
+above written.
+
+(l) We may now, in extension of the statement in \S 29, make
+the obvious remark that
+$$\rho = \sum p\alpha$$
+(where, as in \S 23, the number of vectors, $\alpha$, can always be reduced
+to {\sl three}, at most) is the equation of a curve in space, if the
+numbers $p_1$, $p_2$, \&c. are functions of one indeterminate. In such
+a case the equation is sometimes written
+$$\rho=\phi(t)$$
+But, if $p_1$, $p_2$, \&c. be functions of {\sl two} indeterminates,
+the locus of the extremity of $\rho$ is a {\sl surface};
+whose equation is sometimes written
+$$\rho = \phi(t,u)$$
+
+[It may not be superfluous to call the reader's attention to the
+fact that, in these equations, $\phi(t)$ or $\phi(t, u)$
+is necessarily a vector expression, since it is equated to a vector, $\rho$.]
+
+(m) Thus the equation
+$$\rho = \alpha\cos t+\beta\sin t + \gamma t\eqno(1)$$
+belongs to a helix,
+\boxed{4.6in}{
+\vskip 0.1cm
+In Axiom we can draw this with the commands:
+\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
+tpdhere
+}
+while
+$$\rho = \alpha\cos t+\beta\sin t + \gamma u\eqno(2)$$
+represents a cylinder whose generating lines are parallel to $\gamma$,
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
+tpdhere
+}
+and
+whose base is the ellipse
+$$\rho=\alpha\cos t + \beta\sin t$$
+The helix above lies wholly on this cylinder.
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
+tpdhere
+}
+
+Contrast with (2) the equation
+$$\rho = u(\alpha\cos t + \beta\sin t + \gamma)\eqno(3)$$
+which represents a cone of the second degree
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
+tpdhere
+}
+made up, in fact,
+of all lines drawn from the origin to the ellipse
+$$\rho=\alpha\cos t + \beta\sin t + \gamma$$
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
+tpdhere
+}
+
+If, however, we write
+$$\rho = u(\alpha\cos t + \beta\sin t + \gamma t)$$
+we form the equation of the transcendental cone whose vertex is
+at the origin, and on which lies the helix (1).
+\boxed{4.6in}{
+\vskip 0.1cm
+\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
+tpdhere
+}
+
+In general
+$$\rho=u\phi(t)$$
+is the cone whose vertex is the origin, and on which lies the curve
+$$\rho=\phi(t)$$
+while\hbox{\hskip 4cm}$\rho=\phi(t)+u\alpha$\\
+is a cylinder, with generating lines parallel to $\alpha$, standing on the
+same curve as base.
+
+Again,\hbox{\hskip 3cm}$\rho=p\alpha+q\beta+r\gamma$\\
+with a condition of the form
+$$ap^2+bq^2+cr^2=1$$
+belongs to a central surface of the second order, of which
+$\alpha$, $\beta$, $\gamma$
+are the directions of conjugate diameters. If $a$, $b$, $c$ be all positive,
+the surface is an ellipsoid.
+
+{\bf 32}. In Example ($f$) above we performed an operation equivalent
+to the differentiation of a vector with reference to a single
+{\sl numerical} variable of which it was given as an explicit function.
+As this process is of very great use, especially in quaternion
+investigations connected with the motion of a particle or point; and as it
+will afford us an opportunity of making a preliminary step towards
+overcoming the novel difficulties which arise in quaternion differentiation;
+we will devote a few sections to a more careful, though
+very elementary, exposition of it.
+
+{\bf 33}. It is a striking circumstance, when we consider the way
+in which Newton's original methods in the Differential Calculus
+have been decried, to find that Hamilton was {\sl obliged} to employ
+them, and not the more modern forms, in order to overcome the
+characteristic difficulties of quaternion differentiation. Such a thing
+as {\sl a differential coefficient has absolutely no meaning in quaternions},
+except in those special cases in which we are dealing with degraded
+quaternions, such as numbers, Cartesian coordinates, \&c. But a
+quaternion expression has always a {\sl differential}, which is, simply,
+what Newton called a {\sl fluxion}.
+
+As with the Laws of Motion, the basis of Dynamics, so with the
+foundations of the Differential Calculus ; we are gradually coming
+to the conclusion that Newton s system is the best after all.
+
+{\bf 34}. Suppose $\rho$ to be the vector of a curve in space. Then,
+generally, $\rho$ may be expressed as the sum of a number of terms,
+each of which is a multiple of a constant vector by a function of some
+{\sl one} indeterminate; or, as in \S 31 ($l$),
+if $P$ be a point on the curve,
+$$\overline{OP}=\rho=\phi(t)$$
+
+And, similarly, if $Q$ be {\sl any other} point on the curve,
+$$\overline{OQ}=\rho_1=\rho+\delta\rho=\phi(t_1)=\phi(t+\delta t)$$
+where $\delta t$ is {\sl any number whatever}.
+
+The vectorchord $\overline{PQ}$ is therefore, rigorously,
+$$\delta p = \rho_1\rho = \phi(t+\delta t)\phi t$$
+
+{\bf 35}. It is obvious that, in the present case, {\sl because the vectors
+involved in $\phi$ are constant, and their numerical multipliers alone vary},
+the expression $\phi(t+\delta t)$ is, by Taylor's Theorem, equivalent to
+$$\phi(t)+\frac{d\phi(t)}{dt}\delta t+
+\frac{d^2\phi(t)}{dt^2}\frac{(\delta t)^2}{1\textrm{ . }2}+\ldots$$
+
+Hence,
+$$\delta \rho=\frac{d\phi(t)}{dt}\delta t+
+\frac{d^2\phi(t)}{dt^2}\frac{(\delta t)^2}{1\textrm{ . }2}+\textrm{\&c.}$$
+And we are thus entitled to write, when $\delta t$ has been made
+indefinitely small,
+$$\textrm{Limit}\left(
+\begin{array}{c}
+\delta p\\
+\delta t
+\end{array}
+\right)_{\delta t=0}
+=\frac{d\rho}{dt}
+=\frac{d\phi(t)}{dt}
+=\phi^{'}(t)$$
+
+In such a case as this, then, we are permitted to differentiate,
+or to form the differential coefficient of, a vector, according to the
+ordinary rules of the Differential Calculus. But great additional
+insight into the process is gained by applying Newton's method.
+
+{\bf 36}. Let $\overline{OP}$ be
+$$\rho=\phi(t)$$
+and $overline{OQ}_1$
+$$\rho_1=\phi(t+dt)$$
+where $dt$ is any number whatever.
+
+\includegraphics{ps/quat5.ps}
+
+The number $t$ may here be taken
+as representing {\sl time}, i.e. we may
+suppose a point to move along the
+curve in such a way that the value
+of $t$ for the vector of the point $P$ of
+the curve denotes the interval which
+has elapsed (since a fixed epoch) when the moving point has
+reached the extremity of that vector. If, then, $dt$ represent any
+interval, finite or not, we see that
+$$\overline{OQ}_1=\phi(t+dt)$$
+will be the vector of the point after the additional interval $dt$.
+
+But this, in general, gives us little or no information as to the
+velocity of the point at $P$. We shall get a better approximation
+by halving the interval $dt$, and finding $Q_2$ ,
+where $\overline{OQ}_2 = \phi(t + \frac{1}{2}dt)$,
+as the position of the moving point at that time. Here the vector
+virtually described in $\frac{1}{2}dt$ is
+$\overline{PQ}_2$ . To find, on this supposition,
+the vector described in $dt$, we must double
+$\overline{PQ}_2$ , and we find, as a
+second approximation to the vector which the moving point would
+have described in time $dt$, if it had moved for that period in the
+direction and with the velocity it had at $P$,
+$$
+\begin{array}{rcl}
+\overline{Pq}_2=2\overline{PQ}_2 & = & 2(\overline{OQ}_2\overline{OP})\\
+& = & 2\{\phi(t+\frac{1}{2}dt)\phi(t)\}
+\end{array}
+$$
+The next approximation gives
+$$
+\begin{array}{rcl}
+\overline{Pq}_3=3\overline{PQ}_3 & = & 3(\overline{OQ}_3\overline{OP})\\
+& = & 3\{\phi(t+\frac{1}{3}dt)\phi(t)\}
+\end{array}
+$$
+And so on, each step evidently leading us nearer the sought truth.
+Hence, to find the vector which would have been described in time
+$dt$ had the circumstances of the motion at $P$ remained undisturbed,
+we must find the value of
+$$d\rho=\overline{Pq}=L_{x=\infty}x\left\{\phi\left(t+\frac{1}{x}dt\right)
+\phi(t)\right\}$$
+
+We have seen that in this particular case we may use Taylor's
+Theorem. We have, therefore,
+$$
+\begin{array}{rcl}
+d\rho & = & L_{x=\infty}x
+\left\{
+\phi^{'}(t)\frac{1}{x}dt+
+\phi^{''}(t)\frac{1}{x^2}\frac{(dt)^2}{1\textrm{ . }2}+
+\textrm{\&c}
+\right\}\\
+&&\\
+& = & \phi^{'}(t)dt
+\end{array}
+$$
+And, if we choose, we may now write
+$$\frac{d\rho}{dt}=\phi^{'}(t)$$
+
+{\bf 37}. But it is to be most particularly remarked that in the
+whole of this investigation no regard whatever has been paid to
+the magnitude of $dt$. The question which we have now answered
+may be put in the form  {\sl A point describes a given curve in a given
+manner. At any point of its path its motion suddenly ceases to be
+accelerated. What space will it describe in a definite interval?} As
+Hamilton well observes, this is, for a planet or comet, the case
+of a 'celestial Atwood's machine'.
+
+{\bf 38}. If we suppose the variable, in terms of which $\rho$ is expressed,
+to be the arc, $s$, of the curve measured from some fixed point, we
+find as before
+$$d\rho = \phi^{'}(x)ds$$
+From the very nature of the question it is obvious that the length
+of $dp$ must in this case be $ds$, so that $\phi^{'}(s)$
+is necessarily a unitvector.
+This remark is of importance, as we shall see later; and
+it may therefore be useful to obtain afresh the above result without
+any reference to time or velocity.
+
+{\bf 39}. Following strictly the process of Newton s VIIth Lemma,
+let us describe on $Pq_2$ an arc similar to $PQ_2$, and so on. Then
+obviously, as the subdivision of $ds$ is carried farther, the new arc
+(whose length is always $ds$) more and more nearly (and without
+limit) coincides with the line which expresses the corresponding
+approximation to $dp$.
+
+{\bf 40}. As additional examples let us take some wellknown
+{\sl plane} curves; and first the hyperbola (\S 31 ($k$))
+$$\rho=\alpha t + \frac{\beta}{t}$$
+Here
+$$d\rho=\left(\alpha\frac{\beta}{t^2}\right)dt$$
+This shows that the tangent is parallel to the vector
+$$\alpha t  \frac{\beta}{t}$$
+In words, {\sl if the vector (from the centre) of a point in a hyperbola
+be one diagonal of a parallelogram, two of whose sides coincide with
+the asymptotes, the other diagonal is parallel to the tangent at the
+point, and cuts off a constant area from the space between the
+asymptotes}. (For the sides of this triangular area are $t$ times the
+length of $\alpha$, and $1/t$
+times the length of $\beta$, respectively; the angle
+between them being constant.)
+
+Next, take the cycloid, as in \S 31 ($k$),
+$$\rho=\alpha(t+\sin t)+\beta\cos t$$
+We have
+$$d\rho=\{\alpha(1+\cos t)\beta\sin t\}dt$$
+At the vertex
+$$t=0,\hbox{\hskip 0.5cm}
+\cos t=1,\hbox{\hskip 0.5cm}
+\sin t=0,\hbox{\hskip 0.5cm}
+\textrm{ and }d\rho=2\alpha dt$$
+At a cusp
+$$t=\pi,\hbox{\hskip 0.5cm}
+\cos t=1,\hbox{\hskip 0.5cm}
+\sin t=0,\hbox{\hskip 0.5cm}
+\textrm{ and }d\rho = 0$$
+
+This indicates that, at the cusp, the tracing point is (
+instantaneously) at rest. To find the direction of the tangent, and the
+form of the curve in the vicinity of the cusp, put $t=\pi+\tau$,
+where powers of $\tau$ above the second are omitted. We have
+$$d\rho=\beta\tau dt + \frac{\alpha\tau^2}{2}dt$$
+so that, at the cusp, the tangent is parallel to $\beta$. By making the
+same substitution in the expression for $\rho$, we find that the part of
+the curve near the cusp is a semicubical parabola,
+$$\rho=\alpha(\pi+\tau^3/6)\beta(1\tau^2/2)$$
+or, if the origin be shifted to the cusp ($\rho=\pi\alpha\beta$),
+$$\rho=\alpha\tau^3/6+\beta\tau^2/2$$
+
+{\bf 41}. Let us reverse the first of these questions, and {\sl seek the
+envelope of a line which cuts off from two fixed axes a triangle of
+constant area}.
+
+If the axes be in the directions of $\alpha$ and $\beta$, the intercepts may
+evidently be written $\alpha t$ and $\frac{\beta}{t}$.
+Hence the equation of the line is (\S 30)
+$$\rho=\alpha t + x\left(\frac{\beta}{t}\alpha t\right)$$
+
+The condition of envelopment is, obviously, (see Chap. IX.)
+$$d\rho = 0$$
+This gives
+$0 = \left\{\alphax\left(\frac{\beta}{t^2}+\alpha\right)\right\}dt+
+\left(\frac{\beta}{t}\alpha t\right)dx$
+%tpdhere  this should use an asterisk, a number looks like a superscript
+\setcounter{footnote}{1}%% use a dagger because number looks like superscript
+\footnote{
+Here we have opportunity for a remark (very simple indeed, but)
+of the utmost importance. {\sl We are not to equate separately to zero the
+coefficients of dt and dx}; for we must remember that this equation is
+of the form
+$$0=p\alpha + q\beta$$
+where $p$ and $q$ are numbers; and that, so long as $\alpha$ and $\beta$
+are actual and nonparallel vectors, the existence of such an equation
+requires (\S 24)}
+
+%tpdhere this is a cheesy use of \leqno. figure out how to do it right.
+\noindent
+$$\leqno{\textrm{Hence}}\hbox{\hskip 4cm}(1x)dttdx=0$$
+%tpdhere  this text comes out little.
+$$\leqno{\textrm{and}}\hbox{\hskip 4cm}\frac{x}{t^2}dt+\frac{dx}{t}=0$$
+From these, at once, $x = \frac{1}{2}$, since $dx$ and $dt$ are indeterminate.
+Thus the equation of the envelope is
+$$\begin{array}{rcl}
+\rho & = & \alpha t + \frac{1}{2}\left(\frac{\beta}{t}\alpha t\right)\\
+ & = & \frac{1}{2}\left(\alpha t + \frac{\beta}{t}\right)
+\end{array}$$
+the hyperbola as before; $\alpha$, $\beta$ being portions of its asymptotes.
+
+{\bf 42}. It may assist the student to a thorough comprehension
+of the above process, if we put it in a slightly different form.
+Thus the equation of the enveloping line may be written
+$$\rho=\alpha t(1x)+\beta\frac{x}{t}$$
+which gives
+$$d\rho = 0 = \alpha d\{t(1x)\}+\beta d\left(\frac{x}{t}\right)$$
+Hence, as $\alpha$ is not parallel to $\beta$, we must have
+$$d\{t(1x)\}=0,\hbox{\hskip 1cm}d\left(\frac{x}{t}\right)=0$$
+and these are, when expanded, the equations we obtained in the
+preceding section.
+
+{\bf 43}. For farther illustration we give a solution not directly
+employing the differential calculus. The equations of any two of
+the enveloping lines are
+$$\rho=\alpha t + x\left(\frac{\beta}{t}\alpha t\right)$$
+$$\rho=\alpha t_1 + x_1\left(\frac{\beta}{t_1}\alpha t_1\right)$$
+$t$ and $t_1$ being given, while $x$ and $x_1$ are indeterminate.
+
+At the point of intersection of these lines we have (\S 26),
+$$
+\left.
+\begin{array}{rcl}
+t(1x) & = & t_1(1x_1)\\
+\frac{x}{t} & = & \frac{x_1}{t_1}
+\end{array}
+\right\}$$
+These give, by eliminating $x_1$
+$$t(1x)=t_1\left(1\frac{t_1}{t}x\right)$$
+$$\leqno{\textrm{or}}\hbox{\hskip 4cm}x=\frac{t}{t_1+t}$$
+Hence the vector of the point of intersection is
+$$\rho=\frac{\alpha tt_1+\beta}{t_1+t}$$
+and thus, for the ultimate intersections, where $L\frac{t_1}{t}=1$,
+$$\rho=\frac{1}{2}\left(\alpha t+\frac{\beta}{t}\right)
+\textrm{ as before }$$
+COR. If. instead of the {\sl ultimate} intersections, we consider
+the intersections of pairs of these lines related by some law, we
+obtain useful results. Thus let
+$$tt_1 = 1$$
+$$\rho=\frac{\alpha+\beta}{t+\frac{1}{t}}$$
+or the intersection lies in the diagonal of the parallelogram on
+$\alpha$, $\beta$.
+
+If $t_1=mt$, where $m$ is constant,
+$$\rho=\frac{mt\alpha+\frac{\beta}{t}}{m+1}$$
+But we have also $x=\frac{1}{m+1}$
+
+Hence {\sl the locus of a point which divides in a given ratio a line
+cutting off a given area from two fixed axes, is a hyperbola of which
+these axes are the asymptotes}.
+
+If we take either
+$$tt_1(t+t_1)=\textrm{constant, or }
+\frac{t^2t_1^2}{t+t_1}=\textrm{constant}$$
+the locus is a parabola; and so on.
+
+It will be excellent practice for the student, at this stage, to
+work out in detail a number of similar questions relating to the
+envelope of, or the locus of the intersection of selected pairs from, a
+series of lines drawn according to a given law. And the process
+may easily be extended to planes. Thus, for instance, we may
+form the general equation of planes which cut off constant tetrahedra
+from the axes of coordinates. Their envelope is a surface of
+the third degree whose equation may be written
+$$\rho=x\alpha+y\beta+z\gamma$$
+$$\leqno{\textrm{where}}\hbox{\hskip 4cm}xyz=\alpha^3$$
+
+Again, find the locus of the point of intersection of three of
+this group of planes, such that
+the first intercepts on $\beta$ and $\gamma$,
+the second on $\gamma$ and $\alpha$,
+the third on $\alpha$ and $\beta$, lengths all equal to one
+another, \&c. But we must not loiter with such simple matters as
+these.
+
+{\bf 44}. The reader who is fond of Anharmonic Ratios and Trans
+versals will find in the early chapters of Hamilton's {\sl Elements of
+Quaternions} an admirable application of the composition of vectors
+to these subjects. The Theory of Geometrical Nets, in a plane,
+and in space, is there very fully developed; and the method is
+shown to include, as particular cases, the corresponding processes of
+Grassmann's {\sl Ausdehnungslehre} and M\"obius' {\sl Barycentrische Calcul}.
+Some very curious investigations connected with curves and surfaces
+of the second and third degrees are also there founded upon the
+composition of vectors.
+
+\section{Examples To Chapter 1.}
+
+1. The lines which join, towards the same parts, the extremities
+of two equal and parallel lines are themselves equal and parallel.
+({\sl Euclid}, I. xxxiii.)
+
+2. Find the vector of the middle point of the line which joins
+the middle points of the diagonals of any quadrilateral, plane or
+gauche, the vectors of the corners being given; and so prove that
+this point is the mean point of the quadrilateral.
+
+If two opposite sides be divided proportionally, and two new
+quadrilaterals be formed by joining the points of division, the mean
+points of the three quadrilaterals lie in a straight line.
+
+Show that the mean point may also be found by bisecting the
+line joining the middle points of a pair of opposite sides.
+
+3. Verify that the property of the coefficients of three vectors
+whose extremities are in a line (\S 30) is not interfered with by
+altering the origin.
+
+4. If two triangles $ABC$, $abc$, be so situated in space that $Aa$,
+$Bb$, $Cc$ meet in a point, the intersections of $AB$, $ab$, of $BC$, $bc$,
+and of $CA$, $ca$, lie in a straight line.
+
+5. Prove the converse of 4, i.e. if lines be drawn, one in each
+of two planes, from any three points in the straight line in which
+these planes meet, the two triangles thus formed are sections of a
+common pyramid.
+
+6. If five quadrilaterals be formed by omitting in succession
+each of the sides of any pentagon, the lines bisecting the diagonals
+of these quadrilaterals meet in a point. (H. Fox Talbot.)
+
+7. Assuming, as in \S 7, that the operator
+$$\cos\theta + \sqrt{1}\sin\theta$$
+turns any radius of a given circle through an angle $\theta$ in the
+positive direction of rotation, without altering its length, deduce
+the ordinary formulae for $\cos(A+B)$, $\cos(AB)$, $\sin(A+B)$, and
+$\sin(AB)$, in terms of sines and cosines of $A$ and $B$.
+
+8. If two tangents be drawn to a hyperbola, the line joining
+the centre with their point of intersection bisects the lines join
+ing the points where the tangents meet the asymptotes : and the
+secant through the points of contact bisects the intercepts on
+the asymptotes.
+
+9. Any two tangents, limited by the asymptotes, divide each
+other proportionally.
+
+10. If a chord of a hyperbola be one diagonal of a parallelogram
+whose sides are parallel to the asymptotes, the other diagonal passes
+through the centre.
+
+11. Given two points $A$ and $B$, and a plane, $C$. Find the
+locus of $P$, such that if $AP$ cut $C$ in $Q$, and $BP$ cut $C$ in $R$,
+$\overline{QR}$ may be a given vector.
+
+12. Show that\hbox{\hskip 1cm} $\rho = x^2\alpha+y^2\beta+(x+y)^2\gamma$\\
+is the equation of a cone of the second degree, and that its section
+by the plane
+$$\rho=\frac{p\alpha+q\beta+r\gamma}{p+q+r}$$
+is an ellipse which touches, at their middle points, the sides of the
+triangle of whose corners $\alpha$, $\beta$, $\gamma$
+are the vectors. (Hamilton, {\sl Elements}, p. 96.)
+
+13. The lines which divide, proportionally, the pairs of opposite
+sides of a gauche quadrilateral, are the generating lines of a
+hyperbolic paraboloid. ({\sl Ibid}. p. 97.)
+
+14. Show that\hbox{\hskip 2cm} $\rho=x^3\alpha+y^3\beta+z^3\gamma$\\
+where\hbox{\hskip 4cm} $x+y+z=0$\\
+represents a cone of the third order, and that its section by the plane
+$$\rho=\frac{p\alpha+q\beta+r\gamma}{p+q+r}$$
+is a cubic curve, of which the lines
+$$\rho=\frac{p\alpha+q\beta}{p+q},\textrm{ \&c}$$
+are the asymptotes and the three (real) tangents of inflection. Also
+that the mean point of the triangle formed by these lines is a
+conjugate point of the curve. Hence that the vector $\alpha+\beta+\gamma$
+is a conjugate ray of the cone. ({\sl Ibid}. p. 96.)
+
+\section{Products And Quotients of Vectors}
+
+{\bf 45}. We now come to the consideration of questions in which
+the Calculus of Quaternions differs entirely from any previous
+mathematical method; and here we shall get an idea of what a
+Quaternion is, and whence it derives its name. These questions
+are fundamentally involved in the novel use of the symbols of
+multiplication and division. And the simplest introduction to
+the subject seems to be the consideration of the quotient, or ratio,
+of two vectors.
+
+{\bf 46}. If the given vectors be parallel to each other, we have
+already seen (\S 22) that either may be expressed as a numerical
+multiple of the other; the multiplier being simply the ratio of
+their lengths, taken positively if they have similar currency,
+negatively if they run opposite ways.
+
+{\bf 47}. If they be not parallel, let $\overline{OA}$ and
+$\overline{OB}$ be drawn parallel
+and equal to them from any point $O$; and the question is reduced
+to finding the value of the ratio of two vectors drawn from the
+same point. Let us first find {\sl upon how many distinct numbers this
+ratio depends}.
+
+We may suppose $\overline{OA}$ to be changed into
+$\overline{OB}$ by the following successive processes.
+
+1st. Increase or diminish the length of $\overline{OA}$ till it becomes
+equal to that of $\overline{OB}$. For this only one number is required, viz.
+the ratio of the lengths of the two vectors. As Hamilton remarks,
+this is a positive, or rather a {\sl signless}, number.
+
+2nd. Turn $\overline{OA}$ about $O$, in the common plane of the two
+vectors, until its direction coincides with that of $\overline{OB}$, and
+(remembering the effect of the first operation) we see that the two vectors
+now coincide or become identical. To specify this operation three
+numbers are required, viz. two angles (such as node and inclination
+in the case of a planet's orbit) to fix the plane in which the rotation
+takes place, and {\sl one} angle for the amount of this rotation.
+
+Thus it appears that the ratio of two vectors, or the multiplier
+required to change one vector into another, in general depends upon
+{\sl four} distinct numbers, whence the name QUATERNION.
+
+A quaternion q is thus {\sl defined} as expressing a relation
+$$\beta=q\alpha$$
+between two vectors $\alpha$, $\beta$.
+By what precedes, the vectors $\alpha$, $\beta$,
+which serve for the definition of a given quaternion, must be in a
+given plane, at a given inclination to each other, and with their
+lengths in a given ratio ; but it is to be noticed that they may be
+{\sl any} two such vectors. [{\sl Inclination} is understood to include sense,
+or currency, of rotation from $\alpha$ to $\beta$.]
+
+The particular case of perpendicularity of the two vectors, where
+their quotient is a vector perpendicular to their plane, is fully
+considered below; \S\S 64, 65, 72, \&c.
+
+{\bf 48}. It is obvious that the operations just described may be
+performed, with the same result, in the opposite order, being perfectly
+independent of each other. Thus it appears that a quaternion,
+considered as the factor or agent which changes one definite vector
+into another, may itself be decomposed into two factors of which
+the order is immaterial.
+
+The {\sl stretching} factor, or that which performs the first operation
+in \S 47, is called the TENSOR, and is denoted by prefixing $T$ to the
+quaternion considered.
+
+The {\sl turning factor}, or that corresponding to the second operation
+in \S 47, is called the VERSOR, and is denoted by the letter $U$ prefixed
+to the quaternion.
+
+{\bf 49}. Thus, if $\overline{OA} = \alpha$,
+$\overline{OB} = \beta$, and if $q$ be the quaternion
+which changes $\alpha$ to $\beta$, we have
+$$\beta = q\alpha$$
+which we may write in the form
+$$\frac{\beta}{\alpha} = q\textrm{, or }\beta\alpha^{1}=q$$
+if we agree to {\sl define} that
+$$\frac{\beta}{\alpha}\alpha = \beta\alpha^{1}\alpha = \beta$$
+Here it is to be particularly noticed that we write $q$
+{\sl before} $\alpha$ to
+signify that $\alpha$ is multiplied by (or operated on by) $q$, not $q$
+multiplied by $\alpha$.
+
+This remark is of extreme importance in quaternions, for, as we
+shall soon see, the Commutative Law does not generally apply to
+the factors of a product.
+
+We have also, by \S\S 47, 48,
+$$q=TqUq=UqTq$$
+where, as before, $Tq$ depends merely on the relative lengths of
+$\alpha$ and $\beta$, and $Uq$ depends solely on their directions.
+
+Thus, if $\alpha_1$ and $\beta_1$
+be vectors of unit length parallel to $\alpha$ and $\beta$
+respectively,
+$$T\frac{\beta_1}{\alpha_1} = T\beta_1/T\alpha_1 = 1\textrm{, }
+U\frac{\beta_1}{\alpha_1} = U\beta_1/U\alpha_1 = U\frac{\beta}{\alpha}$$
+As will soon be shown, when $\alpha$ is perpendicular to $\beta$,
+i.e. when the versor of the quotient is quadrantal, it is a unitvector.
+
+{\bf 50}. We must now carefully notice that the quaternion which
+is the quotient when $\beta$ is divided by $\alpha$ in no way depends upon
+the {\sl absolute} lengths, or directions, of these vectors. Its value
+will remain unchanged if we substitute for them any other pair
+of vectors which
+
+\noindent
+\hbox{\hskip 2cm}(1) have their lengths in the same ratio,
+
+\noindent
+\hbox{\hskip 2cm}(2) have their common plane the same or parallel,
+
+\noindent
+and\hbox{\hskip 1.32cm} (3) make the same angle with each other.
+
+Thus in the annexed figure
+
+\includegraphics{ps/quat6.ps}
+
+$$\frac{O_1B_1}{O_1A_1} = \frac{\overline{OB}}{\overline{OA}}$$
+if, and only if,
+
+\noindent
+\hbox{\hskip 2cm}$(1)\textrm{ }\frac{O_1B_1}{O_1A_1} = \frac{OB}{OA}$
+
+\noindent
+\hbox{\hskip 2cm}$(2)\textrm{ plane }AOB\textrm{ parallel to plane }
+A_1O_1B_1$
+
+\noindent
+\hbox{\hskip 2cm}$(3)\textrm{ }\angle{}AOB = \angle A_1O_1B_1$
+
+[Equality of angles is understood to include
+concurrency of rotation. Thus in the annexed
+figure the rotation about an axis drawn upwards
+from the plane is negative (or clock wise) from
+$OA$ to $OB$, and also from $O_1A_1$ to $O_1B_1$.]
+
+It thus appears that if
+$$\beta = q\alpha\textrm{, }\delta = q\gamma$$
+the vectors $\alpha$, $\beta$, $\gamma$, $\delta$
+are parallel to one plane, and may be repre
+sented (in a highly extended sense) as {\sl proportional} to one another,
+thus: 
+$$\beta : \alpha = \delta : \gamma$$
+
+And it is clear from the previous part of this section that this
+may be written not only in the form
+$$\alpha : \beta = \gamma : \delta$$
+but also in either of the following forms: 
+$$\gamma : \alpha = \delta : \beta$$
+$$\alpha : \gamma = \beta : \delta$$
+
+While these proportions are true as equalities of ratios, they
+do not usually imply equalities of products.
+
+Thus, as the first of these was equivalent to the equation
+$$\frac{\beta}{\alpha}=\frac{\delta}{\gamma}=q\textrm{, or }
+\beta\alpha^{1}=\delta\gamma^{1}=q$$
+the following three imply separately, (see next section)
+$$\frac{\alpha}{\beta}=\frac{\gamma}{\delta}=q^{1}\textrm{, }
+\frac{\gamma}{\alpha}=\frac{\delta}{\beta}=r\textrm{, }
+\frac{\alpha}{\gamma}=\frac{\beta}{\delta}=r^{1}$$
+or, if we please,
+$$\alpha\beta^{1}=\gamma\delta^{1}=q^{1}\textrm{, }
+\gamma\alpha^{1}=\delta\beta^{1}=r\textrm{, }
+\alpha\gamma^{1}=\beta\delta^{1}=r^{1}$$
+where $r$ is a {\sl new} quaternion, which has not necessarily anything
+(except its plane), in common with $q$.
+
+But here great caution is requisite, for we are {\sl not} entitled to
+conclude from these that
+$$\alpha\delta=\beta\gamma\textrm{, \&c.}$$
+
+This point will be fully discussed at a later stage. Meanwhile
+we may merely {\sl state} that from
+$$\frac{\alpha}{\beta}=\frac{\gamma}{\delta}\textrm{, or }
+\frac{\beta}{\alpha} = \frac{\delta}{\gamma}$$
+we are entitled to deduce a number of equivalents such as
+$$\alpha\beta^{1}\delta=\gamma\textrm{, or }
+\alpha=\gamma\delta^{1}\beta\textrm{, or }
+\beta^{1}\delta=\alpha^{1}\gamma\textrm{, \&c}$$
+
+{\bf 51}. The {\sl Reciprocal} of a quaternion $q$ is defined by the
+equation
+$$\frac{1}{q}q=q^{1}=1=q\frac{1}{q}=qqe^{1}$$
+Hence if
+$$\frac{\beta}{\alpha}=q\textrm{, or}$$
+$$\beta=q\alpha$$
+we must have
+$$\frac{\alpha}{\beta}=\frac{1}{q}=q^{1}$$
+For this gives
+$$\frac{\alpha}{\beta}\beta=q^{1}q\alpha$$
+and each member of the equation is evidently equal to $\alpha$.
+Or thus: 
+$$\beta=q\alpha$$
+Operate {\sl by} $q^{1}$
+$$q^{1}\beta = \alpha$$
+Operate {\sl on} $\beta^{1}$
+$$q^{1} = \alpha\beta^{1} = \frac{\alpha}{\beta}$$
+
+Or, we may reason thus:  since $q$ changes $\overline{OA}$ to
+$\overline{OA}$, $q^{1}$ must
+change $\overline{OB}$ to $\overline{OA}$,
+and is therefore expressed by $\frac{\alpha}{\beta}$ (\S 49).
+
+The tensor of the reciprocal of a quaternion is therefore the
+reciprocal of the tensor; and the versor differs merely by the
+{\sl reversal} of its representative angle. The versor, it must be
+remembered, gives the plane and angle of the turning  it has
+nothing to do with the extension.
+
+[{\sl Remark}. In \S\S 4951, above, we had such expressions as
+$\frac{\beta}{\alpha}=\beta\alpha^{1}$.
+We have also met with $\alpha^{1}\beta$. Cayley suggests that this
+also may be written in the ordinary fractional form by employing
+the following distinctive notation: 
+$$\frac{\beta}{\alpha}=\beta\alpha^{1}=\frac{\beta}{\alpha}\textrm{, }
+\alpha^{1}\beta=\frac{\beta}{\alpha}$$
+
+(It might, perhaps, be even simpler to use the {\sl solidus} as
+recommended by Stokes, along with an obviously correlative
+type: thus,
+$$\frac{\beta}{\alpha}=\beta\alpha^{1}=\beta/\alpha\textrm{, }
+\alpha^{1}\beta=\alpha\\ \beta$$
+
+I have found such notations occasionally convenient for private
+work, but I hesitate to introduce changes unless they are abso
+lutely required. See remarks on this point towards the end of the
+{\sl Preface to the Second Edition} reprinted above.]
+
+{\bf 52}. The {\sl Conjugate} of a quaternion $q$, written $Kq$, has the
+same tensor, plane, and angle, only the angle is taken the reverse
+way; or the versor of the conjugate is the reciprocal of the versor
+of the quaternion, or (what comes to the same thing) the versor of
+the reciprocal.
+
+\includegraphics{ps/quat7.ps}
+
+Thus, if $OA$, $OB$, $OA^{'}$ , lie in one plane, and if
+$OA^{'} = OA$, and $\angle A^{'}OB = \angle BOA$, we have
+$$\frac{\overline{OB}}{\overline{OA}}=q$$,
+and
+$$\frac{\overline{OB}}{\overline{OA^{'}}}=\textrm{ congugate of }q=Kq$$
+
+By last section we see that
+$$Kq=(Tq)^2q^{1}$$
+Hence\hbox{\hskip 4cm}$qKq=Kqq=(Tq)^2$
+
+This proposition is obvious, if we recollect that
+the tensors of $q$ and $Kq$ are equal, and that the
+versors are such that either {\sl annuls} the effect of the other; while
+the order of their application is indifferent. The joint effect of
+these factors is therefore merely to multiply twice over by the
+common tensor.
+
+{\bf 53}. It is evident from the results of \S 50 that, if $\alpha$ and $\beta$
+be of equal length, they may be treated as of unitlength so far as
+their quaternion quotient is concerned. This quotient is therefore
+a versor (the tensor being unity) and may be represented indifferently
+by any one of an infinite number of concurrent arcs of
+given length lying on the circumference of a circle, of which the
+two vectors are radii. This is of considerable importance in the
+proofs which follow.
+
+\includegraphics{ps/quat8.ps}
+
+Thus the versor
+${\displaystyle \frac{\overline{OB}}{\overline{OA}}}$ may be represented
+in magnitude, plane, and currency of rotation (\S 50)
+by the arc $AB$, which may in this extended sense be written
+${\stackrel{\frown}{AB}}$.
+
+And, similarly, the versor
+${\displaystyle \frac{\overline{OB_1}}{\overline{OA_1}}}$
+may be represented by
+${\stackrel{\frown}{A_1B_1}}$
+which is equal to (and concurrent with)
+${\stackrel{\frown}{AB}}$ if
+$$\angle A_1OB_1 = \angle AOB$$
+i.e. if the versors are {\sl equal}, in the quaternion meaning of the
+word.
+
+{\bf 54}. By the aid of this process, when a versor is represented as
+an arc of a great circle on the unitsphere, we can easily prove
+that {\sl quaternion multiplication is not generally commutative}.
+
+\includegraphics{ps/quat9.ps}
+
+Thus let $q$ be the versor ${\stackrel{\frown}{AB}}$ or
+${\displaystyle \frac{\overline{OB}}{\overline{OA}}}$,
+where $O$ is the centre of the sphere.
+
+Take ${\stackrel{\frown}{BC}} = {\stackrel{\frown}{AB}}$,
+(which, it must be remembered, makes the points $A$, $B$, $C$, lie
+in one great circle), then $q$ may also be
+represented by ${\displaystyle \frac{\overline{OC}}{\overline{OB}}}$.
+
+In the same way any other versor $r$ may be represented by
+${\stackrel{\frown}{DB}}$ or ${\stackrel{\frown}{BE}}$ and by
+${\displaystyle \frac{\overline{OB}}{\overline{OD}}}$ or
+${\displaystyle \frac{\overline{OE}}{\overline{OB}}}$.
+
+[The line $OB$ in the figure is definite, and is given by the
+intersection of the planes of the two versors.]
+
+Now $r\overline{OD} = \overline{OB}$, and $q\overline{OB}=\overline{OC}$.
+
+Hence $qr\overline{OD} = \overline{OC}$,
+
+or $qr = {\displaystyle \frac{\overline{OC}}{\overline{OD}}}$,
+and may therefore be represented by the arc ${\stackrel{\frown}{DC}}$ of
+a great circle.
+
+But $rq$ is easily seen to be represented by the arc
+${\stackrel{\frown}{AE}}$.
+
+For $q\overline{OA}=\overline{OB}$, and $r\overline{OB}=\overline{OE}$,
+
+whence $rq\overline{OA} = \overline{OE}$. and
+$rq = {\displaystyle \frac{\overline{OE}}{\overline{OA}}}$.
+
+Thus the versors $rq$ and $qr$, though represented by arcs of equal
+length, are not generally in the same plane and are therefore
+unequal: unless the planes of $q$ and $r$ coincide.
+
+Remark. We see that we have assumed, or defined, in the
+above proof, that $q . r\alpha = qr . \alpha$.
+and $r.q\alpha = rq.\alpha$ in the special case
+when $q\alpha$, $r\alpha$, $q.r\alpha$ and $r.q\alpha$ are all {\sl vectors}.
+
+{\bf 55}. Obviously ${\stackrel{\frown}{CB}}$ is $Kq$,
+${\stackrel{\frown}{BD}}$ is $Kr$, and
+${\stackrel{\frown}{CD}}$ is $K (qr)$. But
+${\stackrel{\frown}{CD}} = {\stackrel{\frown}{BD}}.{\stackrel{\frown}{CB}}$
+as we see by applying both to OC. This gives us
+the very important theorem
+$$K (qr) = Kr . Kq$$
+i.e. {\sl the conjugate of the product of two versors is the product of their
+conjugates in inverted order}. This will, of course, be extended to
+any number of factors as soon as we have proved the associative
+property of multiplication. (\S 58 below.)
+
+{\bf 56}. The propositions just proved are, of course, true of quater
+nions as well as of versors; for the former involve only an additional
+numerical factor which has reference to the length merely, and not
+the direction, of a vector (\S 48), and is therefore commutative with
+all other factors.
+
+{\bf 57}. Seeing thus that the commutative law does not in general
+hold in the multiplication of quaternions, let us enquire whether
+the Associative Law holds generally. That is if $p$, $q$, $r$ be three
+quaternions, have we
+$$p.qr = pq.r?$$
+
+This is, of course, obviously true if $p$, $q$, $r$ be numerical quantities,
+or even any of the imaginaries of algebra. But it cannot be con
+sidered as a truism for symbols which do not in general give
+
+$$pq = qp$$
+
+We have assumed it, in definition, for the special case when $r$,
+$qr$, and $pqr$ are all vectors. (\S 54.) But we are not entitled to
+assume any more than is absolutely required to make our
+definitions complete.
+
+{\bf 58}. In the first place we remark that $p$, $q$, and $r$ may be
+considered as versors only, and therefore represented by arcs of
+great circles on the unit sphere, for their tensors may obviously
+(\S 48) be divided out from both sides, being commutative with the
+versors.
+
+Let ${\stackrel{\frown}{AB}}=p$,
+${\stackrel{\frown}{ED}} = {\stackrel{\frown}{CA}} = q$, and
+${\stackrel{\frown}{FE}} = r$.
+
+Join $BC$ and produce the great circle till it meets $EF$ in $H$, and
+make ${\stackrel{\frown}{KH}}={\stackrel{\frown}{FE}} = r$,
+and ${\stackrel{\frown}{HG}} = {\stackrel{\frown}{CB}} =pq$ (\S 54).
+
+\includegraphics{ps/quat10.ps}
+
+Join $GK$. Then
+${\stackrel{\frown}{KG}} =
+{\stackrel{\frown}{HG}} . {\stackrel{\frown}{KH}} = pq . r$.
+
+Join $FD$ and produce it to meet $AB$ in $M$. Make
+$${\stackrel{\frown}{LM}} = {\stackrel{\frown}{FD}}
+\textrm{, and }{\stackrel{\frown}{MN}} = {\stackrel{\frown}{AB}}$$
+
+and join $NL$. Then
+$${\stackrel{\frown}{LN}}= {\stackrel{\frown}{MN}} .
+{\stackrel{\frown}{LM}}=p.qr$$.
+
+Hence to show that $p . qr = pq . r$
+
+all that is requisite is to prove that $LN$, and $KG$, described as
+above, are {\sl equal arcs of the same great circle}, since, by the figure,
+they have evidently similar currency. This is perhaps most easily
+effected by the help of the fundamental properties of the curves
+known as {\sl Spherical Conics}. As they are not usually familiar to
+students, we make a slight digression for the purpose of proving
+these fundamental properties ; after Chasles, by whom and Magnus
+they were discovered. An independent proof of the associative
+principle will presently be indicated, and in Chapter VIII. we shall
+employ quaternions to give an independent proof of the theorems
+now to be established.
+
+{\bf 59}.* DEF. {\sl A spherical conic is the curve of intersection of a
+cone of the second degree with a sphere, the vertex of the cone being
+the centre of the sphere}.
+
+LEMMA. If a cone have one series of circular sections, it has
+another series, and any two circles belonging to different series lie
+on a sphere. This is easily proved as follows.
+
+Describe a sphere, $A$, cutting the cone in one circular section,
+$C$, and in any other point whatever, and let the side $OpP$ of the
+cone meet $A$ in $p$, $P$ ; $P$ being a point in $C$. Then $PO.Op$ is
+constant, and, therefore, since $P$ lies in a plane, $p$ lies on a sphere,
+$a$, passing through $0$. Hence the locus, $c$, of $p$ is a circle, being
+the intersection of the two spheres $A$ and $a$.
+
+Let $OqQ$ be any other side of the cone, $q$ and $Q$ being points in
+$c$, $C$ respectively. Then the quadrilateral $qQPp$ is inscribed in a
+circle (that in which its plane cuts the sphere $A$) and the exterior
+
+\includegraphics{ps/quat11.ps}
+
+angle at $p$ is equal to the interior angle at $Q$. If $OL$, $OM$ be the
+lines in which the plane $POQ$ cuts the {\sl cyclic planes} (planes through
+$O$ parallel to the two series of circular sections) they are obviously
+parallel to $pq$, $QP$, respectively; and therefore
+
+$$\angle LOp = \angle Opq = \angle OQP = \angle MOQ$$
+
+Let any third side, $OrR$, of the cone be drawn, and let the
+plane $OPR$ cut the cyclic planes in $0l$, $Om$ respectively. Then,
+evidently,
+$$\angle lOL = \angle qpr$$
+$$\angle MOm = \angle QPR$$
+
+and these angles are independent of the position of the points $p$
+and $P$, if $Q$ and $R$ be fixed points.
+
+\includegraphics{ps/quat12.ps}
+
+In the annexed section of the above spacediagram by a sphere
+whose centre is $O$, $lL$, $Mm$ are the great circles which represent
+the cyclic planes, $PQR$ is the spherical conic which represents the
+cone. The point $P$ represents the line $OpP$, and so with the
+others. The propositions above may now be stated thus,
+
+$$\textrm{Arc } PL = \textrm{arc } MQ$$
+
+and, if $Q$ and $R$ be fixed, $Mm$ and $lL$ are constant arcs whatever be
+the position of $P$.
+
+{\bf 60}. The application to \S 58 is now obvious. In the figure of
+that article we have
+$$
+{\stackrel{\frown}{FE}}={\stackrel{\frown}{KH}}\textrm{, }
+{\stackrel{\frown}{ED}}={\stackrel{\frown}{CA}}\textrm{, }
+{\stackrel{\frown}{HG}}={\stackrel{\frown}{CB}}\textrm{, }
+{\stackrel{\frown}{LM}}={\stackrel{\frown}{FD}}
+$$
+
+Hence $L$, $C$, $G$, $D$ are points of a spherical conic whose cyclic
+planes are those of $AB$, $FE$. Hence also $KG$ passes through $L$,
+and with $LM$ intercepts on $AB$ an arc equal to
+${\stackrel{\frown}{AB}}$. That is, it
+passes through $N$, or $KG$ and $LN$ are arcs of the same great circle :
+and they are equal, for $G$ and $L$ are points in the spherical
+conic.
+
+Also, the associative principle holds for any number of
+quaternion factors. For, obviously,
+
+$$qr . st = qrs . t = \textrm{\&c., \&c.,}$$
+
+since we may consider $qr$ as a single quaternion, and the above
+proof applies directly.
+
+{\bf 61}. That quaternion addition, and therefore also subtraction,
+is commutative, it is easy to show.
+
+\includegraphics{ps/quat13.ps}
+
+For if the planes of two quaternions,
+$q$ and $r$, intersect in the line $OA$, we
+may take any vector $\overline{OA}$ in that line,
+and at once find two others, $\overline{OB}$ and
+$\overline{OC}$, such that
+
+$$\overline{OB} = q\overline{OA}$$
+and\hbox{\hskip 4cm}$\overline{CO} = r\overline{OA}$
+
+And\hbox{\hskip 2cm}$(q + r)\overline{OA}
+\overline{OB}+\overline{OC}=\overline{OC}+\overline{OB}=
+(r + q) \overline{OA}$
+
+since vector addition is commutative (\S 27).
+
+Here it is obvious that $(q + r) \overline{OA}$, being the diagonal of the
+parallelogram on $\overline{OB}$, $\overline{OC}$,
+divides the angle between $OB$ and $OC$
+in a ratio depending solely on the ratio of the lengths of these
+lines, i.e. on the ratio of the tensors of $q$ and $r$. This will be useful
+to us in the proof of the distributive law, to which we proceed.
+
+{\bf 62}. Quaternion multiplication, and therefore division, is
+distributive. One simple proof of this depends on the possibility,
+shortly to be proved, of representing {\sl any} quaternion as a linear
+function of three given rectangular unit vectors. And when the
+proposition is thus established, the associative principle may readily
+be deduced from it.
+
+[But Hamilton seems not to have noticed that we may employ
+for its proof the properties of Spherical Conies already employed
+
+\includegraphics{ps/quat14.ps}
+
+in demonstrating the truth of the associative principle. "For
+continuity we give an outline of the proof by this process.
+
+Let ${\stackrel{\frown}{BA}}$,
+${\stackrel{\frown}{CA}}$
+represent the versors of $q$ and $r$, and be the great
+circle whose plane is that of $p$.
+
+Then, if we take as operand the vector $\overline{OA}$, it is obvious that
+$U (q + r)$ will be represented by some such arc as
+${\stackrel{\frown}{DA}}$ where
+$B$, $D$, $C$ are in one great circle;
+for $(q + r) \overline{OA}$ is in the same plane
+as $q\overline{OA}$ and $r\overline{OA}$,
+and the relative magnitude of the arcs $BD$ and
+$DC$ depends solely on the tensors of $q$ and $r$. Produce $BA$, $DA$,
+$CA$ to meet be in $b$, $d$, $c$ respectively, and make
+
+$${\stackrel{\frown}{Eb}} = {\stackrel{\frown}{BA}},
+{\stackrel{\frown}{Fd}} = {\stackrel{\frown}{DA}},
+{\stackrel{\frown}{Gc}} = {\stackrel{\frown}{CA}}$$
+
+Also make
+${\stackrel{\frown}{b\beta}} =
+{\stackrel{\frown}{d\delta}} =
+{\stackrel{\frown}{c\gamma}}=p$. Then $E$, $F$, $G$, $A$ lie on a spherical
+conic of which $BC$ and $bc$ are the cyclic arcs. And, because
+${\stackrel{\frown}{b\beta}} =
+{\stackrel{\frown}{d\delta}} =
+{\stackrel{\frown}{c\gamma}}$,
+${\stackrel{\frown}{\beta E}}$,
+${\stackrel{\frown}{\delta F}}$,
+${\stackrel{\frown}{\gamma G}}$, when produced, meet in a point $H$
+which is also on the spherical conic (\S 59*). Let these arcs meet $BC$
+in $J$, $L$, $K$ respectively. Then we have
+$${\stackrel{\frown}{JH}}={\stackrel{\frown}{E\beta}}=pUq$$
+$${\stackrel{\frown}{LH}}={\stackrel{\frown}{F\delta}}=pU(q+r)$$
+$${\stackrel{\frown}{KH}}={\stackrel{\frown}{G\gamma}}=pUr$$
+Also\hbox{\hskip 4cm}${\stackrel{\frown}{LJ}}={\stackrel{\frown}{DB}}$\\
+and\hbox{\hskip 4cm}${\stackrel{\frown}{KL}}={\stackrel{\frown}{CD}}$
+
+And, on comparing the portions of the figure bounded respectively
+by $HKJ$ and by $ACB$ we see that (when considered with reference
+to their effects as factors multiplying
+$\overline{OH}$ and $\overline{OA}$ respectively)
+
+\hbox{\hskip 2cm}$pU(q4+r)$ bears the same relation to $pUq$ and $pUr$\\
+that\hbox{\hskip 1cm}$U(q+r)$ bears to $Uq$ and $Ur$.\\
+But\hbox{\hskip 1cm}$T(q+r)U(q+r)=q+r=TqUq+TrUr$. \\
+Hence\hbox{\hskip 1cm}$T(q+r).pU(q+r)=Tq .pUq+Tr.pUr$;\\
+or, since the tensors are mere numbers and commutative with all
+other factors,
+$$p(q+r)=pq+pr$$
+In a similar manner it may be proved that
+$$(q+ )p=qp+rp$$
+And then it follows at once that
+$$(p + q) (r + s) = pr + ps + qr + qs$$
+where, by \S 61, the order of the partial products is immaterial.]
+
+{\bf 63}. By similar processes to those of \S 53 we see that versors,
+and therefore also quaternions, are subject to the indexlaw
+$$q^m.q^n=q^{m+n}$$
+at least so long as $m$ and $n$ are positive integers.
+
+The extension of this property to negative and fractional
+exponents must be deferred until we have defined a negative or
+fractional power of a quaternion.
+
+{\bf 64}. We now proceed to the special case of {\sl quadrantal} versors,
+from whose properties it is easy to deduce all the foregoing
+results of this chapter. It was, in fact, these properties whose
+invention by Hamilton in 1843 led almost intuitively to the
+establishment of the Quaternion Calculus. We shall content
+ourselves at present with an assumption, which will be shown
+to lead to consistent results ; but at the end of the chapter we
+shall show that no other assumption is possible, following for this
+purpose a very curious quasimetaphysical speculation of Hamilton.
+
+{\bf 65}. Suppose we have a system of three mutually perpendicular
+unitvectors, drawn from one point, which we may call for shortness
+{\bf i}, {\bf j}, {\bf k}.
+Suppose also that these are so situated that a positive
+(i.e. {\sl lefthanded}) rotation through a right angle about {\bf i}
+as an axis
+brings {\bf j} to coincide with {\bf k}. Then it is obvious that positive
+quadrantal rotation about {\bf j} will make {\bf k} coincide with {\bf i};
+and, about {\bf k}, will make {\bf i} coincide with {\bf j}.
+
+For defniteness we may suppose {\bf i} to be drawn {\sl eastwards}, {\bf j}
+{\sl northwards}, and {\bf k} {\sl upwards}.
+Then it is obvious that a positive
+(lefthanded) rotation about the eastward line ({\bf i}) brings the northward
+line ({\bf j}) into a vertically upward position ({\bf k}) ; and so of the
+others.
+
+{\bf 66}. Now the operator which turns {\bf j} into {\bf k} is a quadrantal
+versor (\S 53) ; and, as its axis is the vector {\bf i},
+we may call it {\sl i}.
+
+Thus $$\frac{{\rm {\bf k}}}{{\rm {\bf j}}}=
+i\textrm{, or }{\rm {\bf k}}=i{\rm {\bf j}}\eqno{(1)}$$
+
+Similary we may put$$\frac{{\rm {\bf i}}}{{\rm {\bf k}}}=
+j\textrm{, or }{\rm {\bf i}}=j{\rm {\bf k}}\eqno{(2)}$$
+
+and $$\frac{{\rm {\bf j}}}{{\rm {\bf i}}}=
+k\textrm{, or }{\rm {\bf j}}=k{\rm {\bf i}}\eqno{(3)}$$
+
+[It may be here noticed, merely to show the symmetry of the
+system we arc explaining, that if the three mutually perpendicular
+vectors {\bf i}, {\bf j}, {\bf k}
+be made to revolve about a line equally inclined to
+all, so that {\bf i} is brought to coincide with {\bf j},
+{\bf j} will then coincide
+with {\bf k}, and {\bf k} with {\bf i}:
+and the above equations will still hold good,
+only (1) will become (2), (2) will become (3), and (3) will become (1).]
+
+{\bf 67}. By the results of \S 50 we see that
+$$\frac{{\rm {\bf j}}}{\rm {\bf k}}=\frac{{\rm {\bf k}}}{\rm {\bf j}}$$
+i.e. a southward unit vector bears the same ratio to an upward
+unitvector that the latter does to a northward one; and therefore
+we have
+
+Thus $$\frac{{\rm {\bf j}}}{{\rm {\bf k}}}=
+i\textrm{, or }{\rm {\bf j}}=i{\rm {\bf k}}\eqno{(4)}$$
+
+Similary t$$\frac{{\rm {\bf k}}}{{\rm {\bf i}}}=
+j\textrm{, or }{\rm {\bf k}}=j{\rm {\bf i}}\eqno{(5)}$$
+
+and $$\frac{{\rm {\bf i}}}{{\rm {\bf j}}}=
+k\textrm{, or }{\rm {\bf i}}=k{\rm {\bf j}}\eqno{(6)}$$
+
+{\bf 68}. By (4) and (1) we have
+
+$$j = ik = i(ij)\textrm{ (by the assumption in \S 54) }= i^2j$$
+
+Hence
+$$i^2 =  1\eqno{(7)}$$
+
+Arid in the same way, (5) and (2) give
+$$j^2=1\eqno{(8)}$$
+and (6) and (3)
+$$k^2=1\eqno{(9)}$$
+
+Thus, as the directions of {\bf i}, {\bf j}, {\bf k}
+are perfectly arbitrary, we see that
+{\sl the square of every quadrantal versor is negative unity}.
+
+[Though the following proof is in principle exactly the same as
+the foregoing, it may perhaps be of use to the student, in showing
+him precisely the nature as well as the simplicity of the step we
+have taken.
+
+\includegraphics{ps/quat15.ps}
+
+Let $ABA^{'}$ be a semicircle, whose centre
+is $0$, and let $OB$ be perpendicular to $AOA^{'}$.
+
+Then ${\displaystyle\frac{\overline{OB}}{\overline{OA^{'}}}}=q$
+suppose, is a quadrantal versor, and is evidently equal to
+${\displaystyle\frac{\overline{OA^{'}}}{\overline{OB}}}$ ;
+
+\S\S 50, 53. Hence
+
+$$q^2=\frac{\overline{OA^{'}}}{\overline{OB}}.
+\frac{\overline{OB}}{\overline{OA}}=
+\frac{\overline{OA^{'}}}{\overline{OA}}=1]$$
+
+{\bf 69}.
+Having thus found that the squares of {\sl i}, {\sl j}, {\sl k} are each
+equal to negative unity ; it only remains that we find the values of
+their products two and two. For, as we shall see, the result is such
+as to show that the value of any other combination whatever of
+{\sl i},{\sl j}, {\sl k}
+(as factors of a product) may be deduced from the values of
+these squares and products.
+
+Now it is obvious that
+$$\frac{{\rm {\bf k}}}{\rm {\bf i}}=
+\frac{{\rm {\bf i}}}{\rm {\bf k}}=j$$
+(i.e. the versor which turns a westward unitvector into an upward
+one will turn the upward into an eastward unit) ; or
+$${\rm {\bf k}}=j({\rm {\bf i}}) = j{\rm {\bf i}}\eqno{(10)}$$
+
+Now let us operate on the two equal vectors in (10) by the
+same versor, {\sl i}, and we have
+$$i{\rm {\bf k}} = i(j{\rm {\bf i}}) = j{\rm {\bf i}}$$
+But by (4) and (3)
+$$i{\rm {\bf k}}={\rm {\bf j}}=k{\rm {\bf i}}$$
+
+Comparing these equations, we have
+$$ij{\rm {\bf i}}=k{\rm {\bf i}}$$
+$$
+\left.
+\begin{array}{lr}
+\textrm{or, \S 54 (end), } &ij=k\\
+\textrm{and symmetry gives}&jk=i\\
+ &ki=j\\
+\end{array}
+\right\}\eqno{(11)}
+$$
+
+The meaning of these important equations is very simple ; and
+is, in fact, obvious from our construction in \S 54 for the multiplication
+of versors ; as we see by the annexed figure, where we must
+remember that {\sl i}, {\sl j}, {\sl k}
+are quadrantal versors whose planes are at
+right angles, so that the figure represents
+a hemisphere divided into quadrantal
+triangles. [The arrowheads indicate the
+direction of each vector arc.]
+
+\includegraphics{ps/quat16.ps}
+
+Thus, to show that $ij = k$, we have,
+$O$ being the centre of the sphere, $N$, $E$,
+$S$, $W$ the north, east, south, and west,
+and $Z$ the zenith (as in \S 65) ;
+
+$$j\overline{OW}=\overline{OZ}$$
+whence\hbox{\hskip 2cm}$ij\overline{OW}=i\overline{OZ}=
+\overline{OS} = k\overline{OW}$
+
+* The negative sign, being a mere numerical factor, is evidently commutative
+with $j$ indeed we may, if necessary, easily assure ourselves of the fact
+that to turn the negative (or reverse) of a vector through a right
+(or indeed any) angle, is the same thing as to turn the vector through
+that angle and then reverse it.
+
+{\bf 70}. But, by the same figure,
+$$i\overline{ON}=\overline{OZ}$$
+whence\hbox{\hskip 1cm}$ji\overline{ON} =j\overline{OZ}
+= \overline{OE} = \overline{OW} =  k\overline{ON}$.
+
+{\bf 71}. From this it appears that
+$$
+\left.
+\begin{array}{c}
+ji = k\\
+kj = i\\
+ik = j\\
+\end{array}
+\right\}\eqno{(12)}
+$$
+and thus, by comparing (11),
+$$
+\left.
+\begin{array}{c}
+ij=ji=k\\
+jk=kj=i\\
+ki=ik=j\\
+\end{array}
+\right\}\eqno{(11),(12)}
+$$
+
+These equations, along with
+$$i^2=j^2=k^2=1\eqno{((7),(8),(9))}$$
+contain essentially the whole of Quaternions. But it is easy to see
+that, for the first group, we may substitute the single equation
+$$ijk=1\eqno{(13)}$$
+since from it, by the help of the values of the squares of
+{\sl i}, {\sl j}, {\sl k}, all
+the other expressions may be deduced. We may consider it proved
+in this way, or deduce it afresh from the figure above, thus
+$$k\overline{ON}=\overline{OW}$$
+$$jk\overline{ON}=j\overline{OW}=\overline{OZ}$$
+$$ijk\overline{ON}=ij\overline{OW}=i\overline{OZ}=
+\overline{OS}=\overline{ON}$$
+
+{\bf 72}. One most important step remains to be made, to wit the
+assumption referred to in \S 64. We have treated
+{\sl i}, {\sl j}, {\sl k} simply as
+quadrantal versors ; and
+{\bf i}, {\bf j}, {\bf k} as unitvectors at right angles to
+each other, and coinciding with the axes of rotation of these versors.
+But if we collate and compare the equations just proved we have
+
+\hbox{\hskip 4cm {\Huge \{}
+\vbox{
+\hbox{$i^2=1$\hbox{\hskip 5cm}(7)}
+\hbox{${\rm {\bf i}}^2=1$\hbox{\hskip 5cm}(\S 9)}}}
+
+\hbox{\hskip 4cm {\Huge \{}
+\vbox{
+\hbox{$ij=k$\hbox{\hskip 5.3cm}(11)}
+\hbox{$i{\rm {\bf j}}={\rm {\bf k}}$\hbox{\hskip 5.3cm}(1)}}}
+
+\hbox{\hskip 4cm {\Huge \{}
+\vbox{
+\hbox{$ji=k$\hbox{\hskip 5cm}(11)}
+\hbox{$j{\rm {\bf i}}={\rm {\bf k}}$\hbox{\hskip 5cm}(1)}}}
+
+with the other similar groups symmetrically derived from them.
+
+Now the meanings we have assigned to {\sl i}, {\sl j}, {\sl k} are quite
+independent of, and not inconsistent with, those assigned to
+{\rm {\bf i}}, {\rm {\bf j}}, {\rm {\bf k}}.
+And it is superfluous to use two sets of characters when one will
+suffice. Hence it appears that {\sl i}, {\sl j}, {\sl k}
+may be substituted for {\rm {\bf i}}, {\rm {\bf j}}, {\rm {\bf k}};
+in other words, {\sl a unitvector when employed as a factor may be
+considered as a quadrantal versor whose plane is perpendicular to the
+vector}. (Of course it follows that every vector can be treated as the
+product of a number and a quadrantal versor.) This is one of the
+main elements of the singular simplicity of the quaternion calculus.
+
+{\bf 73}. Thus {\sl the product, and therefore the quotient, of two
+perpendicular vectors is a third vector perpendicular to both}.
+
+Hence the reciprocal (\S 51) of a vector is a vector which has
+the {\sl opposite} direction to that of the vector, arid its length is the
+reciprocal of the length of the vector.
+
+The conjugate (\S 52) of a vector is simply the vector reversed.
+
+Hence, by \S 52, if $\alpha$ be a vector
+$$(Ta)^2 = \alpha K\alpha = \alpha (  \alpha) = \alpha{}^2$$
+
+{\bf 74}. We may now see that {\sl every versor may be represented by
+a power of a unitvector}.
+
+For, if $\alpha$ be any vector perpendicular to $i$ (which is
+{\sl any} definite unitvector),
+$i\alpha = \beta$ is a vector equal in length to $\alpha$,
+but perpendicular to both $i$ and $\alpha$
+$$
+\begin{array}{ccl}
+i^2\alpha & = & \alpha\\
+\i^3\alpha & = & i\alpha = \beta\\
+\i^4\alpha & = & i\beta = i^2\alpha = \alpha
+\end{array}
+$$
+Thus, by successive applications of $i$, $\alpha$.
+is turned round $i$ as an axis
+through successive right angles. Hence it is natural to {\sl define}
+$i^m$ {\sl as
+a versor which turns any vector perpendicular to i through m right
+angles in the positive direction of rotation about i as an axis}. Here
+$m$ may have any real value whatever, whole or fractional, for it is
+easily seen that analogy leads us to interpret a negative value of $m$
+as corresponding to rotation in the negative direction.
+
+{\bf 75}. From this again it follows that {\sl any quaternion may be
+expressed as a power of a vector}. For the tensor and versor
+elements of the vector may be so chosen that, when raised to the
+same power, the one may be the tensor and the other the versor
+of the given quaternion. The vector must be, of course, perpen
+dicular to the plane of the quaternion.
+
+{\bf 76}. And we now see, as an immediate result of the last two
+sections, that the indexlaw holds with regard to powers of a
+quaternion (\S 63).
+
+{\bf 77}. So far as we have yet considered it, a quaternion has been
+regarded as the {\sl product} of a tensor and a versor: we are now to
+consider it as a {\sl sum}. The easiest method of so analysing it seems
+to be the following.
+
+\includegraphics{ps/quat17.ps}
+
+Let ${\displaystyle\frac{\overline{OB}}{\overline{OA}}}$
+represent any quaternion. Draw $BC$ perpendicular to $OA$,
+produced if necessary.
+
+Then, \S 19, $\overline{OB} = \overline{OC} + \overline{CB}$\\
+
+But, \S 22, $\overline{OC}=x\overline{OA}$\\
+where $x$ is a number, whose sign is the same
+as that of the cosine of $\angle AOB$.
+
+Also, \S 73, since $CB$ is perpendicular to $OA$,
+$$\overline{CB}=\gamma\overline{OA}$$
+where $\gamma$ is a vector perpendicular to $OA$ and $CB$, i.e. to the plane
+of the quaternion; and, as the figure is drawn, directed {\sl towards} the
+reader.
+
+Hence
+$$\frac{\overline{OB}}{\overline{OA}}=
+\frac{x\overline{OA}+\gamma\overline{OA}}{\overline{OA}}=
+x+\gamma$$
+
+Thus a quaternion, in general, may be decomposed into the sum
+of two parts, one numerical, the other a vector. Hamilton calls
+them the SCALAR, and the VECTOR, and denotes them respectively
+by the letters $S$ and $V$ prefixed to the expression for the
+quaternion.
+
+{\bf 78}. Hence $q = Sq+ Vq$, and if in the above example
+$$\frac{\overline{OB}}{\overline{OA}}=q$$
+then
+$$\overline{OB}=\overline{OC}+\overline{CB}=
+Sq.\overline{OA}+Vq.\overline{OA}
+\setcounter{footnote}{1}%% use a dagger because number looks like superscript
+\footnote{
+The points are inserted to show that $S$ and $V$
+apply only to $q$, and not to $q\overline{OA}$.}
+$$
+
+The equation above gives
+$$\overline{OC}=Sq.\overline{OA}$$
+$$\overline{CB}=Vq.\overline{OA}$$
+
+{\bf 79}. If, in the last figure, we produce $BC$ to $D$, so as to double
+its length, and join $OD$, we have, by \S 52,
+$$\frac{\overline{OD}}{\overline{OA}}=Kq=SKq+VKq$$
+so that\hbox{\hskip 1cm}
+$\overline{OD}=\overline{OC}+\overline{CD}=
+SKq.\overline{OA}+VKq.\overline{OA}$\\
+Hence\hbox{\hskip 3.6cm}$\overline{OC}=SKq.\overline{OA}$\\
+and\hbox{\hskip 4cm}$\overline{CD}=VKq.\overline{OA}$\\
+Comparing this value of $\overline{OC}$ with that in last section, we find
+$$SKq=Sq\eqno{(1)}$$
+or {\sl the scalar of the conjugate of a quaternion is equal to the scalar of
+the quaternion}.
+
+Again, $\overline{CD} = \overline{CB}$
+by the figure, and the substitution of their values gives
+$$VKq=Vq\eqno{(2)}$$
+or {\sl the vector of the conjugate of a quaternion is the vector of the
+quaternion reversed}.
+
+We may remark that the results of this section are simple con
+sequences of the fact that the symbols $S$, $V$, $K$ are commutative
+\setcounter{footnote}{1}%% use a dagger because number looks like superscript
+\footnote{
+It is curious to compare the properties of these quaternion symbols with those
+of the Elective Symbols of Logic, as given in BOOLE'S
+wonderful treatise on the {\sl Laws of Thought}; and to think that the
+same grand science of mathematical analysis, by processes remarkably
+similar to each other, reveals to us truths in the science of position
+far beyond the powers of the geometer, and truths of deductive
+reasoning to which unaided thought could never have led the logician. }.
+
+Thus\hbox{\hskip 3cm}$SKq = KSq = Sq$, \\
+since the conjugate of a number is the number itself; and
+$$VKq=KVq=Vq (\S 73)$$
+
+Again, it is obvious that,
+$$\sum{Sq}=S\sum{q},\;\;\;\;\sum{Vq}=V\sum{q}$$
+and thence\hbox{\hskip 3cm}$\sum{Kq}=K\sum{q}$
+
+{\bf 80}. Since any vector whatever may be represented by
+$$xi+yj+zk$$
+where $x$, $y$, $z$ are numbers (or Scalars),
+and $i$, $j$, $k$ may be any three
+noncoplanar vectors, \S\S 23, 25 though they are usually understood
+as representing a rectangular system of unitvectors and
+since any scalar may be denoted by $w$; we may write, for any
+quaternion $q$, the expression
+$$q=w+xi+yj+zk (\S 78)$$
+
+Here we have the essential dependence on four distinct numbers,
+from which the quaternion derives its name, exhibited in the most
+simple form.
+
+And now we see at once that an equation such as
+$$q^{'}=q$$
+where\hbox{\hskip 3cm}$q^{'}=w^{'}+x^{'}i+y^{'}j+z^{'}k$\\
+involves, of course, the {\sl four} equations
+$$
+w^{'}=w\textrm{, }
+x^{'}=x\textrm{, }
+y^{'}=y\textrm{, }
+z^{'}=z
+$$
+
+{\bf 81}. We proceed to indicate another mode of proof of the distributive
+law of multiplication.
+
+We have already defined, or assumed (\S 61), that
+$$\frac{\beta}{\alpha}+\frac{\gamma}{\alpha}=\frac{\beta+\gamma}{\alpha}$$
+or\hbox{\hskip 3cm}$\beta\alpha^{1}+\gamma\alpha^{1}=
+(\beta+\gamma)\alpha^{1}$\\
+and have thus been able to understand what is meant by adding
+two quaternions.
+
+But, writing $\alpha$ for $\alpha^{1}$,
+we see that this involves the equality
+$$(\beta+\gamma)\alpha = \beta\alpha+\gamma\alpha$$
+from which, by taking the conjugates of both sides, we derive
+$$\alpha^{'}(\beta^{'}+\gamma^{'})=\alpha^{'}\beta^{'}+\alpha^{'}\gamma^{'}
+(\S 55)$$
+And a combination of these results (putting
+$\beta+\gamma$ for $\alpha^{'}$ in the latter, for instance) gives
+$$
+\begin{array}{lcr}
+(\beta+\gamma)(\beta^{'}+\gamma^{'}) & = &
+(\beta+\gamma)\beta^{'}+(\beta+\gamma)\gamma^{'}\\
+& = & \beta\beta^{'}+\gamma\beta^{'}+\beta\gamma^{'}+\gamma\gamma^{'}
+\end{array}
+$$
+by the former.
+
+Hence the {\sl distributive principle is true in the multiplication of
+vectors}.
+
+It only remains to show that it is true as to the scalar and
+vector parts of a quaternion, and then we shall easily attain the
+general proof.
+
+Now, if $a$ be any scalar, $\alpha$ any vector, and $q$ any quaternion,
+$$(a+\alpha)q=aq+\alpha q$$
+
+For, if $\beta$ be the vector in which the plane of $q$ is intersected by
+a plane perpendicular to $\alpha$, we can find other two vectors,
+$\gamma$ and $\delta$ one in each of these planes such that
+$$\alpha=\frac{\gamma}{\beta},\;\;\;\;\;q=\frac{\beta}{\delta}$$
+And, of course, $a$ may be written
+${\displaystyle\frac{a\beta}{\beta}}$; so that
+$$
+\begin{array}{ccl}
+(a+\alpha)q & = & \frac{a\beta+\gamma}{\beta}.\frac{\beta}{\delta}
+=\frac{a\beta+\gamma}{\delta}\\
+& & \\
+& = & a\frac{\beta}{\delta}+\frac{\gamma}{\delta}=
+a\frac{\beta}{\delta}+\frac{\gamma}{\beta}.\frac{\beta}{\delta}\\
+& & \\
+& = & aq + \alpha q
+\end{array}
+$$
+And the conjugate may be written
+$$q^{'}(a^{'}+\alpha^{'})=q^{'}a^{'}+q^{'}\alpha^{'} (\S 55)$$
+Hence, generally,
+$$(a+\alpha)(b+\beta)=ab+a\beta+b\alpha+\alpha\beta$$
+or, breaking up $a$ and $b$ each into the sum of two scalars, and
+$\alpha$, $\beta$ each into the sum of two vectors, \\
+$(a_1+a_2+\alpha_1+\alpha_2)(b_1+b_2+\beta_1+\beta_2)$
+$$=(a_1+a_2)(b_1+b_2)
++(a_1+a_2)(\beta_1+\beta_2)
++(b_1+b_2)(\alpha_1+\alpha_2)
++(\alpha_1+\alpha_2)(\beta_1+\beta_2)
+$$
+(by what precedes, all the factors on the right are distributive, so
+that we may easily put it in the form)
+$$=(a_1+\alpha_1)(b_1+\beta_1)
++(a_1+\alpha_1)(b_2+\beta_2)
++(a_2+\alpha_2)(b_1+\beta_1)
++(a_2+\alpha_2)(b_2+\beta_2)
+$$
+
+Putting $a_1+\alpha_1=p,\;\;\;$
+$a_2+\alpha_2=q,\;\;\;$
+$b_1+\beta_1=r,\;\;\;$ $b_2+\beta_2=s$,\\
+we have $(p+q)(r+s)=pr+ps+qr+qs$
+
+{\bf 82}. Cayley suggests that the laws of quaternion multiplication
+may be derived more directly from those of vector multiplication,
+supposed to be already established. Thus, let $\alpha$ be the unit vector
+perpendicular to the vector parts of $q$ and of $q^{'}$. Then let
+$$\rho=q.\alpha,\;\;\;\sigma=\alpha .q^{'}$$
+as is evidently permissible, and we have
+$$p\alpha=q.\alpha\alpha=q;\;\;\;\alpha\sigma=\alpha\alpha.q^{'}=q^{'}$$
+so that\hbox{\hskip 4cm}$q.q^{'}=\rho\alpha.\alpha\sigma=\rho.\sigma$
+
+The student may easily extend this process.
+
+For variety, we shall now for a time forsake the geometrical
+mode of proof we have hitherto adopted, and deduce some of our
+next steps from the analytical expression for a quaternion given in
+\S 80, and the properties of a rectangular system of unitvectors as
+in \S 71.
+
+We will commence by proving the result of \S 77 anew.
+
+{\bf 83}. Let
+$$\alpha=xi+yj+zk$$
+$$\beta=x^{'}i+y^{'}j+z^{'}k$$
+Then, because by \S 71 every product or quotient of $i$, $j$, $k$ is reducible
+to one of them or to a number, we are entitled to assume
+$$q=\frac{\beta}{\alpha}=\omega+\xi i+\eta j +\zeta k$$
+where $\omega$, $\xi$, $\eta$, $\zeta$ are numbers.
+This is the proposition of \S 80.
+
+[Of course, with this expression for a quaternion, there is no
+necessity for a formal proof of such equations as
+$$p + (q+r) = (p + q) + r$$
+where the various sums are to be interpreted as in \S 61.
+
+All such things become obvious in view of the properties of $i$, $j$ ,$k$.]
+
+{\bf 84}. But it may be interesting to find $\omega$, $\xi$, $\eta$, $\zeta$
+in terms of $x$, $y$, $z$, $x^{'}$, $y^{'}$ , $z^{'}$ .
+
+We have
+$$\beta=q\alpha$$
+or
+$$x^{'}i+y^{'}j+z^{'}k=(\omega+\xi i+\eta j+\zeta k)(xi+yj+zk)$$
+$$=(\xi x+\eta y+\zeta z)
++(\omega x+\eta z\zeta y)i
++(\omega y+\zeta x\xi z)j
++(\omega z+\xi y\eta x)k
+$$
+as we easily see by the expressions for the powers and products of
+$i$, $j$, $k$ given in \S 71. But the student must pay particular attention
+to the {\sl order} of the factors, else he is certain to make mistakes.
+
+This (\S 80) resolves itself into the four equations
+$$
+\begin{array}{lllllllll}
+0 & = & & & \xi x & + & \eta y & + & \zeta z\\
+x^{'} & = & \omega x & & & + & \eta z &  & \zeta y\\
+y^{'} & = & \omega y &  & \xi z & & & + & \zeta x\\
+z^{'} & = & \omega z & + & \xi y &  & \eta x\\
+\end{array}
+$$
+The three last equations give
+$$xx^{'}+yy^{'}+zz^{'}=\omega(x^2+y^2+z^2)$$
+which determines $\omega$.
+
+Also we have, from the same three, by the help of the first,
+$$\xi x^{'}+\eta y^{'}+\zeta z^{'} = 0$$
+which, combined with the first, gives
+$$\frac{\xi}{yz^{'}zy^{'}}
+=\frac{\eta}{zx^{'}xz^{'}}
+=\frac{\zeta}{xy^{'}yx^{'}}
+$$
+and the common value of these three fractions is then easily seen
+to be
+$$\frac{1}{x^2+y^2+z^2}$$
+
+It is easy enough to interpret these expressions by means of
+ordinary coordinate geometry : but a much simpler process will
+be furnished by quaternions themselves in the next chapter, and, in
+giving it, we shall refer back to this section.
+
+{\bf 85}. The associative law of multiplication is now to be proved
+by means of the distributive (\S 81). We leave the proof to the
+student. He has merely to multiply together the factors
+$$w+xi+yj+zk,\;\;\;\;
+w+x^{'}i+y^{'}j+z^{'}k,\;\;\;\;\textrm{ and }
+w^{''} + x^{''}i + y^{''}j + z^{''}k$$
+
+as follows :
+
+First, multiply the third factor by the second, and then multiply
+the product by the first; next, multiply the second factor by the
+first and employ the product to multiply the third: always remembering
+that the multiplier in any product is placed {\sl before} the
+multiplicand. He will find the scalar parts and the coefficients of
+$i$, $j$, $k$, in these products, respectively equal, each to each.
+
+{\bf 86}.
+With the same expressions for $\alpha$, $\beta$, as in section 83, we
+have
+$$\alpha\beta=(xi+yj+zk)(x^{'}i+y^{'}j+z^{'}k)$$
+$$\;\;=(xx^{'}+yy^{'}+zz^{'})
++(yz^{'}zy^{'})i
++(zx^{'}xz^{'})j
++(xy^{'}yx^{'})k
+$$
+
+But we have also
+$$\beta\alpha=
+(xx^{'}+yy^{'}+zz^{'})
+(yz^{'}zy^{'})i
+(zx^{'}xz^{'})j
+(xy^{'}yx^{'})k
+$$
+
+The only difference is in the sign of the vector parts. Hence
+$$S\alpha\beta=S\beta\alpha\eqno{(1)}$$
+$$V\alpha\beta=V\beta\alpha\eqno{(2)}$$
+$$\alpha\beta+\beta\alpha=2S\alpha\beta\eqno{(3)}$$
+$$\alpha\beta\beta\alpha=2V\alpha\beta\eqno{(4)}$$
+$$\alpha\beta=K.\beta\alpha\eqno{(5)}$$
+
+{\bf 87}. If $\alpha=\beta$ we have of course (\S 25)
+$$x=x^{'},\;\;\;\;y=y^{'},\;\;\;\;z=z^{'}$$
+and the formulae of last section become
+$$\alpha\beta=\beta\alpha=\alpha^2=(x^2+y^2+z^2)$$
+which was anticipated in \S 73, where we proved the formula
+$$(T\alpha)^2=\alpha^2$$
+and also, to a certain extent, in \S 25.
+
+{\bf 88}. Now let $q$ and $r$ be any quaternions, then
+$$
+\begin{array}{rcl}
+S.qr & = & S.(Sq+Vq)(Sr+Vr)\\
+ & = & S.(SqSr+Sr.Vq+Sq.Vr+VqVr)\\
+ & = & SqSr+S.VqVr
+\end{array}
+$$
+since the two middle terms are vectors.
+Similarly,
+$$S.rq=SrSq+S.VrVq$$
+Hence, since by (1) of \S 86 we have
+$$S.VqVr=S.VrVq$$
+we see that
+$$S.qr=S.rq\eqno{(1)}$$
+a formula of considerable importance.
+
+It may easily be extended to any number of quaternions,
+because, $r$ being arbitrary, we may put for it $rs$. Thus we have
+$$
+\begin{array}{rcl}
+S.qrs & = & S.rsq\\
+& = & S.sqr
+\end{array}
+$$
+by a second application of the process. In words, we have the
+theorem {\sl the scalar of the product of any number of given
+quaternions depends only upon the cyclical order in which they are
+arranged}.
+
+{\bf 89}. An important case is that of three factors, each a vector.
+The formula then becomes
+$$S.\alpha\beta\gamma=S.\beta\gamma\alpha=S.\gamma\alpha\beta$$
+But
+$$\begin{array}{rcll}
+S.\alpha\beta\gamma & = & S\alpha(S\beta\gamma+V\beta\gamma) &\\
+ & = & S\alpha V\beta\gamma & \textrm{since }\alpha S\beta\gamma
+\textrm{ is a vector}\\
+ & = & S\alpha V\gamma\beta & \textrm{by (2) of \S 86}\\
+ & = & S\alpha(S\gamma\beta+V\gamma\beta) &\\
+ & = & S.\alpha\gamma\beta
+\end{array}
+$$
+Hence {\sl the scalar of the product of three vectors changes sign when
+the cyclical order is altered.}
+
+By the results of \S\S 55, 73, 79 we see that, for any number
+of vectors, we have
+$$K.\alpha\beta\gamma\ldots\phi\chi=
+\pm\chi\phi\ldots\gamma\beta\alpha$$
+(the positive sign belonging to the product of an even number of
+vectors) so that
+$$S.\alpha\beta\ldots\phi\chi=\pm S.\chi\phi\ldots\beta\alpha$$
+
+Similarly
+$$V.\alpha\beta\ldots\phi\chi=\mp V.\chi\phi\ldots\beta\alpha$$
+Thus we may generalize (3) and (4) of \S 86 into
+$$2S.\alpha\beta\ldots\phi\chi=
+\alpha\beta\ldots\chi\phi\pm\phi\chi\ldots\beta\alpha$$
+$$2V.\alpha\beta\ldots\phi\chi=
+\alpha\beta\ldots\chi\phi\mp\phi\chi\ldots\beta\alpha$$
+the upper sign still being used when the number of factors is
+even.
+
+Other curious propositions connected with this will be given
+later (some, indeed, will be found in the Examples appended to
+this chapter), as we wish to develop the really fundamental
+formulae in as compact a form as possible.
+
+{\bf 90}. By (4) of \S 86,
+$$2V\beta\gamma=\beta\gamma\gamma\beta$$
+Hence
+$$2V.\alpha V\beta\gamma=V.\alpha(\beta\gamma\gamma\beta)$$
+(by multiplying both by $\alpha$, and taking the vector parts of each side)
+$$=V(\alpha\beta\gamma+\beta\alpha\gamma\beta\alpha\gamma\alpha\gamma\beta)$$
+(by introducing the null term $\beta\alpha\gamma\beta\alpha\gamma$).
+
+\noindent
+That is
+$$2V.\alpha V\beta\gamma=V.(\alpha\beta+\beta\alpha)\gamma
+V(\beta S\alpha\gamma+\beta V\alpha\gamma+S\alpha\gamma .\beta+
+V\alpha\gamma .\beta$$
+$$=V.(2S\alpha\beta)\gamma2V\beta S\alpha\gamma$$
+(if we notice that $V(V\alpha\gamma .\beta)=V.\beta V\alpha\gamma$
+by (2) of \S 86).
+Hence
+$$V.\alpha V\beta\gamma=\gamma S\alpha\beta\beta S\gamma\alpha\eqno{(1)}$$
+a formula of constant occurrence.
+
+Adding $\alpha S\beta\gamma$ to both sides, we get another most valuable
+formula
+$$V.\alpha\beta\gamma
+=\alpha S\beta\gamma
+\beta S\gamma\alpha
++\gamma S\alpha\beta\eqno{(2)}
+$$
+and the form of this shows that we may interchange $\gamma$ and $\alpha$
+without altering the righthand member. This gives
+$$V.\alpha\beta\gamma = V.\gamma\beta\alpha$$
+a formula which may be greatly extended. (See \S89, above.)
+
+Another simple mode of establishing (2) is as follows :
+$$
+\begin{array}{rcl}
+K.\alpha\beta\gamma & = & \gamma\beta\alpha\\
+\therefore 2V.\alpha\beta\gamma & = &
+\alpha\beta\gammaK.\alpha\beta\gamma\textrm{ (by \S 79(2))}\\
+& = & \alpha\beta\gamma + \gamma\beta\alpha\\
+& = & \alpha(\beta\gamma+\gamma\beta)
+(\alpha\gamma+\gamma\alpha)\beta
++\gamma(\alpha\beta+\beta\alpha)\\
+& = & 2\alpha S\beta\gamma2\beta S\alpha\gamma+2\gamma S\alpha\beta
+\end{array}
+$$
+
+{\bf 91}. We have also
+$$VV\alpha\beta V\gamma\delta = VV\gamma\delta V\alpha\beta\;\;\;\;
+\textrm{ by (2) of \S 86}$$
+$$=\delta S\gamma V\alpha\beta\gamma S\delta V\alpha\beta
+=\delta S.\alpha\beta\gamma\gamma S.\alpha\beta\delta$$
+$$=\beta S\alpha V\gamma\delta+\alpha S\beta V\gamma\delta
+=\beta S.\alpha\gamma\delta+\alpha S.\beta\gamma\delta$$
+all of these being arrived at by the help of \S 90 (1) and of \S 89;
+and by treating alternately $V\alpha\beta$ and
+$V\gamma\delta$ as {\sl simple} vectors.
+
+Equating two of these values, we have
+$$\delta S.\alpha\beta\gamma
+=\alpha S.\beta\gamma\delta
++\beta S.\gamma\alpha\delta
++\gamma S.\alpha\beta\delta\eqno{(3}
+$$
+a very useful formula, expressing any vector whatever in terms
+of three given vectors. [This, of course, presupposes that
+$\alpha$, $\beta$, $\gamma$
+are not coplanar, \S 23. In fact, if they be coplanar, the factor
+$S.\alpha\beta\gamma$ vanishes,
+and thus (3) does not give an expression for $\delta$.
+This will be shown in \S 101 below.]
+
+{\bf 92}. That such an expression as (3) is possible we knew already
+by \S 23. For variety we may seek another expression of a similar
+character, by a process which differs entirely from that employed
+in last section.
+
+$\alpha$, $\beta$, $\gamma$
+being any three noncoplanar vectors, we may derive
+from them three others $V\alpha\beta$, $V\beta\gamma$, $V\gamma\alpha$
+and, as these will not be
+coplanar, any other vector $\delta$ may be expressed as the sum of the
+three, each multiplied by some scalar. It is required to find this
+expression for $\delta$.
+
+Let
+$$\delta=xV\alpha\beta+yV\beta\gamma+zV\gamma\alpha$$
+Then
+$$S\gamma\delta=xS.\gamma\alpha\beta =xS.\alpha\beta\gamma$$
+the terms in y and z going out, because
+$$S\gamma V\beta\gamma = S.\gamma\beta\gamma=S\beta\gamma^2
+=\gamma^2 S\beta=0$$
+for $\gamma^2$ is (\S 73) a number.
+
+Similarly
+$$S\beta\delta=zS.\beta\gamma\alpha=zS.\alpha\beta\gamma$$
+and
+$$S\alpha\delta=qS.\alpha\beta\gamma$$
+Thus
+$$\delta S.\alpha\beta\gamma=V\alpha\beta S\gamma\delta
++V\beta\gamma S\alpha\delta
++V\gamma\alpha S\beta\delta\eqno{(4)}
+$$
+
+{\bf 93}. We conclude the chapter by showing (as promised in \S 64)
+that the assumption that the product of two parallel vectors is
+a number, and the product of two perpendicular vectors a third
+vector perpendicular to both, is not only useful and convenient,
+but absolutely inevitable, if our system is to deal indifferently with
+all directions in space. We abridge Hamilton s reasoning.
+
+Suppose that there is no direction in space preeminent, and
+that the product of two vectors is something which has quantity,
+so as to vary in amount if the factors are changed, and to have its
+sign changed if that of one of them is reversed ; if the vectors be
+parallel, their product cannot be, in whole or in part, a vector
+{\sl inclined} to them, for there is nothing to determine the direction in
+which it must lie. It cannot be a vector {\sl parallel} to them; for by
+changing the signs of both factors the product is unchanged,
+whereas, as the whole system has been reversed, the product
+vector ought to have been reversed. Hence it must be a number.
+Again, the product of two perpendicular vectors cannot be wholly
+or partly a number, because on inverting one of them the sign of
+that number ought to change; but inverting one of them is simply
+equivalent to a rotation through two right angles about the other,
+and (from the symmetry of space) ought to leave the number
+unchanged. Hence the product of two perpendicular vectors must
+be a vector, and a simple extension of the same reasoning shows
+that it must be perpendicular to each of the factors. It is easy to
+carry this farther, but enough has been said to show the character
+of the reasoning.
+
+\section{Examples To Chapter 2.}
+
+{\bf 1}. It is obvious from the properties of polar triangles that any
+mode of representing versors by the {\sl sides} of a spherical triangle
+must have an equivalent statement in which they are represented
+by {\sl angles} in the polar triangle.
+
+Show directly that the product of two versors represented
+by two angles of a spherical triangle is a third versor represented
+by the {\sl supplement} of the remaining angle of the triangle ; and
+determine the rule which connects the {\sl directions} in which these
+angles are to be measured.
+
+{\bf 2}. Hence derive another proof that we have not generally
+$$pq=qp$$
+
+{\bf 3}. Hence show that the proof of the associative principle,
+\S 57, may be made to depend upon the fact that if from any point
+of the sphere tangent arcs be drawn to a spherical conic, and also
+arcs to the foci, the inclination of either tangent arc to one of the
+focal arcs is equal to that of the other tangent arc to the other
+focal arc.
+
+{\bf 4}. Prove the formulae
+$$2S.\alpha\beta\gamma = \alpha\beta\gamma\gamma\beta\alpha$$
+$$2V.\alpha\beta\gamma = \alpha\beta\gamma+\gamma\beta\alpha$$
+
+{\bf 5}. Show that, whatever odd number of vectors be represented
+by $\alpha$, $\beta$, $\gamma$ \&c., we have always
+$$
+V.\alpha\beta\gamma\delta\epsilon=V.\epsilon\delta\gamma\beta\alpha
+$$
+$$
+V.\alpha\beta\gamma\delta\epsilon\zeta\eta
+=V.\eta\zeta\epsilon\delta\gamma\beta\alpha,\textrm{ \&c.}
+$$
+
+{\bf 6}. Show that
+$$
+S.V\alpha\beta V\beta\gamma V\gamma\alpha=(S.\alpha\beta\gamma)^2
+$$
+$$
+V.V\alpha\beta V\beta\gamma V\gamma\alpha=
+V\alpha\beta(\gamma^2S\alpha\betaS\beta\gamma S\gamma\alpha)+\ldots
+$$
+and
+$$
+V(V\alpha\beta V.V\beta\gamma V\gamma\alpha)
+=(\beta S\alpha\gamma\alpha S\beta\gamma)S.\alpha\beta\gamma
+$$
+
+{\bf 7}. If $\alpha$, $\beta$, $\gamma$
+be any vectors at right angles to each other, show that
+$$
+(\alpha^3+\beta^3+\gamma^3)S.\alpha\beta\gamma
+=\alpha^4V\beta\gamma
++\beta^4V\gamma\alpha
++\gamma^4V\alpha\beta
+$$
+$$
+(\alpha^{2n1}+\beta^{2n1}+\gamma^{2n1})S.\alpha\beta\gamma
+=\alpha^{2n}V\beta\gamma
++\beta^{2n}V\gamma\alpha
++\gamma^{2n}V\alpha\beta
+$$
+
+{\bf 8}. If $\alpha$, $\beta$, $\gamma$
+be noncoplanar vectors, find the relations among
+the six scalars, $x$, $y$, $z$ and $\xi$, $\eta$, $\zeta$
+which are implied in the
+equation
+$$
+x\alpha+y\beta+z\gamma
+=\xi V\beta\gamma+\eta V\gamma\alpha+\zeta V\alpha\beta
+$$
+
+{\bf 9}. If $\alpha$, $\beta$, $\gamma$
+be any three noncoplanar vectors, express any
+fourth vector, $\delta$, as a linear function of each of the following sets of
+three derived vectors.
+$$
+V.\gamma\alpha\beta,\;\;\;\;V.\alpha\beta\gamma,\;\;\;\;
+V.\beta\gamma\alpha
+$$
+and
+$$
+V.V\alpha\beta V\beta\gamma V\gamma\alpha,\;\;\;\;
+V.V\beta\gamma V\gamma\alpha V\alpha\beta,\;\;\;\;
+V.V\gamma\alpha V\alpha\beta V\beta\gamma
+$$
+
+{\bf 10}. Eliminate $\rho$ from the equations
+$$
+S\alpha\rho=a,\;\;\;\;
+S\beta\rho=b,\;\;\;\;
+S\gamma\rho=c,\;\;\;\;
+S\delta\rho=d
+$$
+where $\alpha$, $\beta$, $\gamma$, $\delta$
+are vectors, and $a$, $b$, $c$, $d$ scalars.
+
+{\bf 11}. In any quadrilateral, plane or gauche, the sum of the
+squares of the diagonals is double the sum of the squares of the
+lines joining the middle points of opposite sides.
+
+\section{Interpretations And Transformations}
+
+{\bf 94}. Among the most useful characteristics of the Calculus of
+Quaternions, the ease of interpreting its formulae geometrically,
+and the extraordinary variety of transformations of which the
+simplest expressions are susceptible, deserve a prominent place.
+We devote this Chapter to some of the more simple of these,
+together with a few of somewhat more complex character but of
+constant occurrence in geometrical and physical investigations.
+Others will appear in every succeeding Chapter. It is here,
+perhaps, that the student is likely to feel most strongly the
+peculiar difficulties of the new Calculus. But on that very account
+he should endeavour to master them, for the variety of forms
+which any one formula may assume, though puzzling to the
+beginner, is of the utmost advantage to the advanced student, not
+alone as aiding him in the solution of complex questions, but
+as affording an invaluable mental discipline.
+
+{\bf 95}. If we refer again to the figure of \S 77 we see that
+$$OC=OB\cos AOB$$
+$$CB=OB \sin AOB$$
+Hence if
+$$\overline{AB}=\alpha,\;\;\;\;
+\overline{OB}=\beta,\;\;\;\;\textrm{ and }
+\angle AOB=\theta
+$$
+we have
+$$OB=T\beta,\;\;\;\;OA=T\alpha$$
+$$OC=T\beta\cos\theta,\;\;\;\;CB=T\beta\sin\theta
+$$
+Hence
+$$S\frac{\beta}{\alpha}=
+\frac{OC}{OA}=
+\frac{T\beta}{T\alpha}\cos\theta
+$$
+Similarly,
+$$
+TV\frac{\beta}{\alpha}=\frac{CB}{OA}=\frac{T\beta}{T\alpha}\sin\theta
+$$
+
+Hence, if $\eta$ be a unitvector perpendicular to
+$\alpha$ and $\beta$, and such
+that positive rotation about it, through the angle $\theta$, turns $\alpha$
+towards $\beta$ or
+$$
+\eta=
+\frac{U\overline{CB}}{U\overline{OA}}=
+U\frac{\overline{CB}}{\overline{OA}}=
+UV\frac{\beta}{\alpha}
+$$
+we have
+$$
+V\frac{\beta}{\alpha}=
+\frac{T\beta}{T\alpha}\sin\theta .\eta\;\;\;\;\;\textrm{ (See, again, \S 84)}
+$$
+
+{\bf 96}. In the same way, or by putting
+$$
+\begin{array}{rcl}
+\alpha\beta & = & S\alpha\beta+V\alpha\beta\\
+ & = & S\beta\alpha  V\beta\alpha\\
+ & = & \alpha^2\left(S\frac{\beta}{\alpha}V\frac{\beta}{\alpha}\right)\\
+ & = & T\alpha^2\left(S\frac{\beta}{\alpha}+V\frac{\beta}{\alpha}\right)
+\end{array}
+$$
+we may show that
+$$S\alpha\beta=T\alpha T\beta\cos\theta$$
+$$TV\alpha\beta = T\alpha T\beta\sin\theta$$
+and
+$$V\alpha\beta=T\alpha T\beta\sin\theta . \eta$$
+where
+$$\eta=UV\alpha\beta = U(V\beta\alpha)=UV\frac{\beta}{\alpha}$$
+
+Thus {\sl the scalar of the product of two vectors is the continued
+product of their tensors and of the cosine of the supplement of the
+contained angle}.
+
+{\sl The tensor of the vector of the product of two vectors is the con
+tinued product of their tensors and the sine of the contained angle ;
+and the versor of the same is a unitvector perpendicular to both,
+and such that the rotation about it from the first vector (i. e. the
+multiplier) to the second is lefthanded or positive}.
+
+{\sl Hence also $TV\alpha\beta$
+is double the area of the triangle two of whose
+sides are $\alpha$, $\beta$.}
+
+{\bf 97}. (a) In any plane triangle $ABC$ we have
+$$\overline{AC}=\overline{AB}+\overline{BC}$$
+Hence,
+$$
+\overline{AC}^2=S.\overline{AC}\overline{AC}=
+S.\overline{AC}(\overline{AB}+\overline{BC})
+$$
+
+With the usual notation for a plane triangle the interpretation
+of this formula is
+$$b^2 = bc\cos Aab\cos C$$
+or
+$$b=c\cos C+c\cos A$$
+
+(b) Again we have, obviously,
+$$
+\begin{array}{rcl}
+V.\overline{AB}\;\overline{AC}&=&V.\overline{AB}(\overline{AB}+\overline{BC})\\
+&=&V.\overline{AB}\;\overline{BC}
+\end{array}
+$$
+or
+$$cb\sin A = ca\sin B$$
+whence
+$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$
+
+These are truths, but not truisms, as we might have been led
+to fancy from the excessive simplicity of the process employed.
+
+{\bf 98}.
+From \S 96 it follows that, if $\alpha$ and $\beta$ be both actual (i. e.
+real and nonevanescent) vectors, the equation
+$$S\alpha\beta = 0$$
+
+
+shows that $\cos\theta=0$, or that
+$\alpha$ is {\sl perpendicular} to $\beta$. And, in fact,
+we know already that the product of two perpendicular vectors is
+a vector.
+
+Again if
+$$V\alpha\beta=0$$
+we must have $\sin\theta=0$, or
+$\alpha$ is {\sl parallel} to $\beta$. We know already
+that the product of two parallel vectors is a scalar.
+
+Hence we see that
+$$S\alpha\beta=0$$
+is equivalent to
+$$\alpha=V\gamma\beta$$
+where $\gamma$ is an undetermined vector; and that
+$$V\alpha\beta=0$$
+is equivalent to
+$$\alpha=x\beta$$
+where $x$ is an undetermined scalar.
+
+{\bf 99}. If we write, as in \S\S 83, 84,
+$$\alpha=ix+jy+kz$$
+$$\beta=ix^{'}+jy^{'}+kz^{'}$$
+we have, at once, by \S 86,
+$$\begin{array}{rcl}
+S\alpha\beta&=&xx^{'}yy^{'}zz^{'}\\
+&=&rr^{'}\left(
+\frac{x}{r}\frac{x^{'}}{r^{'}}+
+\frac{y}{r}\frac{y^{'}}{r^{'}}+
+\frac{z}{r}\frac{z^{'}}{r^{'}}
+\right)
+\end{array}
+$$
+where
+$$
+r=\sqrt{x^2+y^2+z^2},\;\;\;\;
+r^{'}=\sqrt{x^{'2}+y^{'2}+z^{'2}}
+$$
+Also
+$$
+V\alpha\beta=rr^{'}\left\{
+\frac{yz^{'}zy^{'}}{rr^{'}}i+
+\frac{zx^{'}xz^{'}}{rr^{'}}j+
+\frac{xy^{'}=yx^{'}}{rr^{'}}k
+\right\}
+$$
+
+These express in Cartesian coordinates the propositions we have
+just proved. In commencing the subject it may perhaps assist
+the student to see these more familiar forms for the quaternion
+expressions ; and he will doubtless be induced by their appearance
+to prosecute the subject, since he cannot fail even at this stage to
+see how much more simple the quaternion expressions are than
+those to which he has been accustomed.
+
+{\bf 100}. The expression
+$$S.\alpha\beta\gamma$$
+may be written
+$$SV(\alpha\beta)\gamma$$
+because the quaternion $\alpha\beta\gamma$ may be broken up into
+$$S(\alpha\beta)\gamma+V(\alpha\beta)\gamma$$
+of which the first term is a vector.
+
+But, by \S 96,
+$$SV(\alpha\beta)\gamma=T\alpha T\beta\sin\theta S\eta\gamma$$
+Here $T\eta=1$, let $\phi$ be the angle between $\eta$ and $\gamma$,
+then finally
+$$S.\alpha\beta\gamma = T\alpha T\beta T\gamma\sin\theta\cos\phi$$
+
+But as $\eta$ is perpendicular to $\alpha$ and $\beta$,
+$T\gamma\cos\phi$ is the length of the
+perpendicular from the extremity of $\gamma$
+upon the plane of $\alpha$, $\beta$. And
+as the product of the other three factors is (\S 96) the area of the
+parallelogram two of whose sides are $\alpha$, $\beta$, we see that the
+magnitude of $S.\alpha\beta\gamma$,
+independent of its sign, is {\sl the volume of the
+parallelepiped of which three coordinate edges
+are $\alpha$, $\beta$, $\gamma$};
+or six times the volume of the pyramid which has
+$\alpha$, $\beta$, $\gamma$ for edges.
+
+{\bf 101}. Hence the equation
+$$S.\alpha\beta\gamma=0$$
+if we suppose $\alpha\beta\gamma$ to be actual vectors, shows either that
+$$\sin\theta=0$$
+or
+$$\cos\phi=0$$
+i. e. {\sl two of the three vectors are parallel},
+or {\sl all three are parallel to one plane}.
+
+This is consistent with previous results, for if $\gamma=p\beta$ we have
+$$S.\alpha\beta\gamma=pS.\alpha\beta^2=0$$
+and, if $\gamma$ be coplanar with $\alpha$,$\beta$, we have
+$\gamma=p\alpha+q\beta$ and
+$$S.\alpha\beta\gamma=S.\alpha\beta(p\alpha+q\beta)=0$$
+
+{\bf 102}.
+This property of the expression $S.\alpha\beta\gamma$ prepares us to
+find that it is a determinant. And, in fact, if we take $\alpha$,$\beta$ as in
+\S 83, and in addition
+$$\gamma=ix^{''}+jy^{''}+kz^{''}$$
+we have at once
+$$S.\alpha\beta\gamma=x^{''}(yz^{'}zy^{'})
+y^{''}(zx^{'}xz^{'})
+z^{''}(xy^{'}yx^{'})$$
+$$
+=\left\vert
+\begin{array}{ccc}
+x & y & z\\
+x^{'} & y^{'} & z^{'}\\
+x^{''}&y^{''}&z^{''}
+\end{array}
+\right\vert
+$$
+The determinant changes sign if we make any two rows change
+places. This is the proposition we met with before (\S 89) in the
+form
+$$S.\alpha\beta\gamma=S.\beta\alpha\gamma=S.\beta\gamma\alpha
+\textrm{, \&c}$$
+
+If we take three new vectors
+$$\alpha_1=ix+jx^{'}+kx^{''}$$
+$$\beta_1 =iy+jy^{'}+ky^{''}$$
+$$\gamma_1=iz+jz^{'}+kz^{''}$$
+we thus see that they are coplanar if $\alpha$, $\beta$, $\gamma$ are so.
+That is, if
+$$S.\alpha\beta\gamma=0$$
+then
+$$S.\alpha_1\beta_1\gamma_1=0$$
+
+{\bf 103}. We have, by \S 52,
+$$
+\begin{array}{rcl}
+(Tq)^2 &=&qKq = (Sq+Vq)(SqVq)\;\;\;\;\textrm{(\S 79)}\\
+&=&(Sq)^2(Vq)^2\;\;\;\;\;\textrm{by algebra}\\
+&=&(Sq)^2+(TVq)^2\;\;\;\;\textrm{(\S 73)}\\
+\end{array}
+$$
+If $q=\alpha\beta$, we have $Kq = \beta\alpha$, and the formula becomes
+$$\alpha\beta . \beta\alpha = \alpha^2\beta^2=
+(S\alpha\beta)^2(V\alpha\beta)^2
+$$
+
+In Cartesian coordinates this is\\
+\vskip 0.1cm
+$(x^2+y^2+z^z)(x^{'2}+y^{'2}+z^{'2})$
+$$
+=(xx^{'}+yy^{'}+zz^{'})^2+(yz^{'}zy^{'})^2+
+(zx^{'}xz^{'})^2+(xy^{'}yx^{'})^2
+$$
+More generally we have
+$$
+\begin{array}{rcl}
+(T(qr))^2&=&(Tq)^2(Tr)^2\\
+&=&(S.qr)^2(V.qr)^2
+\end{array}
+$$
+If we write
+$$q=w+\alpha=w+ix+jy+kz$$
+$$r=w^{'}+\beta=w^{'}+ix^{'}+jy^{'}+kz^{'}$$
+this becomes
+$$(w^2+x^2+y^2+z^2)(w^{'2}+x^{'2}+y^{'2}+z^{'2})$$
+$$=(ww^{'}xx^{'}yy^{'}zz^{'})^2+(wx^{'}+w^{'}x+yz^{'}zy^{'})^2$$
+$$=(xy^{'}+w^{'}y+zx^{'}xz^{'})^2+(wz^{'}+w^{'}z+xy^{'}yx^{'})^2$$
+a formula of algebra due to Euler.
+
+{\bf 104}. We have, of course, by multiplication,
+$$
+(\alpha+\beta)^2=
+\alpha^2+\alpha\beta+\beta\alpha+\beta^2=
+\alpha^2+2S\alpha\beta+\beta^2\;\;\;\;\;\textrm{(\S 86 (3))}
+$$
+Translating into the usual notation of plane trigonometry, this
+becomes
+$$c^2=a^22ab\cos C+b^2$$
+the common formula.
+
+Again,
+$$
+V.(\alpha+\beta)(\alpha\beta)=
+V\alpha\beta+V\beta\alpha=
+2V\alpha\beta\;\;\;\;\;\textrm{(\S 86 (2)}
+$$
+Taking tensors of both sides we have the theorem, {\sl the parallelogram
+whose sides are parallel and equal to the diagonals of a
+given parallelogram, has double its area} (\S 96).
+
+Also
+$$S(\alpha+\beta)(\alpha\beta)=\alpha^2\beta^2$$
+and vanishes only when $\alpha^2=\beta^2$,
+or $T\alpha=T\beta$; that is, {\sl the diagonals
+of a parallelogram are at right angles to one another, when, and
+only when, it is a rhombus}.
+
+Later it will be shown that this contains a proof that the angle
+in a semicircle is a right angle.
+
+{\bf 105}. The expression\hbox{\hskip 1cm}$\rho=\alpha\beta\alpha^{1}$\\
+obviously denotes a vector whose tensor is equal to that of $\beta$.
+
+But we have\hbox{\hskip 2cm}$S.\beta\alpha\rho=0$\\
+so that $\rho$ is in the plane of $\alpha$, $\beta$
+
+Also we have\hbox{\hskip 2cm}$S\alpha\rho=S\alpha\beta$\\
+so that $\beta$ and $\rho$ make equal angles with $\alpha$,
+evidently on opposite
+sides of it. Thus if $\alpha$ be the perpendicular to a reflecting surface
+and $\beta$ the path of an incident ray, $\rho$ will be the path of the
+reflected ray.
+
+Another mode of obtaining these results is to expand the above
+expression, thus, \S 90 (2),
+$$
+\begin{array}{rcl}
+\rho&=&2\alpha^{1}S\alpha\beta\beta\\
+&=&2\alpha^{1}S\alpha\beta\alpha^{1}(S\alpha\beta+V\alpha\beta)\\
+&=&\alpha^{1}(S\alpha\betaV\alpha\beta)
+\end{array}
+$$
+so that in the figure of \S 77 we see that if $\overline{OA}=\alpha$,
+and $\overline{OB}=\beta$, we
+have $\overline{OD} = \rho = \alpha\beta\alpha^{1}$
+
+Or, again, we may get the result at once by transforming the
+equation to $\frac{\rho}{\alpha}=K(\alpha^{1}\rho)=K\frac{\beta}{\alpha}$
+
+{\bf 106}. For any three coplanar vectors the expression
+$$\rho=\alpha\beta\gamma$$
+is (\S 101) a vector. It is interesting to determine what this vector
+is. The reader will easily see that if a circle be described about
+the triangle, two of whose sides are (in order) $\alpha$ and $\beta$,
+and if from
+the extremity of $\beta$ a line parallel to $\gamma$ be drawn,
+again cutting the
+circle, the vector joining the point of intersection with the origin
+of $\alpha$ is the direction of the vector $\alpha\beta\gamma$.
+For we may write it in the form
+$$
+\rho=\alpha\beta^2\beta^{1}\gamma=
+(T\beta)^2\alpha\beta^{1}\gamma=
+(T\beta)^2\frac{\alpha}{\beta}\gamma
+$$
+which shows that the versor $\displaystyle\left(\frac{\alpha}{\beta}\right)$
+which turns $\beta$ into a direction
+parallel to $\alpha$, turns $\gamma$ into a direction parallel to $\rho$.
+And this expresses the longknown property of opposite angles of a
+quadrilateral inscribed in a circle.
+
+Hence if $\alpha$, $\beta$, $\gamma$
+be the sides of a triangle taken in order, the
+tangents to the circumscribing circle at the angles of the triangle
+are parallel respectively to
+$$
+\alpha\beta\gamma,\;\;\;\;
+\beta\gamma\alpha,\;\;\;\;\textrm{ and }
+\gamma\alpha\beta
+$$
+
+Suppose two of these to be parallel, i. e. let
+$$\alpha\beta\gamma=x\beta\gamma\alpha=x\alpha\gamma\beta\;\;\;\;(\S 90)$$
+since the expression is a vector. Hence
+$$\beta\gamma=x\gamma\beta$$
+which requires either
+$$x=1,\;\;\;\;V\gamma\beta=0\;\;\;\;\textrm{ or }\gamma \vert\vert \beta$$
+a case not contemplated in the problem; or
+$$x=1,\;\;\;\;S\beta\gamma=0$$
+i. e. the triangle is rightangled. And geometry shows us at once
+that this is correct.
+
+Again, if the triangle be isosceles, the tangent at the vertex is
+parallel to the base. Here we have
+$$x\beta=\alpha\beta\gamma$$
+or
+$$x(\alpha+\gamma)=\alpha(\alpha+\gamma)\gamma$$
+whence $x=\gamma^2=\alpha^2$, or $T\gamma=T\alpha$, as required.
+
+As an elegant extension of this proposition the reader may
+prove that the vector of the continued product $\alpha\beta\gamma\delta$
+of the vectorsides of any quadrilateral inscribed in a sphere
+is parallel to the radius drawn to the corner ($\alpha$, $\delta$).
+[For, if $\epsilon$ be the vector from $\delta$,
+$\alpha$ to $\beta$, $\gamma$, $\alpha\beta\epsilon$ and
+$\epsilon\gamma\delta$ are (by what precedes) vectors {\sl touching} the
+sphere at $\alpha$, $\delta$. And their product (whose vector part must be
+parallel to the radius at $\alpha$, $\delta$) is
+$$\alpha\beta\epsilon . \epsilon\gamma\delta=\epsilon^2 .
+\alpha\beta\gamma\delta]$$
+
+{\bf 107}. To exemplify the variety of possible transformations
+even of simple expressions, we will take cases which are of
+frequent occurrence in applications to geometry.
+
+Thus $$T(\rho+\alpha)=T(\rho\alpha)$$
+[which expresses that if
+$$
+\overline{OA}=\alpha\;\;\;\;
+\overline{OA^{'}}=\alpha\;\;\;\;\textrm{ and }\;\;\;\;
+\overline{OP}=\rho
+$$
+we have\hbox{\hskip 4cm}$AP=A^{'}P$\\
+and thus that $P$ is any point equidistant from two fixed points,]
+may be written $$(\rho+\alpha)^2=(\rho\alpha)^2$$
+or\hbox{\hskip 3cm}$\rho^2+2S\alpha\rho+\alpha^2=
+\rho^22S\alpha\rho+\alpha^2\;\;\;\;\textrm{(\S 104)}$\\
+whence\hbox{\hskip 4cm}$S\alpha\rho=0$\\
+This may be changed to
+$$\alpha\rho+\rho\alpha=0$$
+or
+$$\alpha\rho+K\alpha\rho=0$$
+$$SU\frac{\rho}{\alpha}=0$$
+or finally,
+$$TVU\frac{\rho}{\alpha}=1$$
+all of which express properties of a plane.
+
+Again,\hbox{\hskip 4cm}$T\rho=T\alpha$\\
+may be written\hbox{\hskip 3.2cm}$\displaystyle T\frac{\rho}{\alpha}=1$
+$$\left(S\frac{\rho}{\alpha}\right)^2\left(V\frac{\rho}{\alpha}\right)^2=1$$
+$$(\rho+\alpha)^22S\alpha(\rho+\alpha)=0$$
+$$\rho=(\rho+\alpha)^{1}\alpha(\rho+\alpha)$$
+$$S(\rho+\alpha)(\rho\alpha)=0$$
+or finally,
+$$T.(\rho+\alpha)(\rho\alpha)=2TV\alpha\rho$$
+
+All of these express properties of a sphere. They will be
+interpreted when we come to geometrical applications.
+
+{\bf 108}. {\sl To find the space relation among five points.}
+
+A system of five points, so far as its internal relations are
+concerned, is fully given by the vectors from one to the other four.
+If three of these be called $\alpha$, $\beta$, $\gamma$, the fourth,
+$\delta$, is necessarily expressible as
+$x\alpha+y\beta+z\gamma$. Hence the relation required must be
+independent of x, y, z.
+
+But
+$$
+\left.
+\begin{array}{rlll}
+S\alpha\delta &=\;x\alpha^2 &+\;yS\alpha\beta &+\;zS\alpha\gamma\\
+S\beta\delta &=\;xS\beta\alpha &+\;y\beta^2 &+\;zS\beta\gamma\\
+S\gamma\delta &=\;xS\gamma\alpha &+\;yS\gamma\beta &+\;z\gamma^2\\
+S\delta\delta=\delta^2 &=\;xS\delta\alpha &+\;yS\delta\beta &+\;zS\delta\gamma
+\end{array}
+\right\}\eqno{(1)}
+$$
+The elimination of $x$, $y$, $z$ gives a determinant of the fourth order,
+which may be written
+$$
+\left\vert
+\begin{array}{cccc}
+S\alpha\alpha & S\alpha\beta & S\alpha\gamma & S\alpha\delta\\
+S\beta\alpha & S\beta\beta & S\beta\gamma & S\beta\delta\\
+S\gamma\alpha & S\gamma\beta & S\gamma\gamma & S\gamma\delta\\
+S\delta\alpha & S\delta\beta & S\delta\gamma & S\delta\delta
+\end{array}
+\right\vert=0
+$$
+Now each term may be put in either of two forms, thus
+$$S\beta\gamma=\frac{1}{2}\left\{\beta^2+\gamma^2(\beta\gamma)^2\right\}=
+T\beta T\gamma\cos\widehat{\beta\gamma}$$
+
+
+If the former be taken we have the expression connecting the
+distances, two and two, of five points in the form given by Muir
+(Proc. R. S. E. 1889) ; if we use the latter, the tensors divide out
+(some in rows, some in columns), and we have the relation among
+the cosines of the sides and diagonals of a spherical quadrilateral.
+
+We may easily show (as an exercise in quaternion manipulation
+merely) that this is the {\sl only} condition, by showing that from it
+we can get the condition when any other of the points is taken as
+origin. Thus, let the origin be at $\alpha$, the vectors are
+$\alpha$, $\beta\alpha$, $\gamma\alpha$, $\delta\alpha$.
+But, by changing the signs of the first row, and first
+column, of the determinant above, and then adding their values
+term by term to the other rows and columns, it becomes
+$$
+\left\vert
+\begin{array}{cccc}
+S(\;\;\;\alpha)(\alpha) & S(\;\;\;\alpha)(\beta\alpha)
+& S(\;\;\;\alpha)(\gamma\alpha) & S(\;\;\;\alpha)(\delta\alpha)\\
+S(\beta\alpha)(\alpha) & S(\beta\alpha)(\beta\alpha)
+& S(\beta\alpha)(\gamma\alpha) & S(\beta\alpha)(\delta\alpha)\\
+S(\gamma\alpha)(\alpha) & S(\gamma\alpha)(\beta\alpha)
+& S(\gamma\alpha)(\gamma\alpha) & S(\gamma\alpha)(\delta\alpha)\\
+S(\delta\alpha)(\alpha) & S(\delta\alpha)(\beta\alpha)
+& S(\delta\alpha)(\gamma\alpha) & S(\delta\alpha)(\delta\alpha)
+\end{array}
+\right\vert
+$$
+which, when equated to zero, gives the same relation as before.
+[See Ex. 10 at the end of this Chapter.]
+
+An additional point, with $\epsilon=x^{'}\alpha+y^{'}\beta+z^{'}\gamma$
+gives six additional equations like (1) ; i. e.
+$$
+\begin{array}{rlll}
+S\alpha\epsilon&=x^{'}\alpha^2&+y^{'}S\alpha\beta&+z^{'}S\alpha\gamma\\
+S\beta\epsilon&=x^{'}S\beta\alpha&+y^{'}\beta^2&+z^{'}S\beta\gamma\\
+S\gamma\epsilon&=x^{'}S\gamma\alpha&+y^{'}S\gamma\beta&+z^{'}\gamma^2\\
+S\delta\epsilon&=x^{'}S\delta\alpha&+y^{'}S\delta\beta&+z^{'}S\delta\gamma\\
+&=xS\epsilon\alpha&+yS\epsilon\beta&+zS\epsilon\gamma\\
+\epsilon^2&=x^{'}S\alpha\epsilon&+y^{'}S\beta\epsilon&+z^{'}S\gamma\epsilon
+\end{array}
+$$
+from which corresponding conclusions may be drawn.
+
+Another mode of solving the problem at the head of this
+section is to write the {\sl identity}
+$$
+\sum m(\alpha\theta)^2=\sum m\alpha^2sS.\theta\sum m\alpha+\theta^2\sum m
+$$
+where the $m$s are undetermined scalars, and the $\alpha$s are given
+vectors, while $\theta$ is any vector whatever.
+
+Now, {\sl provided that the number of given vectors exceeds four}, we
+do not completely determine the ms by imposing the conditions
+$$\sum m=0,\;\;\;\;\sum m\alpha=0$$
+Thus we may write the above identity, for each of five vectors
+successively, as
+$$
+\begin{array}{rcl}
+\sum m(\alpha\alpha_1)^2 &=& \sum m\alpha^2\\
+\sum m(\alpha\alpha_2)^2 &=& \sum m\alpha^2\\
+\ldots\ldots &=& \ldots\\
+\sum m(\alpha\alpha_n)^2 &=& \sum m\alpha^2\\
+\end{array}
+$$
+Take, with these,\hbox{\hskip 3cm}$\sum m = 0$\\
+and we have six linear equations from which to eliminate the $m$s.
+The resulting determinant is
+$$
+\left\vert
+\begin{array}{cccccc}
+\overline{\alpha_1\alpha_1^2} & \overline{\alpha_1\alpha_s^2} &
+\overline{\alpha_1\alpha_3^2} & . &
+\overline{\alpha_1\alpha_5^2} & 1\\
+\overline{\alpha_2\alpha_1^2} & \overline{\alpha_2\alpha_s^2} &
+\overline{\alpha_2\alpha_3^2} & . &
+\overline{\alpha_2\alpha_5^2} & 1\\
+. & . & . & & . & \\
+. & . & . & & . & \\
+\overline{\alpha_5\alpha_1^2} & \overline{\alpha_5\alpha_s^2} &
+\overline{\alpha_5\alpha_3^2} & . &
+\overline{\alpha_5\alpha_5^2} & 1\\
+1 & 1 & . & . & 1 & 0\\
+\end{array}
+\right\vert
+\sum m\alpha^2=0
+$$
+
+This is equivalent to the form in which Cayley gave the
+relation among the mutual distances of five points. (Camb. Math.
+Journ. 1841.)
+
+{\bf 109}. We have seen in \S 95 that a quaternion may be divided
+into its scalar and vector parts as follows:
+$$
+\frac{\beta}{\alpha}=S\frac{\beta}{\alpha}+V\frac{\beta}{\alpha}=
+\frac{T\beta}{T\alpha}(\cos\theta+\epsilon\sin\theta)
+$$
+where $\theta$ is the angle between the directions of
+$\alpha$ and $\beta$ and $\displaystyle \epsilon=UV\frac{\beta}{\alpha}$
+is the unitvector perpendicular to the plane of $\alpha$
+and $\beta$ so situated
+that positive (i.e. lefthanded) rotation about it turns
+$\alpha$ towards $\beta$
+
+Similarly we have (\S 96)
+$$
+\begin{array}{rl}
+\alpha\beta&=S\alpha\beta + V\alpha\beta\\
+&=T\alpha T\beta(\cos\theta +\epsilon\sin\theta)
+\end{array}
+$$
+$\theta$ and $\epsilon$ having the same signification as before.
+
+{\bf 110}. Hence, considering the versor parts alone, we have
+$$U\frac{\beta}{\alpha}=\cos\theta+\epsilon\sin\theta$$
+Similarly
+$$U\frac{\gamma}{\beta}=\cos\phi+\epsilon\sin\phi$$
+$\phi$ being the positive angle between the directions of
+$\gamma$ and $\beta$, and $\epsilon$
+the same vector as before, if $\alpha$, $\beta$, $\gamma$ be coplanar.
+
+Also we have
+$$U\frac{\gamma}{\alpha}=\cos(\theta+\phi)+\epsilon\sin(\theta+\phi)$$
+But we have always
+$$\frac{\gamma}{\beta}.\frac{\beta}{\alpha}=\frac{\gamma}{\alpha}$$
+and therefore
+$$U\frac{\gamma}{\beta}.U\frac{\beta}{\alpha}=U\frac{\gamma}{\alpha}$$
+or
+$$
+\cos(\phi+\theta)+\epsilon\sin(\phi+\theta)=
+(\cos\phi+\epsilon\sin\phi)(\cos\theta+\epsilon\sin\theta)
+$$
+$$
+=\cos\phi\cos\theta\sin\phi\sin\theta+
+\epsilon(\sin\phi\cos\theta+\cos\phi\sin\theta)
+$$
+from which we have at once the fundamental formulae for the
+cosine and sine of the sum of two arcs, by equating separately the
+scalar and vector parts of these quaternions.
+
+And we see, as an immediate consequence of the expressions
+above, that
+$$\cos m\theta+\epsilon\sin m\theta=(\cos\theta+\epsilon\sin\theta)^m$$
+if $m$ be a positive whole number. For the lefthand side is a versor
+which turns through the angle $m\theta$ at once, while the righthand
+side is a versor which effects the same object by $m$ successive turn
+ings each through an angle $\theta$. See \S\S 8, 9.
+
+{\bf 111}. To extend this proposition to fractional indices we have
+only to write $\displaystyle \frac{\theta}{n}$ for $\theta$,
+when we obtain the results as in ordinary trigonometry.
+
+From De Moivre's Theorem, thus proved, we may of course
+deduce the rest of Analytical Trigonometry. And as we have
+already deduced, as interpretations of selfevident quaternion
+transformations (\S\S 97, 104), the fundamental formulae for the solution
+of plane triangles, we will now pass to the consideration of spherical
+trigonometry, a subject specially adapted for treatment by qua
+ternions; but to which we cannot afford more than a very few
+sections. (More on this subject will be found in Chap. XI in
+connexion with the Kinematics of rotation.) The reader is referred to
+Hamilton s works for the treatment of this subject by quaternion
+exponentials.
+
+{\bf 112}. Let $\alpha$, $\beta$, $\gamma$
+be unitvectors drawn from the centr to the
+corners $A$, $B$, $C$ of a triangle on the unitsphere. Then it is evident
+that, with the usual notation, we have (\S 96),
+$$
+S\alpha\beta=\cos c,\;\;\;\;
+S\beta\gamma=\cos a,\;\;\;\;
+S\gamma\alpha=\cos b
+$$
+$$
+TV\alpha\beta=\sin c,\;\;\;\;
+TV\beta\gamma=\sin a,\;\;\;\;
+TV\gamma\alpha=\sin b
+$$
+Also $UV\alpha\beta$, $UV\beta\gamma$, $UV\gamma\alpha$
+are evidently the vectors of the corners of the polar triangle.
+
+Hence
+$$S.UV\alpha\beta UV\beta\gamma=\cos B\textrm{, \&c.}$$
+$$TV.UV\alpha\beta UV\beta\gamma=\sin B\textrm{, \&c.}$$
+
+Now (\S 90 (1)) we have
+$$
+\begin{array}{rcl}
+SV\alpha\beta V\beta\gamma&=&S.\alpha V(\beta V\beta\gamma)\\
+&=&S\alpha\beta S\beta\gamma + \beta^2S\alpha\gamma
+\end{array}
+$$
+Remembering that we have
+$$
+SV\alpha\beta V\beta\gamma=
+TV\alpha\beta TV\beta\gamma S.UV\alpha\beta UV\beta\gamma
+$$
+we see that the formula just written is equivalent to
+$$\sin a \sin c \cos B = \cos a \cos c + \cos b$$
+or
+$$\cos b = \cos a \cos c + \sin a \sin c \cos B$$
+
+{\bf 113}. Again,
+$$V.V\alpha\beta V\beta\gamma=\beta S\alpha\beta\gamma$$
+which gives
+$$
+TV.V\alpha\beta V\beta\gamma=
+TS.\alpha\beta\gamma=
+TS.\alpha V\beta\gamma=
+TS.\beta V\gamma\alpha=
+TS.\gamma V\alpha\beta
+$$
+or
+$$\sin a \sin c \sin B = \sin a \sin p_a = \sin b \sin p_b = \sin c \sin p_c$$
+where $p_a$ is the arc drawn from $A$ perpendicular to $BC$, \&c.
+Hence
+$$\sin p_a = \sin c \sin B$$
+$$\sin p_b = \frac{\sin a \sin c}{\sin b} \sin B$$
+$$\sin p_c = \sin a \sin B$$
+
+{\bf 114}. Combining the results of the last two sections, we have
+$$V\alpha\beta .V\beta\gamma=\sin a \sin c \cos B\beta \sin a \sin c \sin B$$
+$$= \sin a \sin c (\cos B  \beta \sin B)$$
+$$
+\left.
+\begin{array}{lcl}
+\textrm{Hence} & \hbox{\hskip 1cm}
+& U.V\alpha\beta V\beta\gamma=(\cos B\beta\sin B)\\
+\textrm{and} & \hbox{\hskip 1.1cm}
+& U.V\gamma\beta V\beta\alpha=(\cos B+\beta\sin B)
+\end{array}
+\right\}
+$$
+These are therefore versors which turn all vectors perpendicular to
+$OB$ negatively or positively about $OB$ through the angle $B$.
+
+[It will be shown later (\S 119) that, in the combination
+$$(\cos B+\beta\sin B)(\;\;\;\;)(\cos B \beta\sin B)$$
+the system operated on is made to rotate, as if rigid, round the
+vector axis $\beta$ through an angle $2B$.]
+
+As another instance, we have
+$$
+\begin{array}{rl}
+\tan B &=\displaystyle \frac{\sin B}{\cos B}\\
+&\\
+&=\displaystyle
+\frac{TV.V\alpha\beta V\beta\gamma}{S.V\alpha\beta V\beta\gamma}\\
+&\\
+&=\displaystyle
+\beta^{1}\frac{V.V\alpha\beta V\beta\gamma}{S.V\alpha\beta V\beta\gamma}\\
+&\\
+&=\displaystyle
+\frac{S.\alpha\beta\gamma}{S\alpha\gamma+S\alpha\beta S\beta\gamma}
+=\textrm{ \&c}
+\end{array}
+\eqno{(1)}
+$$
+
+The interpretation of each of these forms gives a different theorem
+in spherical trigonometry.
+
+{\bf 115}. Again, let us square the equal quantities
+$$
+V.\alpha\beta\gamma\;\;\;\;\textrm{ and }\;\;\;\;
+\alpha S\beta\gamma\beta S\alpha\gamma+\gamma S\alpha\beta$$
+supposing $\alpha$, $\beta$, $\gamma$ to be any unitvectors whatever. We have
+$$
+(V.\alpha\beta\gamma)^2=
+S^2\beta\gamma + S^2\gamma\alpha +
+S^2\alpha\beta + 2S\beta\gamma S\gamma\alpha S\alpha\beta
+$$
+But the lefthand member may be written as
+$$T^2.\alpha\beta\gammaS^2.\alpha\beta\gamma$$
+whence
+$$
+1S^2.\alpha\beta\gamma=S^2\beta\gamma +S^2\gamma\alpha +
+S^2\alpha\beta+2S\beta\gamma S\gamma\alpha S\alpha\beta
+$$
+or
+$$
+1\cos^2 a  \cos^2 b  \cos^2 c + 2 \cos a \cos b \cos c$$
+$$= \sin^2 a \sin^2 p_a = \textrm{ \&c.}$$
+$$= \sin^2 a \sin^2 b \sin^2 C = \textrm{ \&c.}$$
+all of which are wellknown formulae.
+
+{\bf 116}. Again, for any quaternion,
+$$q=Sq+Vq$$
+so that, if $n$ be a positive integer,
+$$q^n=(Sq)^n+n(Sq)^{n1}Vq+
+\frac{n.\overline{n1}}{1.2}(Sq)^{n2}(Vq)^2+\ldots$$
+From this at once
+$$S.q^n=(Sq)^n\frac{n.\overline{n1}}{1.2}(Sq)^{n2}T^2Vq$$
+$$+\frac{n.\overline{n1}.\overline{n2}.\overline{n3}}{1.2.3.4}
+(Sq)^{n4}T^4(Vq)\textrm{\&c.,}$$
+$$V.q^n=Vq\left[n(Sq)^{n1}
+\frac{n.\overline{n1}.\overline{n2}}{1.2.3}
+(Sq)^{n3}T^2Vq+\textrm{\&c.,}\right]$$
+If $q$ be a versor we have
+$$q=\cos u +\theta\sin u$$
+so that
+$$
+\begin{array}{rl}
+S.q^n & =\displaystyle
+(\cos u)^n\frac{n.\overline{n1}}{1.2}(\cos u)^{n2}(\sin u)^2+\ldots\\
+&\\
+&=\displaystyle\cos nu;\\
+&\\
+V.q^n & =\displaystyle
+\theta\sin u\left[n(\cos u)^{n1}
+\frac{n.\overline{n1}.\overline{n2}}{1.2.3}
+(\cos u)^{n3}(\sin u)^2+\ldots\right]\\
+&\\
+&=\displaystyle\theta\sin nu;
+\end{array}
+$$
+as we might at once have concluded from \S 110.
+
+Such results may be multiplied indefinitely by any one who has
+mastered the elements of quaternions.
+
+{\bf 117}. A curious proposition, due to Hamilton, gives us a
+quaternion expression for the {\sl spherical excess} in any triangle.
+The following proof, which is very nearly the same as one of his,
+though by no means the simplest that can be given, is chosen here
+because it incidentally gives a good deal of other information.
+We leave the quaternion proof as an exercise.
+
+Let the unitvectors drawn from the centre of the sphere to
+$A$, $B$, $C$, respectively, be $\alpha$, $\beta$, $\gamma$.
+It is required to express, as an
+arc and as an angle on the sphere, the quaternion
+$$\beta\alpha^{1}\gamma$$
+
+\includegraphics{ps/quat18.ps}
+
+The figure represents an orthographic projection made on a
+plane perpendicular to $\gamma$. Hence $C$ is the centre of the circle $DEe$.
+Let the great circle through $A$, $B$ meet $DEe$ in $E$, $e$, and let $DE$ be
+a quadrant. Thus
+${\stackrel{\frown}{DE}}$ represents $\gamma$ (\S 72). Also make
+${\stackrel{\frown}{EF}} = {\stackrel{\frown}{AB}}$
+$=\beta\alpha^{1}$ Then, evidently,
+$${\stackrel{\frown}{DF}}=\beta\alpha^{1}\gamma$$
+which gives the arcual representation required.
+
+Let $DF$ cut $Ee$ in $G$. Make $Ca = EG$, and join $D$, $a$, and $a$, $F$.
+Obviously, as $D$ is the pole of $Ee$, $Da$ is a quadrant ; and since
+$EG = Ca$, $Ga = EG$, a quadrant also. Hence $a$ is the pole of $DG$,
+and therefore the quaternion may be represented by the angle
+$DaF$.
+
+Make $Cb = Ca$, and draw the arcs $Pa\beta$, $Pb\alpha$ from $P$, the pole of
+$AB$. Comparing the triangles $Eb\alpha$ and $ea\beta$,
+we see that $E\alpha = e\beta$.
+But, since $P$ is the pole of $AB$, $F\beta a$ is a right angle: and therefore
+as $Fa$ is a quadrant, so is $F\beta$. Thus $AB$ is the complement of $E\alpha$
+or $\beta e$, and therefore
+$$\alpha\beta=2AB$$
+
+Join $bA$. and produce it to $c$ so that $Ac = bA$; join $c$, $P$, cutting
+$AB$ in $o$. Also join $c$, $B$, and $B$, $a$.
+
+Since $P$ is the pole of $AB$, the angles at $o$ are right angles;
+and therefore, by the equal triangles $b\alpha A$, $coA$, we have
+$$\alpha A = Ao$$
+But
+$$\alpha\beta = 2AB$$
+whence
+$$oB=B\beta$$
+and therefore the triangles $coB$ and $Ba\beta$ are equal, and $c$, $B$, $a$
+lie on the same great circle.
+
+Produce $cA$ and $cB$ to meet in $H$ (on the opposite side of the
+sphere). $H$ and $c$ are diametrically opposite, and therefore $cP$,
+produced, passes through $H$.
+
+Now $Pa = Pb = PH$, for they differ from quadrants by the
+equal arcs $a\beta$, $b\alpha$, $oc$. Hence these arcs divide the
+triangle $Hab$ into three isosceles triangles.
+
+But
+$$\angle PHb + \angle PHA = \angle aHb = \angle bca$$
+Also
+$$\angle Pab = \pi  \angle cab  \angle PaH$$
+$$\angle Pba = \angle Pab = \pi  \angle cba  \angle PbH$$
+Adding,
+$$2\angle Pab = 2\pi  \angle cab  \angle cba  \angle bca$$
+$$= \pi  (\textrm{spherical excess of }abc)$$
+But, as $\angle Fa\beta$ and $\angle Dae$ are right angles, we have
+$$
+\textrm{angle of }\beta\alpha^{1}\gamma =
+\angle FaD = \beta ae = \angle Pab
+$$
+$$=\frac{\pi}{2}  \frac{1}{2}(\textrm{spherical excess of }abc)$$
+
+[Numerous singular geometrical theorems, easily proved {\sl ab
+initio} by quaternions, follow from this: e.g. The arc $AB$, which
+bisects two sides of a spherical triangle $abc$, intersects the base at
+the distance of a quadrant from its middle point. All spherical
+triangles, with a common side, and having their other sides
+bisected by the same great circle (i.e. having their vertices in a
+small circle parallel to this great circle) have equal areas, \&c. ]
+
+{\bf 118}. Let $\overline{Oa}=\alpha^{'}$, $\overline{Ob}=\beta^{'}$,
+$\overline{Oc}=\gamma^{'}$, and we have
+$$
+\begin{array}{rcl}
+\left(\frac{\alpha^{'}}{\beta^{'}}\right)^{\frac{1}{2}}
+\left(\frac{\beta^{'}}{\gamma^{'}}\right)^{\frac{1}{2}}
+\left(\frac{\gamma^{'}}{\alpha^{'}}\right)^{\frac{1}{2}}&=&
+{\stackrel{\frown}{Ca}}.{\stackrel{\frown}{cA}}.{\stackrel{\frown}{Bc}}\\
+&=&{\stackrel{\frown}{Ca}}.{\stackrel{\frown}{BA}}\\
+&=&{\stackrel{\frown}{EG}}.{\stackrel{\frown}{FE}}=
+{\stackrel{\frown}{FG}}
+\end{array}
+$$
+
+But $FG$ is the complement of $DF$. Hence the {\sl angle of the
+quaternion}
+$$
+\left(\frac{\alpha^{'}}{\beta^{'}}\right)^{\frac{1}{2}}
+\left(\frac{\beta^{'}}{\gamma^{'}}\right)^{\frac{1}{2}}
+\left(\frac{\gamma^{'}}{\alpha^{'}}\right)^{\frac{1}{2}}
+$$
+{\sl is half the spherical excess of the triangle whose angular points are
+at the extremities of the unitvectors} $\alpha^{'}$, $\beta^{'}$, and
+$\gamma^{'}$.
+
+[In seeking a purely quaternion proof of the preceding proposi
+tions, the student may commence by showing that for any three
+unitvectors we have
+$$
+\frac{\beta}{\alpha}\frac{\gamma}{\beta}\frac{\alpha}{\gamma}=
+(\beta\alpha^{1}\gamma)^2
+$$
+
+The angle of the first of these quaternions can be easily assigned;
+and the equation shows how to find that of $\beta\alpha^{1}\gamma$.
+
+Another easy method is to commence afresh by forming from
+the vectors of the corners of a spherical triangle three new vectors
+thus:
+$$
+\alpha^{'}=\left(\frac{\beta+\gamma}{\alpha}^{2}\right)^2 .\;\alpha,\;\;\;\;\;
+\textrm{\&c.}
+$$
+
+Then the angle between the planes of $\alpha$, $\beta^{'}$ and
+$\gamma^{'}$, $\alpha$; or of $\beta$, $\gamma^{'}$ and $\alpha^{'}$,
+$\beta$; or of $\gamma$, $\alpha^{'}$ and $\beta^{'}$, $\gamma$
+is obviously the spherical excess.
+
+But a still simpler method of proof is easily derived from the
+composition of rotations.]
+
+{\bf 119}. It may be well to introduce here, though it belongs
+rather to Kinematics than to Geometry, the interpretation of the
+operator
+$$q(\;\;\;)q^{1}$$
+
+By a rotation, about the axis of $q$, through double the angle of $q$,
+the quaternion $r$ becomes the quaternion $qrq^{1}$ . Its tensor and
+angle remain unchanged, its plane or axis alone varies.
+
+\includegraphics{ps/quat19.ps}
+
+A glance at the figure is sufficient for
+the proof, if we note that of course
+$T . qrq^{1} = Tr$, and therefore that we need
+consider the {\sl versor} parts only. Let $Q$
+be the pole of $q$.
+$$
+{\stackrel{\frown}{AB}}=q,\;\;\;\;
+{\stackrel{\frown}{AB^{1}}}=q^{1},\;\;\;\;
+{\stackrel{\frown}{B^{'}C^{'}}}=r
+$$
+Join $C^{'}A$, and make
+${\stackrel{\frown}{AC}}={\stackrel{\frown}{C^{'}A}}$. Join $CB$.
+
+Then ${\stackrel{\frown}{CB}}$ is $qrq^{1}$,
+its arc $CB$ is evidently equal in length to that
+of $r$, $B^{'}C^{'}$; and its plane (making the same angle with
+$B^{'}B$ that that of
+$B^{'}C^{'}$ does) has evidently been made to revolve about $Q$, the
+pole of $q$, through double the angle of $q$.
+
+It is obvious, from the nature of the above proof, that this
+operation is distributive; i.e. that
+$$q(r+s)q^{1}=qrq^{1}+qsq^{1}$$
+
+If $r$ be a vector, $=\rho$, then $q\rho q^{1}$
+(which is also a vector) is the
+result of a rotation through double the angle of $q$ about the axis
+of $q$. Hence, as Hamilton has expressed it, if $B$ represent a rigid
+system, or assemblage of vectors,
+$$qBq^{1}$$
+is its new position after rotating through double the angle of $q$
+about the axis of $q$.
+
+{\bf 120}. To compound such rotations, we have
+$$r . qBq^{1} . r^{1} = rq . B . (rq)^{1}$$
+
+To cause rotation through an angle $t$fold the double of the angle
+of $q$ we write
+$$q^{t}Bq^{t}$$
+
+To reverse the direction of this rotation write
+$$q^{t}Bq^{t}$$
+
+To {\sl translate} the body $B$ without rotation, each point of it moving
+through the vector $\alpha$, we write $\alpha + B$.
+
+To produce rotation of the translated body about the same axis,
+and through the same angle, as before,
+$$q(\alpha+B)q^{1}$$
+
+Had we rotated first, and then translated, we should have had
+$$\alpha+qBq^{1}$$
+
+From the point of view of those who do not believe in the
+Moon s rotation, the former of these expressions ought to be
+$$q\alpha q^{1}+B$$
+instead of
+$$q\alpha q^{1}+qBq^{1}$$
+But to such men quaternions are unintelligible.
+
+{\bf 121}. The operator above explained finds, of course, some
+of its most direct applications in the ordinary questions of
+Astronomy, connected with the apparent diurnal rotation of the
+stars. If $\lambda$ be a unitvector parallel to the polar axis, and $h$ the
+hour angle from the meridian, the operator is
+$$
+\left(\cos\frac{h}{2}\lambda\sin\frac{h}{2}\right)
+\left(\;\;\;\;\right)
+\left(\cos\frac{h}{2}+\lambda\sin\frac{h}{2}\right)
+$$
+or
+$$L^{1}\left(\;\;\;\;\right)L$$
+
+the inverse going first, because the {\sl apparent} rotation is negative
+(clockwise).
+
+If the upward line be $i$, and the southward $j$, we have
+$$\lambda = i\sin lj\cos l$$
+where $l$ is the latitude of the observer. The meridian equatorial
+unit vector is
+$$\mu = i\cos l+j\sin l$$
+and $\lambda$, $\mu$, $k$ of course form a rectangular unit system.
+
+The meridian unitvector of a heavenly body is
+$$\delta=i\cos(ld)+j\sin(ld)$$
+$$=\lambda\sin d+\mu\cos d$$
+where $d$ is its declination.
+
+Hence when its hourangle is $h$, its vector is
+$$\delta^{'}=L^{1}\delta L$$
+
+The vertical plane containing it intersects the horizon in
+$$iVi\delta^{'}=jSj\delta^{'}+kSk\delta^{'}$$
+so that
+$$\tan(azimuth)=\frac{Sk\delta^{'}}{Sj\delta^{'}}\eqno{(1)}$$
+
+[This may also be obtained directly from the last formula (1)
+of \S 114.]
+
+To find its Amplitude, i.e. its azimuth at rising or setting,
+the hourangle must be obtained from the condition
+$$Si\delta^{'}=0\eqno{(2)}$$
+
+These relations, with others immediately deducible from them,
+enable us (at once and for ever) to dispense with the hideous
+formulae of Spherical Trigonometry.
+
+{\bf 122}. To show how readily they can be applied, let us
+translate the expressions above into the ordinary notation. This
+is effected at once by means of the expressions for $\lambda$, $\mu$, $L$,
+and $\delta$ above, which give by inspection
+$$\delta^{'}=\lambda\sin d+(\mu\cos hk\sin h)\cos d$$
+= x sin d + (fjb cos h k sin h) cos d,
+and we have from (1) and (2) of last section respectively
+$$
+\tan(azimuth)=
+\frac{\sin h\cos d}{\cos l\sin d\sin l\cos d\cos h}\eqno{(1)}
+$$
+$$
+\cos h+\tan l \tan d=0\eqno{(2)}
+$$
+
+In Capt. Weir s ingenious {\sl Azimuth Diagram}, these equations
+are represented graphically by the rectangular coordinates of a
+system of confocal conics: viz.
+$$
+\left.
+\begin{array}{c}
+x = \sin h \sec l \\
+y = \cos h \tan l
+\end{array}
+\right\}\eqno{(3)}
+$$
+
+The ellipses of this system depend upon $l$ alone, the hyperbolas
+upon $h$. Since (1) can, by means of (3), be written as
+$$\tan(azimuth)=\frac{x}{\tan dy}$$
+we see that the azimuth can be constructed at once by joining
+with the point $0$, $\tan d$, the intersection of the proper ellipse and
+hyperbola.
+
+Equation (2) puts these expressions for the coordinates in the
+form
+$$
+\left.
+\begin{array}{c}
+x=\sec l\sqrt{1\tan^{2} l\tan^{2} d}\\
+y=\tan^{2} l \tan d
+\end{array}
+\right\}
+$$
+
+The elimination of $d$ gives the ellipse as before, but that of $l$
+gives, instead of the hyperbolas, the circles
+$$x^{2}+y^{2}y(\tan d\cot d)=1$$
+
+The radius is
+$$\frac{1}{2}(\tan d+ \cot d)$$
+and the coordinates of the centre are
+$$0,\;\;\;\frac{1}{2}(\tan d \cot d)$$
+
+123. A scalar equation in $\rho$, the vector of an undetermined
+point, is generally the equation of a {\sl surface}; since we may use
+in it the expression
+$$\rho=x\alpha$$
+where $x$ is an unknown scalar, and $\alpha$ any assumed unitvector.
+The result is an equation to determine $x$. Thus one or more
+points are found on the vector $x\alpha$, whose coordinates satisfy the
+equation; and the locus is a surface whose degree is determined
+by that of the equation which gives the values of $x$.
+
+But a {\sl vector} equation in $\rho$, as we have seen, generally leads to
+three scalar equations, from which the three rectangular or other
+components of the sought vector are to be derived. Such a vector
+equation, then, usually belongs to a definite number of {\sl points} in
+space. But in certain cases these may form a {\sl line}, and even a
+{\sl surface}, the vector equation losing as it were one or two of the
+three scalar equations to which it is usually equivalent.
+
+Thus while the equation
+$$\alpha\rho=\beta$$
+gives at once
+$$\rho=\alpha^{1}\beta$$
+which is the vector of a definite point, since by making $\rho$ a {\sl vector}
+we have evidently assumed
+$$S\alpha\beta=0$$
+the closely allied equation
+$$V\alpha\rho=\beta$$
+is easily seen to involve
+$$S\alpha\beta=0$$
+and to be satisfied by
+$$\rho=\alpha^{1}\beta+x\alpha$$
+whatever be $x$. Hence the vector of any point whatever in the
+line drawn parallel to $\alpha$ from the extremity of $\alpha^{1}\beta$
+satisfies the given equation. [The difference between the results depends
+upon the fact that $S\alpha\rho$ is indeterminate in the second form, but
+definite (= 0) in the first.]
+
+{\bf 124}. Again,
+$$V\alpha\rho . V\rho\beta=(V\alpha\beta)^{2}$$
+is equivalent to but two scalar equations. For it shows that $V\alpha\rho$
+and $V\beta\rho$ are parallel, i.e. $\rho$
+lies in the same plane as $\alpha$ and $\beta$, and
+can therefore be written (\S 24)
+$$\rho=x\alpha+y\beta$$
+where $x$ and $y$ are scalars as yet undetermined.
+
+We have now
+$$V\alpha\rho=yV\alpha\beta$$
+$$V\rho\beta=xV\alpha\beta$$
+which, by the given equation, lead to
+$$xy=1,\;\;\;\textrm{or}\;\;\;y=\frac{1}{x}$$
+or finally
+$$\rho=x\alpha+\frac{1}{x}\beta$$
+which (\S 40) is the equation of a hyperbola whose asymptotes are
+in the directions of $\alpha$ and $\beta$.
+
+{\bf 125}. Again, the equation
+$$V . V\alpha\beta V\alpha\rho=0$$
+though apparently equivalent to three scalar equations, is really
+equivalent to one only. In fact we see by \S 91 that it may be
+written
+$$\alpha S.\alpha\beta\rho=0$$
+whence, if $\alpha$ be not zero, we have
+$$S . \alpha\beta\rho=0$$
+and thus (\S 101) the only condition is that $\rho$ is coplanar with
+$\alpha$, $\beta$.
+Hence the equation represents the plane in which
+$\alpha$ and $\beta$ lie.
+
+{\bf 126}. Some very curious results are obtained when we extend
+these processes of interpretation to functions of a {\sl quaternion}
+$$q=w+\rho$$
+instead of functions of a mere {\sl vector} $\rho$.
+
+A scalar equation containing such a quaternion, along with
+quaternion constants, gives, as in last section, the equation of a
+surface, if we assign a definite value to $w$. Hence for successive
+values of $w$, we have successive surfaces belonging to a system ;
+and thus when $w$ is indeterminate the equation represents not a
+{\sl surface}, as before, but a {\sl volume},
+in the sense that the vector of any
+point within that volume satisfies the equation.
+
+Thus the equation
+$$(Tq)^2=a^2$$
+or
+$$w^2\rho^2=a^2$$
+or
+$$(TP)^2=a^2w^2$$
+represents, for any assigned value of $w$, not greater than $a$, a sphere
+whose radius is $\sqrt{a^2w^2}$. Hence the equation is satisfied by the
+vector of any point whatever in the {\sl volume} of a sphere of radius $a$,
+whose centre is origin.
+
+Again, by the same kind of investigation,
+$$(T (q\beta))^2=a^2$$
+where $q=w+\rho$, is easily seen to represent the volume of a sphere
+of radius $a$ described about the extremity of $\beta$ as centre.
+
+Also $S(q^2) = a^2$ is the equation of infinite space less the space
+contained in a sphere of radius $a$ about the origin.
+
+Similar consequences as to the interpretation of vector
+equations in quaternions may be readily deduced by the reader.
+
+{\bf 127}. The following transformation is enuntiated without proof
+by Hamilton ({\sl Lectures}, p. 587, and {\sl Elements}, p. 299).
+$$r^{1}(r^2q^2)^{\frac{1}{2}}q^{1}=U(rq+KrKq)$$
+To prove it, let
+$$r^{1}(r^2q^2)^{\frac{1}{2}}q^{1}=t$$
+then
+$$Tt=1$$
+and therefore
+$$Kt=t^{1}$$
+But
+$$(r^2q^2)^{\frac{1}{2}}=rtq$$
+or
+$$r^2q^2=rtqrtq$$
+or
+$$rq=tqrt$$
+Hence
+$$KqKr=t^{1}KrKqt^{1}$$
+or
+$$KrKq=tKqKrt$$
+Thus we have
+$$U(rq\pm KrKq)=tU(qr\pm KqKr)t$$
+or, if we put
+$$s=U(qr\pm KqKr)$$
+$$Ks=\pm tst$$
+Hence
+$$sKs=(Ts)^2=1=\pm stst$$
+which, if we take the positive sign, requires
+$$st=\pm 1$$
+or
+$$t=\pm s^{1}=\pm UKs$$
+which is the required transformation.
+
+[It is to be noticed that there are other results which might
+have been arrived at by using the negative sign above ; some
+involving an arbitrary unitvector, others involving the imaginary
+of ordinary algebra.]
+
+{\bf 128}. As a final example, we take a transformation of Hamilton's,
+of great importance in the theory of surfaces of the second order.
+
+Transform the expression
+$$(S\alpha\rho)^2+(S\beta\rho)^2+(S\gamma\rho)^2$$
+in which $\alpha$, $\beta$, $\gamma$
+are any three mutually rectangular vectors, into the form
+$$\left(\frac{T(\iota\rho+\rho\kappa)}{\kappa^2\iota^2}\right)^2$$
+which involves only two vectorconstants, $\iota$, $\kappa$.
+
+[The student should remark here that $\iota$, $\kappa$, two undetermined
+vectors, involve six disposable constants : and that $\alpha$, $\beta$,
+$\gamma$, being a {\sl rectangular} system, involve also only six constants.]
+$$
+\begin{array}{rcl}
+\{T(\iota\rho+\rho\kappa)\}^2
+&=& (\iota\rho+\rho\kappa)(\rho\iota+\kappa\rho)\;\;\;\;(\S\S 52,55)\\
+&=& (\iota^2+\kappa^2)\rho^2+(\iota\rho\kappa\rho+\rho\kappa\rho\iota)\\
+&=& (\iota^2+\kappa^2)\rho^2+2S.\iota\rho\kappa\rho\\
+&=& (\iota\kappa)^2\rho^2+4S\iota\rho S\kappa\rho
+\end{array}
+$$
+Hence
+$$
+(S\alpha\rho)^2+(S\beta\rho)^2+(S\gamma\rho)^2=
+\frac{(\iota\kappa)^2}{(\kappa^2\iota^2)^2}\rho^2+
+4\frac{S\iota\rho S\kappa\rho}{(\kappa^2\iota^2)^2}
+$$
+But
+$$
+\alpha^{2}(S\alpha\rho)^2+
+\beta^{2}(S\beta\rho)^2+
+\gamma^{2}(S\gamma\rho)^2=
+\rho^2\;\;\;\;(\S\S 25,73).
+$$
+Multiply by $\beta^2$ and subtract, we get
+$$
+\left(1\frac{\beta^2}{\alpha^2}\right)(S\alpha\rho)^2
+\left(\frac{\beta^2}{\gamma^2}1\right)(S\gamma\rho)^2=
+\left\{\frac{(\iota\kappa)^2}{(\kappa^2\iota^2)^2}\beta^2\right\}\rho^2+
+4\frac{S\iota\rho S\kappa\rho}{(\kappa^2\iota^2)^2}
+$$
+
+The left side breaks up into two real factors if $\beta^2$ be intermediate
+in value to $\alpha^2$ and $\gamma^2$:
+and that the right side may do so the term
+in $\rho^2$ must vanish. This condition gives
+$$\beta^2=\frac{(\iota\kappa)^2}{(\kappa^2\iota^2)^2}$$
+and the identity becomes
+$$
+S\left\{
+\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}+
+\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
+\right\}
+\rho S\left\{
+\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}
+\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
+\right\}\rho=
+4\frac{S\iota\rho S\kappa\rho}{(\kappa^2\iota^2)^2}
+$$
+Hence we must have
+$$
+\frac{2\iota}{\kappa^2\iota^2}=
+p\left\{
+\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}+
+\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
+\right\}
+$$
+$$
+\frac{2\kappa}{\kappa^2\iota^2}=
+\frac{1}{p}\left\{
+\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}
+\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
+\right\}
+$$
+where $\rho$ is an undetermined scalar.
+
+To determine $\rho$, substitute in the expression for $\beta^2$, and we find
+$$
+\begin{array}{rcl}
+4\beta^2=\frac{4(\iota\kappa)^2}{(\kappa^2\iota^2)^2}
+&=&\left(p\frac{1}{p}\right)^2(\alpha^2\beta^2)+
+\left(p+\frac{1}{p}\right)^2(\beta^2\gamma^2)\\
+&=&\left(p^2+\frac{1}{p^2}\right)(\alpha^2\gamma^2)
+2(\alpha^2+\gamma^2)+4\beta^2
+\end{array}
+$$
+
+Thus the transformation succeeds if
+$$p^2+\frac{1}{p^2}=\frac{2(\alpha^2+\gamma^2)}{\alpha^2\gamma^2}$$
+which gives
+$$p+\frac{1}{p}=\pm 2\sqrt{\frac{\alpha^2}{\alpha^2\gamma^2}}$$
+$$p\frac{1}{p}=\pm 2\sqrt{\frac{\gamma^2}{\alpha^2\gamma^2}}$$
+Hence
+$$
+\frac{4(\kappa^2\iota^2)}{(\kappa^2\iota^2)^2}=
+\left(\frac{1}{p^2}p^2\right)(\alpha^2\gamma^2)=
+\pm 4\sqrt{\alpha^2\gamma^2}
+$$
+$$
+(\kappa^2\iota^2)^{1}=\pm T\alpha T\gamma$$
+
+Again
+$$
+p=\frac{T\alpha +T\gamma}{\sqrt{\gamma^2\alpha^2}},\;\;\;\;
+\frac{1}{p}=\frac{T\alpha T\gamma}{\sqrt{\gamma^2\alpha^2}}
+$$
+and therefore
+$$
+2\iota=\frac{T\alpha +T\gamma}{T\alpha T\gamma}
+\left(
+\sqrt{\frac{\beta^2\alpha^2}{\gamma^2\alpha^2}}U\alpha+
+\sqrt{\frac{\gamma^2\beta^2}{\gamma^2\alpha^2}}U\gamma
+\right)
+$$
+$$
+2\kappa=\frac{T\alpha T\gamma}{T\alpha T\gamma}
+\left(
+\sqrt{\frac{\beta^2\alpha^2}{\gamma^2\alpha^2}}U\alpha
+\sqrt{\frac{\gamma^2\beta^2}{\gamma^2\alpha^2}}U\gamma
+\right)
+$$
+Thus we have proved the possibility of the transformation, and
+determined the transforming vectors $\iota$, $\kappa$.
+
+{\bf 129}. By differentiating the equation
+$$
+(S\alpha\rho)^2+(S\beta\rho)^2+(S\gamma\rho)^2=
+\left(\frac{T(\iota\rho+\rho\kappa)}{(\kappa^2\iota^2)}\right)^2
+$$
+we obtain, as will be seen in Chapter IV, the following,
+$$
+S\alpha\rho S\alpha\rho^{'}+
+S\beta\rho S\beta\rho^{'}+
+S\gamma\rho S\gamma\rho^{'}=
+\frac{S.(\iota\rho+\rho\kappa)(\kappa\rho^{'}+\rho^{'}\iota)}
+{(\kappa^2\iota^2)^2}
+$$
+where $\rho$ also may be any vector whatever.
+
+This is another very important formula of transformation ; and
+it will be a good exercise for the student to prove its truth by
+processes analogous to those in last section. We may merely
+observe, what indeed is obvious, that by putting $\rho^{'}=\rho$ it becomes
+the formula of last section. And we see that we may write, with
+the recent values of $\iota$ and $\kappa$ in terms of
+$\alpha$, $\beta$, $\gamma$, the identity
+$$
+\begin{array}{rcl}
+\alpha S\alpha\rho+\beta S\beta\rho+\gamma S\gamma\rho
+&=&\displaystyle
+\frac{(\iota^2+\kappa^2)\rho+2V.\iota\rho\kappa}{(\kappa^2\iota^2)^2}\\
+&&\\
+&=&\displaystyle
+\frac{(\iota\kappa)^2\rho+2(\iota S\kappa\rho+\kappa S\iota\rho)}
+{(\kappa^2\iota^2)^2}
+\end{array}
+$$
+
+{\bf 130}. In various quaternion investigations, especially in such
+as involve {\sl imaginary} intersections of curves and surfaces, the old
+imaginary of algebra of course appears. But it is to be particularly
+noticed that this expression is analogous to a scalar and not to a
+vector, and that like real scalars it is commutative in
+multiplication with all other factors. Thus it appears, by the same proof
+as in algebra, that any quaternion expression which contains this
+imaginary can always be broken up into the sum of two parts, one
+real, the other multiplied by the first power of $\sqrt{1}$. Such an
+expression, viz.
+$$q=q^{'}+\sqrt{1}q^{''}$$
+where $q^{'}$ and $q^{''}$ are real quaternions, is called by Hamilton a
+BIQUATERNION. [The student should be warned that the term
+Biquaternion has since been employed by other writers in the
+sense sometimes of a ``set'' of 8 elements, analogous to the
+Quaternion 4 ; sometimes for an expression $q^{'} + \theta q^{''}$
+where $\theta$ is not
+the algebraic imaginary. By them Hamilton s Biquaternion is
+called simply a quaternion with nonreal constituents.] Some
+little care is requisite in the management of these expressions, but
+there is no new difficulty. The points to be observed are: first,
+that any biquaternion can be divided into a real and an imaginary
+part, the latter being the product of $\sqrt{1}$ by a real quaternion;
+second, that this $\sqrt{1}$ is commutative with all other quantities in
+multiplication; third, that if two biquaternions be equal, as
+$$q^{'}+\sqrt{1}\;q^{''}=r^{'}+\sqrt{1}\;r^{''}$$
+we have, as in algebra,
+$$q^{'}=r^{'},\;\;\;\;q^{''}=r^{''}$$
+so that an equation between biquaternions involves in general
+{\sl eight} equations between scalars. Compare \S 80.
+
+{\bf 131}. We have obviously, since $\sqrt{1}$ is a scalar,
+$$S(q^{'}+\sqrt{1}\;q^{''})=Sq^{'}+\sqrt{1}\;Sq^{''}$$
+$$V(q^{'}+\sqrt{1}\;q^{''})=Vq^{'}+\sqrt{1}\;Vq^{''}$$
+Hence (\S 103)
+$$\{T(q^{'}+\sqrt{1}\;q^{''})\}^2$$
+$$
+=(Sq^{'}+\sqrt{1}\;Sq^{''}+Vq^{'}+\sqrt{1}\;Vq^{''})
+(Sq^{'}+\sqrt{1}\;Sq^{''}Vq^{'}\sqrt{1}\;Vq^{''})
+$$
+$$=(Sq^{'}+\sqrt{1}\;Sq^{''})^2(Vq^{'}+\sqrt{1}\;Vq^{''})^2$$
+$$=(Tq^{'})^2(Tq^{''})^2+2\sqrt{1}\;S.q^{'}Kq^{''}$$
+
+The only remark which need be made on such formulae is this, that
+{\sl the tensor of a biquaternion may vanish while both of the component
+quaternions are finite}.
+
+Thus, if
+$$Tq^{'}=Tq^{''}$$
+and
+$$S.q^{'}Kq^{''}=0$$
+the above formula gives
+$$T(q^{'}+\sqrt{1}\;q^{''})=0$$
+The condition
+$$S.q^{'}Kq^{''}=0$$
+may be written
+$$
+Kq^{''}=q^{'1}\alpha,\;\;\;\textrm{ or }\;\;\;
+q^{''}=\alpha Kq^{'1}=\frac{\alpha q^{'}}{(Tq^{'})^2}
+$$
+where $\alpha$ is any vector whatever.
+
+Hence
+$$Tq^{'}=Tq^{''}=TKq^{''}=\frac{T\alpha}{Tq^{''}}$$
+and therefore
+$$
+Tq^{'}(Uq^{'}\sqrt{1}\;U\alpha . Uq^{'})=
+(1\sqrt{1}\;U\alpha)q^{'}
+$$
+is the general form of a biquaternion whose tensor is zero.
+
+{\bf 132}. More generally we have, $q$, $r$, $q^{'}$, $r^{'}$
+being any four real and nonevanescent quaternions,
+$$
+(q+\sqrt{1}\;q^{'})(r+\sqrt{1}\;r^{'})=
+qrq^{'}r^{'}+\sqrt{1}\;(qr^{'}+q^{'}r)
+$$
+That this product may vanish we must have
+$$qr=q^{'}r^{'}$$
+and
+$$qr^{'}=q^{'}r$$
+Eliminating $r^{'}$ we have
+$$qq^{'1}qr=q^{'}r$$
+which gives
+$$(q^{'1}q)^2=1$$
+i.e.
+$$q=q^{'}\alpha$$
+where $\alpha$ is some unitvector.
+
+And the two equations now agree in giving
+$$r=\alpha r^{'}$$
+so that we have the biquaternion factors in the form
+$$q^{'}(\alpha +\sqrt{1})\;\;\;\textrm{ and }\;\;\;(\alpha\sqrt{1})r^{'}$$
+and their product is
+$$q^{'}(\alpha +\sqrt{1})(\alpha \sqrt{1})r^{'}$$
+which, of course, vanishes.
+
+[A somewhat simpler investigation of the same proposition
+may be obtained by writing the biquaternions as
+$$
+q^{'}(q^{'1}q+\sqrt{1})\;\;\;\textrm{ and }\;\;\;
+(rr^{'1}+\sqrt{1})r^{'}
+$$
+or
+$$
+q^{'}(q^{''}+\sqrt{1})\;\;\;\textrm{ and }\;\;\;
+(r^{''}+\sqrt{1})r^{'}
+$$
+and showing that
+$$q^{''}=r^{''}=\alpha \;\;\;\textrm{ where }\;T\alpha=1]$$
+
+From this it appears that if the product of two {\sl bivectors}
+$$
+\rho+\sigma\sqrt{1}\;\;\;\textrm{ and }\;\;\;
+\rho^{'}+\sigma^{'}\sqrt{1}
+$$
+is zero, we must have
+$$\sigma^{1}\rho=\rho^{'}\sigma^{'1}=U\alpha$$
+where $\alpha$ may be any vector whatever. But this result is still more
+easily obtained by means of a direct process.
+
+{\bf 133}. It may be well to observe here (as we intend to avail our
+selves of them in the succeeding Chapters) that certain abbreviated
+forms of expression may be used when they are not liable to confuse,
+or lead to error. Thus we may write
+$$T^2q\;\;\;\textrm{for}\;\;\;(Tq)^2$$
+just as we write
+$$\cos^2\theta\;\;\;\textrm{for}\;\;\;(\cos\theta)^2$$
+although the true meanings of these expressions are
+$$T(Tq)\;\;\;\textrm{and}\;\;\;\cos(\cos\theta)$$
+
+The former is justifiable, as $T(Tq) = Tq$, and therefore $T^2q$ is not
+required to signify the second tensor (or tensor of the tensor) of $q$.
+But the trigonometrical usage is defensible only on the score of
+convenience, and is habitually violated by the employment of
+$cos^{1}x$ in its natural and proper sense.
+Similarly we may write
+$$S^2q\;\;\;\textrm{for}\;\;\;(Sq)^2,\;\;\;\textrm{\&c.}$$
+but it may be advisable not to use
+$$Sq^2$$
+as the equivalent of either of those just written; inasmuch as it
+might be confounded with the (generally) different quantity
+$$S.q^2\;\;\;\textrm{or}\;\;\;S(q^2)$$
+although this is rarely written without the point or the brackets.
+
+The question of the use of points or brackets is one on which
+no very definite rules can be laid down. A beginner ought to use
+them freely, and he will soon learn by trial which of them are
+absolutely necessary to prevent ambiguity.
+
+In the present work this course has been adopted: the
+earlier examples in each part of the subject being treated with
+a free use of points and brackets, while in the later examples
+superfluous marks of the kind are gradually got rid of.
+
+It may be well to indicate some general principles which
+regulate the omission of these marks. Thus in $S.\alpha\beta$ or
+$V.\alpha\beta$
+the point is obviously unnecessary: because $S\alpha=0$, and
+$V\alpha=\alpha$
+so that the $S$ would annihilate the term if it applied to $\alpha$ alone,
+while in the same case the $V$ would be superfluous. But in $S.qr$
+and $V.qr$, the point (or an equivalent) is indispensable, for $Sq.r$,
+and $Vq.r$ are usually quite different from the first written
+quantities. In the case of $K$, and of $d$ (used for scalar differentiation),
+the {\sl omission} of the point indicates that the operator acts
+{\sl only} on the nearest factor: thus
+$$Kqr=(Kq)r=Kq.r,\;\;\;dqr=(dq)r=dq.r$$
+Kqr = (Kq) r = Kq . r, dqr = (dq) r=dq.r;
+while, if its action extend farther, we write
+$$K.qr=K(qr),\;\;\;d.qr=d(qr)\;\;\;\textrm{\&c.}$$
+
+In more complex cases we must be ruled by the general
+principle of dropping nothing which is essential. Thus, for
+instance
+$$V(pK(dq)V(Vq.r))$$
+may be written without ambiguity as
+$$V(pK(dq)V(Vq.r))$$
+but nothing more can be dropped without altering its value.
+
+Another peculiarity of notation, which will occasionally be
+required, shows {\sl which portions} of a complex product are affected
+by an operator. Thus we write
+$$\nabla S\sigma\tau$$
+if $\nabla$ operates on $\sigma$ and also on $\tau$, but
+$$\nabla_1S\sigma\tau_1$$
+if it operates on $\tau$ alone. See, in this connection, the last Example
+at the end of Chap. IV. below.
+
+{\bf 134}. The beginner may expect to be at first a little puzzled
+with this aspect of the notation; but, as he learns more of the
+subject, he will soon see clearly the distinction between such an
+expression as
+$$S.V\alpha\beta V\beta\gamma$$
+where we may omit at pleasure either the point or the first V
+without altering the value, and the very different one
+$$S\alpha\beta .V\beta\gamma$$
+which admits of no such changes, without alteration of its value.
+
+All these simplifications of notation are, in fact, merely examples
+of the transformations of quaternion expressions to which part of
+this Chapter has been devoted. Thus, to take a very simple ex
+ample, we easily see that
+$$
+\begin{array}{rcl}
+S.V\alpha\beta V\beta\gamma
+&=&SV\alpha\beta V\beta\gamma
+=S.\alpha\beta V\beta\gamma
+=S\alpha V.\beta V\beta\gamma
+=S\alpha V.(V\beta\gamma)\beta\\
+&=&S\alpha V.(V\gamma\beta)\beta
+=S.\alpha V(\gamma\beta)\beta
+=S.V(\gamma\beta)\beta\alpha
+=SV\gamma\beta V\beta\alpha\\
+&=&S.\gamma\beta V\beta\alpha
+=S.K(\beta\gamma)V\beta\alpha
+=S.\beta\gamma KV\beta\alpha
+=S.\beta\gamma V\beta\alpha\\
+&=&S.V\gamma\beta V\beta\alpha,\textrm{\&c., \&c.}
+\end{array}
+$$
+
+The above group does not nearly exhaust the list of even the simpler
+ways of expressing the given quantity. We recommend it to the
+careful study of the reader. He will find it advisable, at first, to
+use stops and brackets pretty freely; but will gradually learn to
+dispense with those which are not absolutely necessary to prevent
+ambiguity.
+
+There is, however, one additional point of notation to which
+the reader s attention should be most carefully directed. A very
+simple instance will suffice. Take the expressions
+$$
+\frac{\beta}{\gamma}.\frac{\gamma}{\alpha}\;\;\;\;\textrm{and}\;\;\;\;
+\frac{\beta\gamma}{\gamma\alpha}
+$$
+
+The first of these is
+$$\beta\gamma^{1}.\gamma\alpha^{1}=\beta\alpha^{1}$$
+and presents no difficulty. But the second, though at first sight
+it closely resembles the first, is in general totally different in
+value, being in fact equal to
+$$\beta\gamma\alpha^{1}\gamma^{1}$$
+
+For the denominator must be treated as {\sl one quaternion}. If,
+then, we write
+$$\frac{\beta\gamma}{\gamma\alpha}=q$$
+we have
+$$\beta\gamma=q\gamma\alpha$$
+so that, as stated above,
+$$q=\beta\gamma\alpha^{1}\gamma^{1}$$
+We see therefore that
+$$
+\frac{\beta}{\gamma}.\frac{\gamma}{\alpha}=
+\frac{\beta}{\alpha}=
+\frac{\beta\gamma}{\alpha\gamma};\;\;\;\textrm{but {\sl not}}\;\;=
+\frac{\beta\gamma}{\gamma\alpha}
+$$
+
+\section{Examples to Chapter 3}
+
+{\bf 1}. Investigate, by quaternions, the requisite formulae for
+changing from any one set of coordinate axes to another ; and
+derive from your general result, and also from special investiga
+tions, the usual expressions for the following cases:
+
+\begin{itemize}
+\item[(a)] Rectangular axes turned about z through any angle.
+
+\item[(b)] Rectangular axes turned into any new position by rota
+tion about a line equally inclined to the three.
+
+\item[(c)] Rectangular turned to oblique, one of the new axes
+lying in each of the former coordinate planes.
+\end{itemize}
+
+{\bf 2}. Point out the distinction between
+$$
+\left(\frac{\alpha+\beta}{\alpha}\right)^2\;\;\;\textrm{and}\;\;\;
+\frac{(\alpha+\beta)^2}{\alpha^2}
+$$
+and find the value of their difference.
+
+If
+$$
+T\beta/\alpha=1\;\;\;\textrm{and}\;\;\;
+U\frac{\alpha+\beta}{\alpha}=\left(\frac{\beta}{\alpha}\right)^{\frac{1}{2}}
+$$
+
+Show also that
+$$
+\frac{\alpha+\beta}{\alpha\beta}=
+\frac{V\alpha\beta}{1+S\alpha\beta^{'}}
+$$
+and
+$$
+\frac{\alpha\beta}{\alpha+\beta}=
+\frac{V\alpha\beta}{1S\alpha\beta^{'}}
+$$
+provided $\alpha$ and $\beta$ be unitvectors. If these conditions are not
+fulfilled, what are the true values ?
+
+{\bf 3}. Show that, whatever quaternion $r$ may be, the expression
+$$\alpha r+r\beta$$
+in which $\alpha$ and $\beta$ are any two unit vectors, is reducible to the
+form
+$$l(\alpha+\beta)+m(\alpha\beta1)$$
+where $l$ and $m$ are scalars.
+
+{\bf 4}. If $Tp=T\alpha=T\beta=1$, and $S.\alpha\beta\rho=0$
+show by direct transformations that
+$$S.U(\rho\alpha)U(\rho\beta)=\pm\sqrt{\frac{1}{2}(1S\alpha\beta)}$$
+Interpret this theorem geometrically.
+
+{\bf 5}. If $S\alpha\beta=0$, $T\alpha=T\beta=1$, show that
+$$
+(1+\alpha^{m})\beta=
+2\cos\frac{m\pi}{4}\alpha^{\frac{m}{2}}\beta=
+2S\alpha^{\frac{m}{2}}.\alpha^{\frac{m}{2}}\beta
+$$
+
+{\bf 6}. Put in its simplest form the equation
+$$
+\rho S.V\alpha\beta V\beta\gamma V\gamma\alpha=
+aV.V\gamma\alpha V\alpha\beta+
+bV.V\alpha\beta V\beta\gamma+
+cV.V\beta\gamma V\gamma\alpha
+$$
+and show that
+$$a=S.\beta\gamma\rho,\;\;\;\textrm{\&c.}$$
+
+{\bf 7}. Show that any quaternion may in general, in one way only,
+be expressed as a homogeneous linear function of four given
+quaternions. Point out the nature of the exceptional cases. Also
+find the simplest form in which any quaternion may generally be
+expressed in terms of two given quaternions.
+
+{\bf 8}. Prove the following theorems, and exhibit them as properties
+of determinants :
+
+\begin{itemize}
+\item[(a)] $S.(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=
+2S.\alpha\beta\gamma$
+\item[(b)]$S.V\alpha\beta V\beta\gamma V\gamma\alpha=
+(S.\alpha\beta\gamma)^2$
+\item[(c)]$S.V(\alpha+\beta)(\beta+\gamma)V(\beta+\gamma)(\gamma+\alpha)
+V(\gamma+\alpha)(\alpha+\beta)=4(S.\alpha\beta\gamma)^2$
+\item[(d)]$S.V(V\alpha\beta V\beta\gamma)V(V\beta\gamma V\gamma\alpha)
+V(V\gamma\alpha V\alpha\beta)=(S.\alpha\beta\gamma)^4$
+\item[(e)]$S.\delta\epsilon\zeta=16(S.\alpha\beta\gamma)^4$\\
+where
+$$\delta=V(V(\alpha+\beta)(\beta+\gamma)V(\beta+\gamma)(\gamma+\alpha))$$
+$$\epsilon=V(V(\beta+\gamma)(\gamma+\alpha)V(\gamma+\alpha)(\alpha+\beta))$$
+$$\zeta=V(V(\gamma+\alpha)(\alpha+\beta)V(\alpha+\beta)(\beta+\gamma))$$
+\end{itemize}
+
+{\bf 9}. Prove the common formula for the product of two determinants
+of the third order in the form
+$$
+S.\alpha\beta\gamma S.\alpha_1\beta_1\gamma_1=
+\left
+\begin{array}{ccc}
+S\alpha\alpha_1 & S\beta\alpha_1 & S\gamma\alpha_1\\
+S\alpha\beta_1 & S\beta\beta_1 & S\gamma\beta_1\\
+S\alpha\gamma_1 & S\beta\gamma_1 & S\gamma\gamma_1
+\end{array}
+\right
+$$
+
+{\bf 10}. Show that, whatever be the eight vectors involved,
+$$
+\left
+\begin{array}{cccc}
+S\alpha\alpha_1 & S\alpha\beta_1 & S\alpha\gamma_1 & S\alpha\delta_1\\
+S\beta\alpha_1 & S\beta\beta_1 & S\beta\gamma_1 & S\beta\delta_1\\
+S\gamma\alpha_1 & S\gamma\beta_1 & S\gamma\gamma_1 & S\gamma\delta_1\\
+S\delta\alpha_1 & S\delta\beta_1 & S\delta\gamma_1 & S\delta\delta_1
+\end{array}
+\right
+=S.\alpha\beta\gamma S. \beta_1\gamma_1\delta_1S\alpha_1(\delta\delta)=0
+$$
+
+If the single term $S\alpha\alpha_1$, be changed to $S\alpha_0\alpha_1$,
+the value of the determinant is
+$$S.\beta\gamma\delta S.\beta_1\gamma_1\delta_1 S\alpha_1(\alpha_0\alpha)$$
+
+State these as propositions in spherical trigonometry.
+
+Form the corresponding null determinant for any two groups
+of five quaternions : and give its geometrical interpretation.
+
+{\bf 11}. If, in \S 102, $\alpha$, $\beta$, $\gamma$ be three mutually
+perpendicular vectors, can anything be predicated as to $\alpha_1$,
+$\beta_1$, $\gamma_1$? If $\alpha$, $\beta$, $\gamma$ be rectangular
+unitvectors, what of $\alpha_1$, $\beta_1$, $\gamma_1$?
+
+{\bf 12}. If $\alpha$, $\beta$, $\gamma$, $\alpha^{'}$, $\beta^{'}$,
+$\gamma^{'}$ be two sets of rectangular unitvectors, show that
+$$
+S\alpha\alpha^{'}=
+S\gamma\beta^{'}S\beta\gamma^{'}=
+S\beta\beta^{'}S\gamma\gamma^{'}\;\;\;\textrm{\&c. \&c.}
+$$
+
+{\bf 13}. The lines bisecting pairs of opposite sides of a quadrilateral
+(plane or gauche) are perpendicular to each other when the
+diagonals of the quadrilateral are equal.
+
+{\bf 14}. Show that
+\begin{itemize}
+\item [(a)]$S.q^2=2S^2qT^2q$
+\item [(b)]$S.q^3=S^3q3SqT^2Vq$
+\item [(c)]$\alpha^2\beta^2\gamma^2+S^2.\alpha\beta\gamma=
+V^2.\alpha\beta\gamma$
+\item [(d)]$S(V.\alpha\beta\gamma V.\beta\gamma\alpha V.\gamma\alpha\beta)=
+4S\alpha\beta S\beta\gamma S\gamma\alpha S.\alpha\beta\gamma$
+\item [(e)]$V.q^3=(2S^2qT^2Vq)Vq$
+\item [(f)]$qUVq^{1}=Sq.UVq+TVq$
+\end{itemize}
+
+and interpret each as a formula in plane or spherical trigonometry.
+
+{\bf 15}. If $q$ be an undetermined quaternion, what loci are represented by
+\begin{itemize}
+\item[(a)]$(q\alpha^{1})^2=a^2$
+\item[(b)]$(q\alpha^{1})^4=a^4$
+\item[(c)]$S.(q\alpha)^2=a^2$
+\end{itemize}
+where $a$ is any given scalar and $\alpha$ any given vector ?
+
+{\bf 16}. If $q$ be any quaternion, show that the equation
+$$Q^2=q^2$$
+is satisfied, not alone by $Q = \pm q$, but also by
+$$Q=\pm \sqrt{1}(Sq.UVqTVq)$$
+
+\begin{flushright}
+(Hamilton, {\sl Lectures}, p. 673.)
+\end{flushright}
+
+{\bf 17}. Wherein consists the difference between the two equations
+$$
+T^2\frac{\rho}{\alpha}=1\;\;\;\textrm{and}\;\;\;
+\left(\frac{\rho}{\alpha}\right)^2=1
+$$
+
+What is the full interpretation of each, $\alpha$ being a given, and p an
+undetermined, vector?
+
+{\bf 18}. Find the {\sl full} consequences of each of the following
+groups of equations, as regards both the unknown vector $\rho$ and
+the given vectors $\alpha$, $\beta$, $\gamma$:
+$$
+\begin{array}{crcrcr}
+ & S.\alpha\beta\rho=0 & & S\alpha\rho=0 & & S\alpha\rho=0\\
+(a) & & (b) & S.\alpha\beta\rho=0 & (c) & S.\alpha\beta\rho=0\\
+ & S.\beta\gamma\rho=0 & & S\beta\rho=0 & & S.\alpha\beta\gamma\rho=0
+\end{array}
+$$
+
+{\bf 19}. From \S\S 74, 110, show that, if $\epsilon$
+be any unitvector, and $m$ any scalar,
+$$\epsilon^{m}=\cos\frac{m\pi}{2}+\epsilon\sin\frac{m\pi}{2}$$
+Hence show that if $\alpha$, $\beta$, $\gamma$
+be radii drawn to the corners of a triangle on the unitsphere,
+whose spherical excess is $m$ right angles,
+$$
+\frac{\alpha+\beta}{\beta+\gamma}.
+\frac{\gamma+\alpha}{\alpha+\beta}.
+\frac{\beta+\gamma}{\gamma+\alpha}=
+\alpha^m
+$$
+Also that, if $A$, $B$, $C$ be the angles of the triangle, we have
+$$
+\gamma^{\frac{2C}{\pi}}
+\beta^{\frac{2B}{\pi}}
+\alpha^{\frac{2A}{\pi}}
+=1
+$$
+
+{\bf 20}. Show that for any three vectors $\alpha$, $\beta$, $\gamma$ we have
+$$
+(U\alpha\beta)^2+(U\beta\gamma)^2+(U\alpha\gamma)^2+(U.\alpha\beta\gamma)^2+
+4U\alpha\gamma .SU\alpha\beta SU\beta\gamma=2
+$$
+
+\begin{flushright}
+(Hamilton, {\sl Elements}, p. 388.)
+\end{flushright}
+
+{\bf 21}. If $a_1$, $a_2$, $a_3$, $x$ be any four scalars,
+and $\rho_1$, $\rho_2$, $\rho_3$ any three vectors, show that
+$$
+(S.\rho_1\rho_2\rho_3)^2+
+(\sum.a_1V\rho_2\rho_3)^2+
+x^2(\sum V\rho_1\rho_2)^2
+$$
+$$
+x^2(\sum.a_1(\rho_2\rho_3))^2
++2\prod(x^2+S\rho_1\rho_2+a_1a_2)
+$$
+$$
+=2\prod(x^2+\rho^2)+
+2\prod a^2+
+$$
+$$
+\sum\{(x^2+a_1^2+\rho_1^2)((V\rho_2\rho_3)^2+
+2a_2a_3(x^2+S\rho_2\rho_3)x^2(\rho_2\rho_3)^2)\}
+$$
+where $\displaystyle \prod a^2=a_1^2a_2^2a_3^2$
+
+Verify this formula by a simple process in the particular case
+$$a_1=a_2=a_3=x=0$$
+
+\begin{flushright}
+({\sl Ibid})
+\end{flushright}
+
+{\bf 22}. Eliminate $p$ from the equations
+$$V.\beta\rho\alpha\rho=0,\;\;\;S\gamma\rho=0$$
+and state the problem and its solution in a geometrical form.
+
+{\bf 23}. If $p$, $q$, $r$, $s$ be four versors, such that
+$$qp=sr=\alpha$$
+$$rq=ps=\beta$$
+where $\alpha$ and $\beta$ are unitvectors; show that
+$$S(V.VsVqV.VrVp)=0$$
+Interpret this as a property of a spherical quadrilateral.
+
+{\bf 24}. Show that, if $pq$, $rs$, $pr$, and $qs$ be vectors, we have
+$$S(V.VpVsV.VqVr)=0$$
+
+{\bf 25}. If $\alpha$, $\beta$, $\gamma$ be unitvectors,
+$$
+V\beta\gamma S.\alpha\beta\gamma=
+\alpha(1S^2\beta\gamma)
+\beta(S\alpha\gamma S\beta r + S\alpha\beta)
+\gamma(S\alpha\beta S\beta\gamma+S\alpha\gamma)
+$$
+
+{\bf 26}. If $i$, $j$, $k$, $i^{'}$, $j^{'}$, $k^{'}$,
+be two sets of rectangular unitvectors, show that
+$$
+\begin{array}{rcl}
+S.Vii^{'}Vjj^{'}Vkk^{'}&=&(Sij^{'})^2(Sji^{'})^2\\
+ &=&(Sjk^{'})^2(Skj^{'})^2=\textrm{\&c.}
+\end{array}
+$$
+and find the values of the vector of the same product.
+
+{\bf 27}. If $\alpha$, $\beta$, $\gamma$
+be a rectangular unitvector system, show that,
+whatever be $\lambda$, $\mu$, $\nu$
+$$\lambda S^2i\alpha +\mu S^2j\gamma +\nu S^2k\beta$$
+$$\lambda S^2k\gamma +\mu S^2i\beta +\nu S^2j\alpha$$
+and
+$$\lambda S^2j\beta +\mu S^2k\alpha +\nu S^2i\gamma$$
+are coplanar vectors. What is the connection between this and
+the result of the preceding example ?
+
+\vfill
+\newpage
+\section{Axiom Examples}
+The basic operation for creating quaternions is {\bf quatern}.
+This is a quaternion over the rational numbers.
+\spadcommand{q:=quatern(2/11,8,3/4,1)}
+$$
+{2 \over {11}} {8 \ i}+{{3 \over 4} \ j}+k
+$$
+\returnType{Type: Quaternion Fraction Integer}
+
+This is a quaternion over the integers.
+\spadcommand{r:=quatern(1,2,3,4)}
+$$
+1+{2 \ i}+{3 \ j}+{4 \ k}
+$$
+\returnType{Type: Quaternion Integer}
+
+We can also construct quaternions with complex components.
+First we construct a complex number.
+\spadcommand{b:=complex(3,4)}
+$$
+3+{4 \ i}
+$$
+\returnType{Type: Complex Integer}
+and then we use it as a component in a quaternion.
+\spadcommand{s:=quatern(3,1/7,b,2)}
+$$
+3+{{1 \over 7} \ i}+{{\left( 3+{4 \ i} \right)}\ j}+{2 \ k}
+$$
+\returnType{Type: Quaternion Complex Fraction Integer}
+Notice that the $i$ component of the complex number has no
+relation to the $i$ component of the quaternion even though
+they use the same symbol by convention.
+
+The four parts of a quaternion are the real part, the $i$ imaginary
+part, the $j$ imaginary part, and the $k$ imaginary part. The
+{\bf real} function returns the real part.
+\spadcommand{real q}
+$$
+2 \over {11}
+$$
+\returnType{Type: Fraction Integer}
+
+The {\bf imagI} function returns the $i$ imaginary part.
+\spadcommand{imagI q}
+$$
+8
+$$
+\returnType{Type: Fraction Integer}
+
+The {\bf imagJ} function returns the $j$ imaginary part.
+\spadcommand{imagJ q}
+$$
+3 \over 4
+$$
+\returnType{Type: Fraction Integer}
+
+The {\bf imagK} function returns the $k$ imaginary part.
+\spadcommand{imagK q}
+$$
+1
+$$
+\returnType{Type: Fraction Integer}
+
+Quaternions satisfy a very fundamental relationship between the parts,
+namely that
+$$i^2 = j^2 = k^2 = ijk = 1$$. This is similar to the requirement
+in complex numbers of the form $a+bi$ that $i^2 = 1$.
+
+The set of quaternions is denoted by $\mathbb{H}$, whereas the integers
+are denoted by $\mathbb{Z}$ and the complex numbers by $\mathbb{C}$.
+
+Quaternions are not commutative which means that in general
+$$AB \ne BA$$
+for any two quaternions, A and B. So, for instance,
+\spadcommand{q*r}
+$$
+{{437} \over {44}} {{{84} \over {11}} \ i}+{{{1553} \over {44}} \ j}
+{{{523} \over {22}} \ k}
+$$
+\returnType{Type: Quaternion Fraction Integer}
+\spadcommand{r*q}
+$$
+{{437} \over {44}} {{{84} \over {11}} \ i} {{{1439} \over {44}} \
+j}+{{{599} \over {22}} \ k}
+$$
+\returnType{Type: Quaternion Fraction Integer}
+and these are clearly not equal.
+
+Complex $2\times2$ matrices form an alternate, equivalent
+representation of quaternions. These matrices have the form:
+$$
+\left[
+\begin{array}{cc}
+u & v \\
+\overline{v} & \overline{u}
+\end{array}
+\right]
+$$
+=
+$$
+\left[
+\begin{array}{cc}
+a+bi & c+di \\
+c+di & abi
+\end{array}
+\right]
+$$
+where $u$ and $v$ are complex, $\overline{u}$ is complex conjugate
+of $u$, $\overline{z}$ is the complex conjugate of $z$, and a,b,c,
+and d are real.
+
+Within the quaternion each component operator represents a basis
+element in $\mathbb{R}^4$ thus:
+$$
+1 =
+\left[
+\begin{array}{cccc}
+1 & 0 & 0 & 0\\
+0 & 1 & 0 & 1\\
+0 & 0 & 1 & 0\\
+0 & 0 & 0 & 1
+\end{array}
+\right]
+$$
+
+$$
+i =
+\left[
+\begin{array}{cccc}
+0 & 1 & 0 & 0\\
+1 & 0 & 0 & 1\\
+0 & 0 & 0 & 1\\
+0 & 0 & 1 & 0
+\end{array}
+\right]
+$$
+
+
\chapter{Groebner Basis}
Groebner Basis
\chapter{Greatest Common Divisor}
@@ 2514,6 +7682,21 @@ Literate Programming''\\
\bibitem{31} Daly, Timothy, "The Axiom Literate Documentation"\\
{\bf http://axiom.axiomdeveloper.org/axiomwebsite/documentation.html}
\bibitem{32} {\bf http://www.puffinwarellc.com/p3a.htm}
+\bibitem{33} Tait, P.G.,
+{\it An Elementary Treatise on Quaternions} \\
+C.J. Clay and Sons, Cambridge University Press Warehouse,
+Ave Maria Lane 1890
+\bibitem{34} Knuth, Donald, {\it The \TeX{}book} \\
+Reading, Massachusetts,
+AddisonWesley Publishing Company, Inc.,
+1984. ISBN 0201134489
+\bibitem{35} Hathway, Arthur S., "A Primer Of Quaternions" (1896)
+\bibitem{36} Conway, John H. and Smith, Derek, A.,
+"On Quaternions and Octonions", A.K Peters, Natick, MA. (2003)
+ISBN 1568811349
+\bibitem{37} http://mathworld.wolfram.com/Quaternion.html
\end{thebibliography}
+\end{document}
+
\printindex
\end{document}
diff git a/books/bookvol10.4.pamphlet b/books/bookvol10.4.pamphlet
index 5f0db1e..d7eaf5c 100644
 a/books/bookvol10.4.pamphlet
+++ b/books/bookvol10.4.pamphlet
@@ 17,8 +17,8 @@
%%
%% pagepic adds an image and an index entry
%%
\newcommand{\pagepic}[2]{% e.g. \pagepic{pathandfile}{abb}
\includegraphics{#1}%
+\newcommand{\pagepic}[3]{% e.g. \pagepic{pathandfile}{abb}{scale}
+\includegraphics[scale=#3]{#1}\\%
\index{images!#2}}
%%
%% pageto is a forward link to a referenced page
@@ 128520,6 +128520,145 @@ QuotientFieldCategoryFunctions2(A, B, R, S): Exports == Impl where
@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\section{package QUATCT2 QuaternionCategoryFunctions2}
+<>=
+)set break resume
+)spool QuaternionCategoryFunctions2.output
+)set message test on
+)set message auto off
+)clear all
+
+(1) > )read af
+S 1 of 4
+q := quatern(2/11,8,3/4,1)
+R
+R
+R 2 3
+R (1)   8i +  j + k
+R 11 4
+R Type: Quaternion Fraction Integer
+E 1
+
+S 2 of 4
+f(a:Fraction Integer):Complex Fraction Integer == a::Complex Fraction Integer
+R
+R Function declaration f : Fraction Integer > Complex Fraction
+R Integer has been added to workspace.
+R Type: Void
+E 3
+
+S 3 of 4
+map(f,q)
+R
+R Compiling function f with type Fraction Integer > Complex Fraction
+R Integer
+R
+R 2 3
+R (3)   8i +  j + k
+R 11 4
+R Type: Quaternion Complex Fraction Integer
+E 3
+
+S 4 of 4
+)show QuaternionCategoryFunctions2
+R
+R QuaternionCategoryFunctions2(QR: QuaternionCategory R,R: CommutativeRing,QS: QuaternionCategory S,S: CommutativeRing) is a package constructor
+R Abbreviation for QuaternionCategoryFunctions2 is QUATCT2
+R This constructor is exposed in this frame.
+R Issue )edit quat.spad.pamphlet to see algebra source code for QUATCT2
+R
+R Operations 
+R map : ((R > S),QR) > QS
+R
+E 4
+
+)spool
+)lisp (bye)
+@
+<>=
+====================================================================
+QuaternionCategoryFunctions2 examples
+====================================================================
+
+The package QuaternionCategoryFunctions2 provides a function map
+to convert an expression in Quaternion(R) to an expression in
+Quaternion(S) using the function f.
+
+q := quatern(2/11,8,3/4,1)
+
+ 2 3
+   8i +  j + k
+ 11 4
+ Type: Quaternion Fraction Integer
+
+f(a:Fraction Integer):Complex Fraction Integer == a::Complex Fraction Integer
+ Function declaration f : Fraction Integer > Complex Fraction
+ Integer has been added to workspace.
+ Type: Void
+
+map(f,q)
+ Compiling function f with type Fraction Integer > Complex Fraction
+ Integer
+
+ 2 3
+   8i +  j + k
+ 11 4
+ Type: Quaternion Complex Fraction Integer
+
+
+See Also:
+o )show QuaternionCategoryFunctions2
+
+@
+\pagehead{QuaternionCategoryFunctions2}{QUATCT2}
+\pagepic{ps/v104quaternioncategoryfunctions2.ps}{QUATCT2}{1.00}
+
+{\bf Exports:}\\
+\cross{QUATCT2}{map}
+
+<>=
+)abbrev package QUATCT2 QuaternionCategoryFunctions2
+++ Author: Robert S. Sutor
+++ Date Created: 23 May 1990
+++ Change History:
+++ 23 May 1990
+++ Basic Operations: map
+++ Related Constructors: QuaternionCategory, Quaternion
+++ Also See:
+++ AMS Classifications: 11R52
+++ Keywords: quaternions, division ring, map
+++ Description:
+++ \spadtype{QuaternionCategoryFunctions2} implements functions between
+++ two quaternion domains. The function \spadfun{map} is used by
+++ the system interpreter to coerce between quaternion types.
+
+QuaternionCategoryFunctions2(QR,R,QS,S) : Exports ==
+ Implementation where
+ R : CommutativeRing
+ S : CommutativeRing
+ QR : QuaternionCategory R
+ QS : QuaternionCategory S
+ Exports == with
+ map: (R > S, QR) > QS
+ ++ map(f,u) maps f onto the component parts of the quaternion u.
+ ++ to convert an expression in Quaterion(R) to Quaternion(S)
+ ++
+ ++X f(a:FRAC(INT)):COMPLEX(FRAC(INT)) == a::COMPLEX(FRAC(INT))
+ ++X q:=quatern(2/11,8,3/4,1)
+ ++X map(f,q)
+
+ Implementation == add
+ map(fn : R > S, u : QR): QS ==
+ quatern(fn real u, fn imagI u, fn imagJ u, fn imagK u)$QS
+
+@
+<>=
+"QUATCT2" [color="#FF4488",href="bookvol10.4.pdf#nameddest=QUATCT2"]
+"ACFS" [color="#4488FF",href="bookvol10.2.pdf#nameddest=QUATCAT"]
+"QUATCT2" > "QUATCAT"
+
+@
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Chapter R}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{package REP RadicalEigenPackage}
diff git a/books/ps/v104quaternioncategoryfunctions2.ps b/books/ps/v104quaternioncategoryfunctions2.ps
new file mode 100644
index 0000000..cf2f1d8
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+%%EOF
diff git a/changelog b/changelog
index 4db9a21..df84830 100644
 a/changelog
+++ b/changelog
@@ 1,3 +1,9 @@
+20090709 tpd src/axiomwebsite/patches.html 20090709.01.tpd.patch
+20090709 tpd src/algebra/Makefile move QUATCT2 to bookvol10.4
+20090709 tpd src/algebra/quat.spad deleted
+20090709 tpd books/ps/v104quaternioncategoryfunctions2.ps added
+20090709 tpd books/bookvol10.4 add QUATCT2, help, regress, example
+20090709 tpd books/bookvol10.1 add chapter on quaternions
20090706 tpd src/axiomwebsite/patches.html 20090706.02.tpd.patch
20090706 tpd src/axiomwebsite/patches.html fixup patch order
20090706 tpd src/axiomwebsite/patches.html 20090706.01.tpd.patch
diff git a/src/algebra/Makefile.pamphlet b/src/algebra/Makefile.pamphlet
index f236d7a..ec15b71 100644
 a/src/algebra/Makefile.pamphlet
+++ b/src/algebra/Makefile.pamphlet
@@ 15790,16 +15790,6 @@ it to interpret special characters, in our case newlines.
SHELL=bash
@
\subsection{The SPADFILES list}
Note that we have excluded {\bf mlift.spad.jhd} from this list.
We need to figure out which mlift.spad to keep.

<>=

SPADFILES= \
 ${OUTSRC}/quat.spad

@
\subsection{The ALDORFILES list}
<>=
@@ 15830,8 +15820,7 @@ DOCFILES= \
${DOC}/ndftip.as.dvi \
${DOC}/nepip.as.dvi \
${DOC}/noptip.as.dvi ${DOC}/nqip.as.dvi \
 ${DOC}/nrc.as.dvi ${DOC}/nsfip.as.dvi \
 ${DOC}/quat.spad.dvi
+ ${DOC}/nrc.as.dvi ${DOC}/nsfip.as.dvi
@
\section{The Makefile Stanzas}
@@ 16507,6 +16496,7 @@ SPADHELP=\
${HELP}/PartialFraction.help \
${HELP}/Permanent.help ${HELP}/Permutation.help \
${HELP}/Polynomial.help ${HELP}/Quaternion.help \
+ ${HELP}/QuaternionCategoryFunctions2.help \
${HELP}/Queue.help \
${HELP}/RadixExpansion.help ${HELP}/RealClosure.help \
${HELP}/RealSolvePackage.help ${HELP}/RegularTriangularSet.help \
@@ 16589,6 +16579,7 @@ REGRESS= Any.regress \
Permanent.regress Permutation.regress \
PlaneAlgebraicCurvePlot.regress Plot.regress \
Polynomial.regress Quaternion.regress \
+ QuaternionCategoryFunctions2.regress \
Queue.regress \
RadixExpansion.regress RealClosure.regress \
RealSolvePackage.regress RegularTriangularSet.regress \
@@ 17612,6 +17603,18 @@ ${HELP}/Quaternion.help: ${BOOKS}/bookvol10.3.pamphlet
>${INPUT}/Quaternion.input
@echo "Quaternion (QUAT)" >>${HELPFILE}
+${HELP}/QuaternionCategoryFunctions2.help: ${BOOKS}/bookvol10.4.pamphlet
+ @echo 7945 create QuaternionCategoryFunctions2.help from \
+ ${BOOKS}/bookvol10.4.pamphlet
+ @${TANGLE} R"QuaternionCategoryFunctions2.help" \
+ ${BOOKS}/bookvol10.4.pamphlet \
+ >${HELP}/QuaternionCategoryFunctions2.help
+ @cp ${HELP}/QuaternionCategoryFunctions2.help ${HELP}/QUATCT2.help
+ @${TANGLE} R"QuaternionCategoryFunctions2.input" \
+ ${BOOKS}/bookvol10.4.pamphlet \
+ >${INPUT}/QuaternionCategoryFunctions2.input
+ @echo "QuaternionCategoryFunctions2 (QUATCT2)" >>${HELPFILE}
+
${HELP}/Queue.help: ${BOOKS}/bookvol10.3.pamphlet
@echo 7950 create Queue.help from ${BOOKS}/bookvol10.3.pamphlet
@${TANGLE} R"Queue.help" ${BOOKS}/bookvol10.3.pamphlet \
@@ 18021,7 +18024,7 @@ everything: check lib db cmd gloss
@ echo OBJ= ${OBJ} MNT= ${MNT} O=${O} LISP=${LISP} BYE=${BYE}
#src: ${AS}
src: ${SPADFILES} ${ORDER}
+src: ${ORDER}
@ echo 4304 Building nrlibS from spad sources
# @ (cd ${MID} ; \
diff git a/src/algebra/quat.spad.pamphlet b/src/algebra/quat.spad.pamphlet
deleted file mode 100644
index 9703b0e..0000000
 a/src/algebra/quat.spad.pamphlet
+++ /dev/null
@@ 1,5294 +0,0 @@
\documentclass{book}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{axiom}
%%% We add some special commands to handle Axiom portions of the document.
\makeindex

%% We need to use a dagger to mark footnotes because numbers can be
%% mistaken as superscripts
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
%% Before each footnote we explicitly set the counter to 1 so that
%% when it increments it prints a dagger
\setcounter{footnote}{1}

\begin{document}
\begin{center}
\bigskip \huge
AXIOM QUATERNIONS

\bigskip\bigskip
\footnotesize BY

\bigskip
\large Peter Guthrie Tait, Robert S. Sutor, Timothy Daly

\end{center}
\section*{Preface}
\addcontentsline{toc}{section}{Preface}
The Theory of Quaternions is due to Sir William Rowan Hamilton,
Royal Astronomer of Ireland, who presented his first paper on the
subject to the Royal Irish Academy in 1843. His Lectures on
Quaternions were published in 1853, and his Elements, in 1866,
shortly after his death. The Elements of Quaternions by Tait\cite{7} is
the accepted textbook for advanced students.

Large portions of this file are derived from a public domain version
of Tait's book combined with the algebra available in Axiom.
The purpose is to develop a tutorial introduction to the Axiom
domain and its uses.
\vfill
\newpage
\tableofcontents
\vfill
\newpage


\chapter{Quaternions}

\section{Vectors, and their Composition}

{\bf 1}. For at least two centuries the geometrical representation
of the negative and imaginary algebraic quantities, $1$ and $\sqrt{1}$
has been a favourite subject of speculation with mathematicians.
The essence of almost all of the proposed processes consists in
employing such expressions to indicate the DIRECTION, not the
{\sl length}, of lines.

{\bf 2}. Thus it was long ago seen that if positive quantities were
measured off in one direction along a fixed line, a useful and lawful
convention enabled us to express negative quantities of the same
kind by simply laying them off on the same line in the opposite
direction. This convention is an essential part of the Cartesian
method, and is constantly employed in Analytical Geometry and
Applied Mathematics.

{\bf 3}. Wallis, towards the end of the seventeenth century, proposed
to represent the impossible roots of a quadratic equation by going
{\sl out} of the line on which, if real, they would have been laid off.
This construction is equivalent to the consideration of $\sqrt{1}$ as a
directed unitline perpendicular to that on which real quantities
are measured.

{\bf 4}. In the usual notation of Analytical Geometry of two
dimensions, when rectangular axes are employed, this amounts
to reckoning each unit of length along $Oy$ as $+\sqrt{1}$, and on
$Oy^{'}$ as $\sqrt{1}$ ; while on $Ox$ each unit is $+1$, and on
$Ox$ it is $1$.

If we look at these four lines in circular order, i.e. in the order of
positive rotation (that of the northern hemisphere of the earth
about its axis, or {\sl opposite} to that of the hands of a watch), they
give
$$ 1, \sqrt{1}, 1, \sqrt{1}$$

\boxed{4.6in}{
\vskip 0.1cm
In Axiom the same elements would be written as complex numbers
which are constructed using the function {\bf complex}:
\spadcommand{complex(1,0)}
$$1$$
\returnType{Type: Complex Integer}
}
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{complex(0,1)}
$$\%i$$
\returnType{Type: Complex Integer}
}
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{complex(1,0)}
$$1$$
\returnType{Type: Complex Integer}
}
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{complex(0,1)}
$$i$$
\returnType{Type: Complex Integer}
}
\boxed{4.6in}{
\vskip 0.1cm
Note that \%i is of type Complex(Integer), that is, the imaginary
part of a complex number. The apparently equivalent expression
\spadcommand{sqrt(1)}
$$\sqrt{1}$$
\returnType{Type: AlgebraicNumber}
has the type AlgebraicNumber which means that it is the root of
a polynomial with rational coefficients.\\
}

In this series each expression is derived from that which precedes
it by multiplication by the factor $\sqrt{1}$. Hence we may consider
$\sqrt{1}$ as an operator, analogous to a handle perpendicular to the
plane of $xy$, whose effect on any line in that plane is to make it
rotate (positively) about the origin through an angle of $90^{\circ}$.

\boxed{4.6in}{
\vskip 0.1cm
In Axiom
\spadcommand{\%i*\%i}
$$1$$
\returnType{Type: Complex Integer}
}

{\bf 5}. In such a system, (which seems to have been first developed,
in 1805, by Bu\'ee) a point in the plane of reference is defined by a
single imaginary expression. Thus $a + b\sqrt{1}$ may be considered
as a single quantity, denoting the point, $P$, whose coordinates are
$a$ and $b$. Or, it may be used as an expression for the line $OP$
joining that point with the origin. In the latter sense, the expression
$a + b\sqrt{1}$ implicitly contains the {\sl direction}, as well as the
{\sl length}, of this line ; since, as we see at once, the direction is
inclined at an angle $\tan^{1}(b/a)$ to the axis of $x$,
and the length is $\sqrt{a^2+b^2}$. Thus, say we have
$$OP = a + b\sqrt{1}$$
the line $OP$ considered as that by which we pass from one
extremity, $O$, to the other, $P$. In this sense it is called a VECTOR.
Considering, in the plane, any other vector,
$$OQ = a^{'}+b^{'}\sqrt{1}$$

\boxed{4.6in}{
\vskip 0.1cm
In order to created superscripted variables we use the superscript
function from the SYMBOL domain. So we can create $a^{'}$ as ``ap''
(that is, ``aprime'') and $b^{'}$ as ``bp'' (``bprime'') thus
(also note that the underscore character is Axiom's escape character
which removes any special meaning of the next character, in this case,
the quote character):
\spadcommand{ap:=superscript(a,[\_'])}
$$a^{'}$$
\returnType{Type: Symbol}
}
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{bp:=superscript(b,[\_'])}
$$b^{'}$$
\returnType{Type: Symbol}
}
\boxed{4.6in}{
\vskip 0.1cm
at this point we can type
\spadcommand{ap+bp*\%i}
$$a^{'}+b^{'}\ \%i$$
\returnType{Type: Complex Polynomial Integer}
}

the addition of these two lines obviously gives
$$OR = a + a^{'} + (b + b^{'})\sqrt{1}$$

\boxed{4.6in}{
\vskip 0.1cm
In Axiom the computation looks like:
\spadcommand{op:=complex(a,b)}
$$a + b\ \%i$$
\returnType{Type: Complex Polynomial Integer}
}
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{oq:=complex(ap,bp)}
$$a^{'} + b^{'}\ \%i$$
\returnType{Type: Complex Polynomial Integer}
}
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{op + oq}
$$a + a^{'} + (b + b^{'})\%i$$
\returnType{Type: Complex Polynomial Integer}
}

and we see that the sum is the diagonal of the parallelogram on
$OP$, $OQ$. This is the law of the composition of simultaneous
velocities; and it contains, of course, the law of subtraction of one
directed line from another.

{\bf 6}. Operating on the first of these symbols by the factor $\sqrt{1}$,
it becomes $ b + a\sqrt{1}$; and now, of course, denotes the point
whose $x$ and $y$ coordinates are $ b$ and $a$; or the line joining this
point with the origin. The length is still $\sqrt{a^2+b^2}$, but the angle
the line makes with the axis of $x$ is $\tan^{1}( a/b)$; which is
evidently greater by $\pi/2$ than before the operation.

\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{op*complex(0,1)}
$$b+a\ i$$
\returnType{Type: Complex Polynomial Integer}
}

{\bf 7}. De Moivre's Theorem tends to lead us still further in the
same direction. In fact, it is easy to see that if we use, instead
of $\sqrt{1}$, the more general factor $\cos \alpha + \sqrt{1} \sin \alpha$,
its effect on
any line is to turn it through the (positive) angle $\alpha$. in the plane
of $x$, $y$. [Of course the former factor, $\sqrt{1}$, is merely the
particular case of this, when $\alpha=\frac{\pi}{2}$].

Thus
$$
\begin{array}{ll}
 &(\cos \alpha + \sqrt{1} \sin \alpha) (a + b \sqrt{1})\\
= & a \cos \alpha  b \sin \alpha + \sqrt{1} (a \sin \alpha + b \cos \alpha)
\end{array}
$$

by direct multiplication. The reader will at once see that the new
form indicates that a rotation through an angle $\alpha$ has taken place,
if he compares it with the common formulae for turning the coordinate
axes through a given angle. Or, in a less simple manner, thus

$$
\begin{array}{rcl}
Length & = & \sqrt{(a \cos \alpha  b \sin \alpha)^2 +
 (a \sin \alpha + b \cos \alpha)^2} \\
 & = & \sqrt{a^2 + b^2}
\end{array}
$$
as before.

Inclination to axis of $x$
$$
\begin{array}{cl}
= & \tan^{1}{\frac{a \sin \alpha + b \cos \alpha}
 {a \cos \alpha  b \sin \alpha}}\\
= & \tan^{1}{\frac{\tan \alpha + \frac{b}{a}}
 {1  \frac{b}{a} \tan \alpha}}\\
= & \alpha + \tan^{1}{\frac{b}{a}}
\end{array}
$$

{\bf 8}. We see now, as it were, why it happens that

$$(\cos \alpha + \sqrt{1} \sin \alpha)^m =
\cos m\alpha + \sqrt{1} \sin m\alpha
$$

In fact, the first operator produces $m$ successive rotations in the
same direction, each through the angle $\alpha$ ; the second, a single
rotation through the angle $m\alpha$.

{\bf 9}. It may be interesting, at this stage, to anticipate so far as to
remark that in the theory of Quaternions the analogue of

$$
\begin{array}{lclr}
 & \textrm{ } & \cos \theta + \sqrt{1} \sin \theta &\\
\textrm{is} & \textrm{ } & \cos \theta + \omega \sin \theta &\\
\textrm{where} & \textrm{ } & \omega^2 = 1 & \\
\end{array}
$$

Here, however, $\omega$ is not the algebraic $\sqrt{1}$, but is
{\sl any directed unitline} whatever in space.

{\bf 10}. In the present century Argand, Warren, Mourey, and
others, extended the results of Wallis and Bu\'ee. They attempted
to express as a line the product of two lines each represented by a
symbol such $a+b\sqrt{1}$. To a certain extent they succeeded,
but all their results remained confined to two dimensions.

The product, $\prod$, of two such lines was defined as the fourth
proportional to unity and the two lines, thus

$$
\begin{array}{lclr}
 & \textrm{ } &
1 : a+b\sqrt{1} :: a^{'}+b^{'}\sqrt{1} : \prod\\
\textrm{or} & \textrm{ } &
\prod = (aa^{'}  bb^{'})+(a^{'}b+b^{'}a)\sqrt{1}
\end{array}
$$

The length of $\prod$ is obviously the product of the lengths of the
factor lines; and its direction makes an angle with the axis of $x$
which is the sum of those made by the factor lines. From this
result the quotient of two such lines follows immediately.

{\bf 11}. A very curious speculation, due to Servois and published
in 1813 in Gergonne's {\sl Annales}, is one of the very few, so far as has
been discovered, in which a wellfounded guess at a possible mode
of extension to three dimensions is contained. Endeavouring to
extend to {\sl space} the form $a+b\sqrt{1}$ for the plane, he is guided by
analogy to write for a directed unitline in space the form

$$p \cos \alpha + q \cos \beta + r \cos \gamma$$

where $\alpha$, $\beta$, $\gamma$
are its inclinations to the three axes. He perceives
easily that $p$, $q$, $r$ must be {\sl nonreals}:
but, he asks, ``seraientelles
{\sl imaginaires} r\'eductibles \`a la forme g\'en\'erale $A+B\sqrt{1}$?''
The $i$,$j$, $k$ of the Quaternion Calculus furnish an answer to this
question. (See Chap. II.) But it may be remarked that, in applying the
idea to lines in a plane, a vector $OP$ will no longer be represented
(as in \S 5) by
$$
\begin{array}{lclr}
 & \textrm{ OP} & = & a + b\sqrt{1}\\
\textrm{but by} & \textrm{ OP} & = & pa + qb\\
\textrm{And if, similarly,} & \textrm{ OQ} & = & pa^{'} + qb^{'}\\
\end{array}
$$

the addition of these two lines gives for $OR$ (which retains its
previous signification)
$$OR = p(a+a^{'} + q(b+b^{'})$$

{\bf 12}. Beyond this, few attempts were made, or at least recorded,
in earlier times, to extend the principle to space of three dimensions;
and, though many such had been made before 1843, none,
with the single exception of Hamilton's, have resulted in simple,
practical methods; all, however ingenious, seeming to lead almost
at once to processes and results of fearful complexity.

For a lucid, complete, and most impartial statement of the
claims of his predecessors in this field we refer to the Preface to
Hamilton's {\sl Lectures on Quaternions}. He there shows how his long
protracted investigations of Sets culminated in this unique system
of tridimensionalspace geometry.

{\bf 13}. It was reserved for Hamilton to discover the use and
properties of a class of symbols which, though all in a certain sense
square roots of 1, may be considered as {\sl real} unit lines, tied down
to no particular direction in space ; the expression for a vector is,
or may be taken to be,
$$ \rho = ix + jy + kz$$

but such vector is considered in connection with an {\sl extraspatial}
magnitude $w$, and we have thus the notion of a QUATERNION

$$w + \rho$$

This is the fundamental notion in the singularly elegant, and
enormously powerful, Calculus of Quaternions.

While the schemes for using the algebraic $\sqrt{1}$ to indicate
direction make one direction in space expressible by real numbers,
the remainder being imaginaries of some kind, and thus lead to
expressions which are heterogeneous ; Hamilton s system makes all
directions in space equally imaginary, or rather equally real, thereby
ensuring to his Calculus the power of dealing with space
indifferently in all directions.

In fact, as we shall see, the Quaternion method is independent
of axes or any supposed directions in space, and takes its reference
lines solely from the problem it is applied to.

{\bf 14}. But, for the purpose of elementary exposition, it is best
to begin by assimilating it as closely as we can to the ordinary
Cartesian methods of Geometry of Three Dimensions, with which
the student is supposed to be, to some extent at least, acquainted.
Such assistance, it will be found, can (as a rule) soon be dispensed
with; and Hamilton regarded any apparent necessity for an oc
casional recurrence to it, in higher applications, as an indication
of imperfect development in the proper methods of the new
Calculus.

We commence, therefore, with some very elementary geometrical
ideas, relating to the theory of vectors in space. It will subsequently
appear how we are thus led to the notion of a Quaternion.

{\bf 15}. Suppose we have two points $A$ and $B$ in {\sl space}, and suppose
$A$ given, on how many numbers does $B$'s relative position depend ?

If we refer to Cartesian coordinates (rectangular or not) we find
that the data required are the excesses of $B$'s three coordinates
over those of $A$. Hence three numbers are required.

Or we may take polar coordinates. To define the moon's
position with respect to the earth we must have its Geocentric
Latitude and Longitude, or its Right Ascension and Declination,
and, in addition, its distance or radiusvector. {\sl Three} again.

{\bf 16}. Here it is to be carefully noticed that nothing has been
said of the {\sl actual} coordinates of either A or B, or of the earth
and moon, in space; it is only the {\sl relative} coordinates that are
contemplated.

Hence any expression, as $\overline{AB}$, denoting a line considered with
reference to direction and currency as well as length, (whatever
may be its actual position in space) contains implicitly {\sl three}
numbers, and all lines parallel and equal to $AB$, and concurrent
with it, depend in the same way upon the same three. Hence, {\sl all
lines which are equal, parallel, and concurrent, may be represented
by a common symbol, and that symbol contains three distinct numbers}.
In this sense a line is called a VECTOR, since by it we pass from
the one extremity, $A$, to the other, $B$, and it may thus be
considered as an instrument which {\sl carries} $A$ to $B$: so that a
vector may be employed to indicate a definite {\sl translation} in space.

[The term " currency " has been suggested by Cayley for use
instead of the somewhat vague suggestion sometimes taken to
be involved in the word "direction." Thus parallel lines have
the same direction, though they may have similar or opposite
currencies. The definition of a vector essentially includes its
currency.]

{\bf 17}. We may here remark, once for all, that in establishing a
new Calculus, we are at liberty to give any definitions whatever
of our symbols, provided that no two of these interfere with, or
contradict, each other, and in doing so in Quaternions {sl simplicity}
and (so to speak) {\sl naturalness} were the inventor's aim.

{\bf 18}. Let $\overline{AB}$ be represented by $\alpha$, we know that
$\alpha$ involves
{\sl three} separate numbers, and that these depend solely upon the
position of $B$ {\sl relatively} to $A$.
Now if $CD$ be equal in length to $AB$
and if these lines be parallel, and have the same currency, we may
evidently write
$$\overline{CD} = \overline{AB} = \alpha$$
where it will be seen that the sign of equality between vectors
contains implicitly {\sl equality in length}, {\sl parallelism in direction},
and {\sl concurrency}. So far we have {\sl extended} the meaning of an
algebraical symbol. And it is to be noticed that an equation
between vectors, as
$$\alpha = \beta$$
contains {\sl three} distinct equations between mere numbers.

{\bf 19}. We must now define $+$ (and the meaning of $$ will follow)
in the new Calculus. Let $A$, $B$, $C$ be any three points, and (with
the above meaning of $=$ ) let
$$\overline{AB} = \alpha, \overline{BC} = \beta, \overline{AC} = \gamma$$
If we define $+$ (in accordance with the idea (\S 16) that a vector
represents a {\sl translation}) by the equation
$$
\begin{array}{lcl}
 & \textrm{ } & \alpha + \beta = \gamma\\
 & & \\
\textrm{or} & \textrm{ } &
\overline{AB} + \overline{BC} = \overline{AC}
\end{array}
$$
we contradict nothing that precedes, but we at once introduce the
idea that {\sl vectors are to be compounded},
{\sl in direction and magnitude},
{\sl like simultaneous velocities}. A reason for this may be seen in
another way if we remember that by {\sl adding} the (algebraic) differences
of the Cartesian coordinates of $B$ and $A$, to those of the
coordinates of $C$ and $B$, we get those of the coordinates of $C$ and
$A$. Hence these coordinates enter {\sl linearly} into the expression for
a vector. (See, again, \S 5.)

{\bf 20}. But we also see that if $C$ and $A$ coincide (and $C$ may be
{\sl any} point)
$$\overline{AC} = 0$$
for no vector is then required to carry $A$ to $C$. Hence the above
relation may be written, in this case,
$$\overline{AB}+\overline{BA} = 0$$
or, introducing, and by the same act defining, the symbol $$,
$$\overline{AB} = \overline{BA}$$

Hence, {\sl the symbol $$, applied to a vector, simply shows that its
currency is to be reversed}.
And this is consistent with all that precedes; for instance,
$$
\begin{array}{lcrcl}
 & \textrm{ } & \overline{AB} + \overline{BC} &=& \overline{AC}\\
\textrm{and} & \textrm{ } & \overline{AB} = \overline{AC} && \overline{BC} \\
\textrm{or} & \textrm{ } & = \overline{AC} &+& \overline{CB} \\
\end{array}
$$
are evidently but different expressions of the same truth.

{\bf 21}. In any triangle, $ABC$, we have, of course,
$$\overline{AB} + \overline{BC} + \overline{CA} = 0$$
and, in any closed polygon, whether plane or gauche,
$$\overline{AB}+\overline{BC}+\ldots+\overline{YZ}+\overline{ZA} = 0$$

In the case of the polygon we have also
$$\overline{AB}+\overline{BC}+\ldots+\overline{YZ} = \overline{AZ}$$

These are the wellknown propositions regarding composition
of velocities, which, by Newton's second law of motion, give us
the geometrical laws of composition of forces acting at one point.

{\bf 22}. If we compound any number of {\bf parallel} vectors, the result
is obviously a numerical multiple of any one of them.
Thus, if $A$, $B$, $C$ are in one straight line,
$$\overline{BC} = x\overline{AB}$$
where $x$ is a number, positive when $B$ lies between $A$ and $C$,
otherwise negative; but such that its numerical value, independent
of sign, is the ratio of the length of $BC$ to that of $AB$. This is
at once evident if $AB$ and $BC$ be commensurable; and is easily
extended to incommensurables by the usual {\sl reductio ad absurdum}.

{\bf 23}. An important, but almost obvious, proposition is that {\sl any
vector may be resolved, and in one way only, into three components
parallel respectively to any three given vectors, no two of which are
parallel, and which are not parallel to one plane}.

\includegraphics{ps/quat1.ps}

Let $OA$, $OB$, $OC$ be the three fixed
vectors, $OP$ any other vector. From $P$ draw
$PQ$ parallel to $CO$, meeting the plane $BOA$
in $Q$. [There must be a definite point $Q$,
else $PQ$, and therefore $CO$, would be parallel
to $BOA$, a case specially excepted.] From $Q$
draw $QR$ parallel to $BO$, meeting $OA$ in $R$.

Then we have $\overline{OP}=\overline{OR} + \overline{RQ} + \overline{QP}$
(\S 21),
and these components are respectively parallel to the three given
vectors. By \S 22 we may express $\overline{OR}$ as a numerical multiple
of $\overline{OA}$, $\overline{RQ}$ of $\overline{OB}$, and
$\overline{QP}$ of $\overline{OC}$. Hence we have, generally, for
any vector in terms of three fixed noncoplanar vectors, $\alpha$,
$\beta$, $\gamma$
$$\overline{OP} = \rho = x\alpha + y\beta + z\gamma$$
which exhibits, in one form, the {\sl three} numbers on which a vector
depends (\S 16). Here $x$, $y$, $z$ are perfectly definite, and can have
but single values.

{\bf 24}. Similarly any vector, as $\overline{OQ}$, in the same plane with
$\overline{OA}$ and $\overline{OB}$,
can be resolved (in one way only) into components $\overline{OR}$,
$\overline{RQ}$,
parallel respectively to $\overline{OA}$ and
$\overline{OB}$; so long, at least, as these
two vectors are not parallel to each other.

{\bf 25}. There is particular advantage, in certain cases, in employing
a series of {\sl three mutually perpendicular unitvectors} as
lines of reference. This system Hamilton denotes by $i$,$j$, $k$.

Any other vector is then expressible as
$$\rho= xi + yj + zk$$
Since $i$, $j$, $k$ are unitvectors, $x$, $y$, $z$ are here the lengths of
conterminous edges of a rectangular parallelepiped of which $\rho$
is the vectordiagonal; so that the length of $\rho$ is, in this case,
$$\sqrt{x^2+y^2+z^2}$$
Let \hbox{\hskip 4cm}$\omega = \xi i + \eta j + \zeta k$\\
be any other vector, then (by the proposition of \S 23) the vector
$$
\begin{array}{lcr}
\textrm{equation} & \textrm{ } & \rho = \omega
\end{array}
$$
obviously involves the following three equations among numbers,
$$x=\xi, y=\eta, z=\zeta$$
Suppose $i$ to be drawn eastwards, $j$ northwards, and $k$ upwards,
this is equivalent merely to saying that {\sl if two points coincide, they
are equally to the east (or west) of any third point, equally to the
north (or south) of it, and equally elevated above (or depressed below)
its level.}

{\bf 26}. It is to be carefully noticed that it is only when
$\alpha$, $\beta$, $\gamma$
are not coplanar that a vector equation such as
$$\rho = \omega$$
or\hbox{\hskip 3cm}
$x\alpha + y\beta + z\gamma = \xi \alpha + \eta \beta + \zeta \gamma$\\
necessitates the three numerical equations
$$x = \xi, y = \eta, z = \zeta$$
For, if $\alpha$, $\beta$, $\gamma$ be coplanar (\S 24),
a condition of the following form must hold
$$\gamma = a\alpha + b\beta$$
Hence,\hbox{\hskip 3cm}$\rho=(x+za)\alpha+(y+zb)\beta$\\
\hbox{\hskip 4cm}$\omega=(\xi+\zeta a)\alpha + (\eta+\zeta b)\beta$\\
and the equation\hbox{\hskip 3cm}$\rho=\omega$\\
now requires only the two numerical conditions
$$x+za=\xi+\zeta a\hbox{\hskip 1cm}y+zb = \eta+\zeta b$$

{\bf 27}. {\sl The Commutative and Associative Laws hold in the combination
of vectors by the signs $+$ and $$}. It is obvious that, if we
prove this for the sign $+$, it will be equally proved for $$, because
$$ before a vector (\S 20) merely indicates that it is to be reversed
before being considered positive.

Let $A$, $B$, $C$, $D$ be, in order, the corners of a parallelogram ; we
have, obviously,
$$\overline{AB} = \overline{DC}\hbox{\hskip 1cm}\overline{AD}=\overline{BC}$$
And \hbox{\hskip 2cm}$\overline{AB}+\overline{BC}=
\overline{AC}=
\overline{AD}+\overline{DC}=
\overline{BC}+\overline{AB}$

Hence the commutative law is true for the addition of any two
vectors, and is therefore generally true.

Again, whatever four points are represented by $A$, $B$, $C$, $D$, we
$$\overline{AD}=\overline{AB}+\overline{BD}=\overline{AC}+\overline{CD}$$
or substituting their values for $\overline{AD}$, $\overline{BD}$,
$\overline{AC}$ respectively, in these three expressions,
$$\overline{AB}+\overline{BC}+\overline{CD}=
\overline{AB}+(\overline{BC}+\overline{CD})=
(\overline{AB}+\overline{BC})+\overline{CD}$$
And thus the truth of the associative law is evident.

{\bf 28}. The equation
$$\rho = x\beta$$
where $\rho$ is the vector connecting a variable point with the origin,
$\beta$ a definite vector, and $x$ an indefinite number, represents the
straight line drawn from the origin parallel to $\beta$ (\S 22).

The straight line drawn from $A$, where $\overline{OA} = \alpha$,
and parallel to $\beta$, has the equation
$$\rho = \alpha + x\beta\eqno(1)$$
In words, we may pass directly from $O$ to $P$ by the vector $\overline{OP}$
or $\rho$; or we may pass first to $A$, by means of $\overline{OA}$ or
$\alpha$, and then to $P$ along a vector parallel to $\beta$ (\S 16).

Equation (1) is one of the many useful forms into which
Quaternions enable us to throw the general equation of a straight
line in space. As we have seen (\S 25) it is equivalent to three
numerical equations; but, as these involve the indefinite quantity
$x$, they are virtually equivalent to but {\sl two}, as in ordinary Geometry
of Three Dimensions.

{\bf 29}. A good illustration of this remark is furnished by the fact
that the equation
$$\rho = y\alpha + x\beta$$
which contains two indefinite quantities, is virtually equivalent to
only one numerical equation. And it is easy to see that it represents
the plane in which the lines $\alpha$ and $\beta$ lie; or the surface
which is formed by drawing, through every point of $OA$, a line
parallel to $OB$. In fact, the equation, as written, is simply \S 24
in symbols.

And it is evident that the equation
$$\rho = \gamma + y\alpha + x\beta$$
is the equation of the plane passing through the extremity of $\gamma$,
and parallel to $\alpha$ and $\beta$.

It will now be obvious to the reader that the equation
$$\rho = p_1\alpha_1+p_2\alpha_2+\ldots=\sum{p\alpha}$$
where $\alpha_1$, $\alpha_2$ , \&c. are given vectors,
and $p_1$, $p_2$, \&c. numerical quantities,
{\sl represents a straight line}
if $p_1$, $p_2$, \&c. be linear functions of
{\sl one} indeterminate number; and a {\sl plane}, if they be linear
expressions containing two indeterminate numbers. Later (\S 31 (l)),
this theorem will be much extended.

Again, the equation
$$\rho = x\alpha + y\beta +z\gamma$$
refers to {\sl any} point whatever in space, provided
$\alpha$, $\beta$, $\gamma$ are not coplanar. (Ante, \S 23)

{\bf 30}. The equation of the line joining any two points $A$ and $B$,
where $\overline{OA} = \alpha$ and $\overline{OB} = \beta$, is obviously
$$\rho = \alpha + x(\beta\alpha)$$
or \hbox{\hskip 4.2cm}$\rho=\beta+y(\alpha\beta)$\\
These equations are of course identical, as may be seen by putting
$1y$ for $x$.

The first may be written
$$\rho+(x1)\alphax\beta = 0$$
or\hbox{\hskip 4cm}$p\rho+q\alpha+r\beta=0$\\
subject to the condition $p + q + r = 0$ identically. That is 
A homogeneous linear function of three vectors, equated to zero,
expresses that the extremities of these vectors are in one straight
line, {\sl if the sum of the coefficients be identically zero}.

Similarly, the equation of the plane containing the extremities
$A$, $B$, $C$ of the three noncoplanar vectors
$\alpha$, $\beta$, $\gamma$ is
$$\rho=\alpha+x(\beta\alpha)+y(\gamma\beta)$$
where $x$ and $y$ are each indeterminate.

This may be written
$$p\rho + q\alpha + r\beta +s\gamma = 0$$
with the identical relation
$$p+q+r+x=0$$
which is one form of the condition that four points may lie in one plane.

{\bf 31}. We have already the means of proving, in a very simple
manner, numerous classes of propositions in plane and solid
geometry. A very few examples, however, must suffice at this
stage; since we have hardly, as yet, crossed the threshold of the
subject, and are dealing with mere linear equations connecting two
or more vectors, and even with them {\sl we are restricted as yet to
operations of mere addition}. We will give these examples with a
painful minuteness of detail, which the reader will soon find to be
necessary only for a short time, if at all.

(a) {\sl The diagonals of a parallelogram bisect each other}.

Let $ABCD$ be the parallelogram, $O$ the point of intersection of
its diagonals. Then
$$\overline{AO}+\overline{OB}=\overline{AB}=\overline{DC}=
\overline{DO}+\overline{OC}$$
which gives\hbox{\hskip 2cm}$\overline{AO}\overline{OC}=
\overline{DO}\overline{OB}$\\
The two vectors here equated are parallel to the diagonals respectively.
Such an equation is, of course, absurd unless
\begin{enumerate}
\item The diagonals are parallel, in which case the figure
is not a parallelogram;
\item $\overline{AO} = \overline{OC}$, and
$\overline{DO} = \overline{OB}$, the proposition.
\end{enumerate}

(b) {\sl To shew that a triangle can be constructed, whose sides
are parallel, and equal, to the bisectors of the sides of any
triangle}.

Let $ABC$ be any triangle, $Aa$, $Bb$, $Cc$ the bisectors of the
sides.

Then
$$
\begin{array}{ccc}
\overline{Aa} & =\overline{AB}+\overline{Ba}
 & =\overline{AB}+\frac{1}{2}\overline{BC}\\
\overline{Bb} & \ldots & = \overline{BC} + \frac{1}{2}\overline{CA}\\
\overline{Cc} & \ldots & = \overline{CA} + \frac{1}{2}\overline{AB}
\end{array}
$$
Hence \hbox{\hskip 2cm}$\overline{Aa}+\overline{Bb}+\overline{Cc}=
\frac{3}{2}(\overline{AB}+\overline{BC}+\overline{CA})=0$\\
which (\S 21) proves the proposition.

Also
$$
\begin{array}{rcl}
\overline{Aa} & = & \overline{AB}+\frac{1}{2}\overline{BC}\\
 & = & \overline{AB}\frac{1}{2}(\overline{CA}+\overline{AB})\\
 & = & \frac{1}{2}(\overline{AB}\overline{CA})\\
 & = & \frac{1}{2}(\overline{AB}+\overline{AC})
\end{array}
$$
results which are sometimes useful. They may be easily verified
by producing $\overline{Aa}$ to twice its length and joining the extremity
with $B$.

($b^{'}$) {\sl The bisectors of the sides of a triangle meet in a point,
which trisects each of them}.

Taking $A$ as origin, and putting $\alpha$, $\beta$, $\gamma$
for vectors parallel, and
equal, to the sides taken in order $BC$, $CA$, $AB$; the equation of
$Bb$ is (\S 28 (1))
$$\rho=\gamma+x(\gamma+\frac{\beta}{2})=(1+x)\gamma+\frac{x}{2}\beta$$
That of $Cc$ is, in the same way,
$$\rho=(1+y)\beta\frac{y}{2}\gamma$$
At the point $O$, where $Bb$ and $Cc$ intersect,
$$\rho=(1+x)\gamma+\frac{x}{2}\beta=(1+y)\beta\frac{y}{2}\gamma$$
Since $\gamma$ and $\beta$ are not parallel, this equation gives
$$1+x=\frac{y}{2}\textrm{\ \ and\ \ }\frac{x}{2}=(1+y)$$
From these\hbox{\hskip 3cm}$x=y=\frac{2}{3}$

Hence\hbox{\hskip 1cm}$\overline{AO}=\frac{1}{3}(\gamma\beta)=
\frac{2}{3}\overline{Aa}$ (See Ex. (b))\\

This equation shows, being a vector one, that $\overline{Aa}$ passes
through $O$, and that $AO$ : $Oa$ :: 2:1.

(c) If
$$\overline{OA}=\alpha$$
$$\overline{OB}=\beta$$
$$\overline{OC}=l\alpha+m\beta$$
{\sl be three given coplanar vectors, $c$ the intersection of $AB$, $OC$, and
if the lines indicated in the figure be drawn, the points
$a_1$,$b_1$,$c_1$ lie in a straight line. }
\includegraphics{ps/quat2.ps}

We see at once, by the process indicated in \S 30, that
$$\overline{Oc}=\frac{l\alpha+m\beta}{l+m},\hbox{\hskip 1cm}
\overline{Ob}=\frac{l\alpha}{1m},\hbox{\hskip 1cm}
\overline{Oa}=\frac{m\beta}{1l}$$
Hence we easily find
$$\overline{Oa_1}=\frac{m\beta}{1l2m},\hbox{\hskip 0.5cm}
\overline{Ob_1}=\frac{l\alpha}{12lm},\hbox{\hskip 0.5cm}
\overline{Oc_1}=\frac{l\alpha+m\beta}{ml}$$
These give
$$(1l2m)\overline{Oa_1}+(12lm)\overline{Ob_1}(ml)\overline{Oc_1}=0$$
But\hbox{\hskip 1cm}$(1l2m)+(12lm)(m  l)=0$ identically.

This, by \S 30, proves the proposition.

(d) Let $\overline{OA} = \alpha$,
$\overline{OB} = \beta$, be any two vectors. If $MP$ be a
given line parallel to $OB$; and $OQ$, $BQ$, be drawn parallel to $AP$,
$OP$ respectively ; the locus of $Q$ is a straight line parallel to $OA$.

\noindent
\includegraphics{ps/quat3.ps}

\noindent
Let \hbox{\hskip 4cm}$\overline{OM}=e\alpha$\\
Then \hbox{\hskip 3cm}$\overline{AP}=\overline{e1}\alpha+x\beta$

Hence the equation of $OQ$ is
$$\rho=y(\overline{e1}\alpha+x\beta)$$
and that of $BQ$ is\hbox{\hskip 1cm}$\rho=\beta+z(e\alpha+x\beta)$\\
At Q we have, therefore,
$$
\left.
\begin{array}{c}
xy=1+zx\\
y(e1)=ze
\end{array}
\right\}
$$
These give $xy = e$, and the equation of the locus of $Q$ is
$$\rho = e\beta+y^{'}\alpha$$
i.e. a straight line parallel to $OA$, drawn through $N$ in $OB$
produced, so that
$$ON : OB :: OM : OA$$

COR. If $BQ$ meet $MP$ in $q$, $\overline{Pq} = \beta$;
and if $AP$ meet $NQ$ in $p$, $\overline{Qp} = \alpha$.

Also, for the point $R$ we have $\overline{pR} = \overline{AP}$,
$\overline{QR} = \overline{Bq}$.

Further, the locus of $R$ is a hyperbola, of which $MP$ and $NQ$
are the asymptotes. See, in this connection, \S 31 (k) below.

Hence, {\sl if from any two points, $A$ and $B$, lines be drawn intercepting
a given length $Pq$ on a given line $Mq$ ; and if, from $R$ their
point of intersection, $Rp$ be laid off $= PA$, and $RQ = qB$ ; $Q$ and $p$
lie on a fixed straight line, and the length of $Qp$ is constant}.

(e) {\sl To find the centre of inertia of any system of masses.}

If $\overline{OA} = \alpha$, $\overline{OB} = \alpha_1$,
be the vector sides of any triangle, the
vector from the vertex dividing the base $AB$ in $C$ so that
$$BC : CA :: m : m_1$$
is \hbox{\hskip 4cm}$\frac{m\alpha+m_1\alpha_1}{m+m_1}$\\

For $AB$ is $\alpha_1\alpha$, and therefore $\overline{AC}$ is
$$\frac{m_1}{m+m_1}(\alpha_1\alpha)$$

Hence\hbox{\hskip 3cm}$\overline{OC}=\overline{OA}+\overline{AC}$
$$=\alpha+\frac{m_1}{m+m_1}(\alpha_1\alpha)$$
$$=\frac{m\alpha +m_1\alpha_1}{m+m_1}$$
This expression shows how to find the centre of inertia of two
masses ; $m$ at the extremity of $\alpha$, $m_1$ at that of $\alpha_1$.
Introduce $m_2$ at the extremity of $a_2$,
then the vector of the centre of inertia of the
three is, by a second application of the formula,
$$\frac{(m+m_1)(\frac{m\alpha+m_1\alpha_1}{m+m_1})+m_2\alpha_2}
{(m+m_1)+m_2}=\frac{m\alpha+m_1\alpha_1+m_2\alpha_2}{m+m_1+m_2}$$
From this it is clear that, for any number of masses, expressed
generally by $m$ at the extremity of the vector $\alpha$, the vector of the
centre of inertia is
$$\beta=\frac{\sum(m\alpha)}{\sum(m)}$$
This may be written\hbox{\hskip 1cm}$\sum m(\alpha\beta)=0$\\
Now a $\alpha_1\beta$
is the vector of $m_1$ with respect to the centre of inertia.
Hence the theorem, {\sl If the vector of each element of a mass, drawn
from the centre of inertia, be increased in length in proportion to the
mass of the element, the sum of all these vectors is zero.}

(f) We see at once that the equation

\noindent
\includegraphics{ps/quat4.ps}

$$\rho=\alpha t +\frac{\beta t^2}{2}$$
where $t$ is an indeterminate
number, and $\alpha$, $\beta$ given vectors,
represents a parabola.
The origin, $O$, is a point on
the curve, $\beta$ is parallel to
the axis, i.e. is the diameter
$OB$ drawn from the origin,
and $\alpha$ is $OA$ the tangent at the origin. In the figure
$$\overline{QP}=\alpha t,\hbox{\hskip 1cm}\overline{OQ}=\frac{\beta t^2}{2}$$

The secant joining the points where $t$ has the values $t$ and $t^{'}$ is
represented by the equation
$$
\begin{array}{rcl}
\rho&=&\alpha t +\frac{\beta t^2}{2}+x\left(\alpha t^{'}+\frac{\beta t^{'2}}{2}
\alpha t\frac{\beta t^2}{2}\right)\hbox{\hskip 1cm}(\S 30)\\
&=&\alpha t+\frac{\beta t^2}{2}+x(t^{'}t)\left\{\alpha+\beta\frac{t^{'}t}{2}
\right\}
\end{array}
$$
Write $x$ for $x(t^{'}t)$ [which may have any value], then put
$t^{'}=t$, and the equation of the tangent at the point ($t$) is
$$\rho=\alpha t + \frac{\beta t^2}{2}+x(\alpha+\beta t)$$
In this put $x = t$, and we have
$$\rho=\frac{\beta t^2}{2}$$
or the intercept of the tangent on the diameter is equal in length
to the abscissa of the point of contact, but has the opposite
currency.

Otherwise: the tangent is parallel to the vector $\alpha+\beta t$ or
$\alpha t + \beta t^2$ or $\frac{\beta t^2}{2}+\alpha t+\frac{\beta t^2}{2}$
or $\overline{OQ}+\overline{OP}$.
But $\overline{TP}=\overline{TO}+\overline{OP}$,
hence $\overline{TO} = \overline{OQ}$.

(g) Since the equation of any tangent to the parabola is
$$\rho=\alpha t + \frac{\beta t^2}{2} + x(\alpha+\beta t)$$
let us find the tangents which can be drawn from a given point.
Let the vector of the point be
$$\rho=p\alpha + q\beta\hbox{\hskip 0.5cm}(\S 24)$$
Since the tangent is to pass through this point, we have, as con
ditions to determine $t$ and $x$,
$$t+x=p$$
$$\frac{t^2}{2}+xt=q$$
by equating respectively the coefficients of $\alpha$ and $\beta$.

Hence\hbox{\hskip 3.5cm}$t=p \pm \sqrt{p^22q}$

Thus, in general, {\sl two} tangents can be drawn from a given point.
These coincide if $$p^2=2q$$
that is, if the vector of the point from which they are to be drawn
is $$\rho=p\alpha+q\beta=p\alpha+\frac{p^2}{2}\beta$$
i.e. if the point lies on the parabola. They are imaginary if
$2q > p^2$, that is, if the point be
$$\rho=p\alpha+\left(\frac{p^2}{2}+r\right)\beta$$
$r$ being {\sl positive}. Such a point is evidently {\sl within} the curve,
as at $R$, where $\overline{OQ}=\frac{p^2}{2}\beta$,
$\overline{QP}=p\alpha$, $\overline{PR} = r\beta$.

(h) Calling the values of $t$ for the two tangents found in (g)
$t_1$ and $t_2$ respectively, it is obvious that the vector joining the
points of contact is
$$\alpha t_1+\frac{\beta t_1^2}{2}\alpha t_2  \frac{\beta t_2^2}{2}$$
which is parallel to\hbox{\hskip 2cm}
$\alpha+\beta\frac{t_1+t_2}{2}$
or, by the values of $t_1$ and $t_2$ in (g),
$$\alpha+p\beta$$
Its direction, therefore, does not depend on $q$. In words, {\sl If pairs of
tangents be drawn to a parabola from points of a diameter produced,
the chords of contact are parallel to the tangent at the vertex of the
diameter.} This is also proved by a former result, for we must have
$\overline{OT}$ for each tangent equal to $\overline{QO}$.

(i) The equation of the chord of contact, for the point whose vector is
$$\rho=p\alpha+q\beta$$
is thus\hbox{\hskip 3cm}
$\rho=\alpha t_1+\frac{\beta t_1^2}{2}+y(\alpha+p\beta)$

Suppose this to pass always through the point whose vector is
$$\rho=a\alpha+b\beta$$
Then we must have
$$
\left.
\begin{array}{rcl}
t_1+y & = & a\\
\frac{t_1^2}{2}+py & = & b
\end{array}
\right\}
$$
or\hbox{\hskip 4cm}$t_1=p\pm\sqrt{p^22p\alpha+2\beta}$

Comparing this with the expression in (g), we have
$$
q = pa  b
$$
that is, the point from which the tangents are drawn has the vector
a straight line (\S 28 (1)).

The mere form of this expression contains the proof of the usual
properties of the pole and polar in the parabola ; but, for the sake
of the beginner, we adopt a simpler, though equally general, process.

Suppose $\alpha = 0$. This merely restricts the pole to the particular
diameter to which we have referred the parabola. Then the pole
is $Q$, where $$\rho = b\beta$$
and the polar is the line $TU$, for which
$$\rho=b\beta+p\alpha$$
{\sl Hence the polar of any point is parallel to the tangent at the
extremity of the diameter on which the point lies, and its intersection
with that diameter is as far beyond the vertex as the pole
is within, and vice versa. }

(j) As another example let us prove the following theorem.
{\sl If a triangle be inscribed in a parabola, the three points in which
the sides are met by tangents at the angles lie in a straight line. }

Since $O$ is any point of the curve, we may take it as one corner
of the triangle. Let $t$ and $t_1$ determine the others. Then, if
$\omega_1$,$\omega_2$,$\omega_3$
represent the vectors of the points of intersection of the
tangents with the sides, we easily find
$$
\begin{array}{rcl}
\omega_1 & = & \frac{t_1^2}{2t_1t}
\left(\alpha+\frac{t}{2}\beta\right)\\
&&\\
\omega_2 & = & \frac{t^2}{2tt_1}
\left(\alpha+\frac{t_1}{2}\beta\right)\\
&&\\
\omega_3 & = & \frac{tt_1}{t_1+t}\alpha
\end{array}
$$
These values give
$$\frac{2t_1t}{t_1}\omega_1 
\frac{2tt_1}{t}\omega_2 
\frac{t_1^2t^2}{tt_1}\omega_3 = 0$$
Also
$$\frac{2t_1t}{t_1} 
\frac{2tt_1}{t} 
\frac{t_1^2t^2}{tt_1} = 0$$
identically.

Hence, by \S 30, the proposition is proved.

(k) Other interesting examples of this method of treating
curves will, of course, suggest themselves to the student. Thus
$$\rho = \alpha\cos t + \beta\sin t$$
or
$$\rho=\alpha x + \beta\sqrt{1x^2}$$
represents an ellipse, of which the given vectors $\alpha$ and $\beta$
are semiconjugate diameters. If $t$ represent time, the radiusvector of this
ellipse traces out equal areas in equal times. [We may anticipate
so far as to write the following :
$$2 \textrm{Area} = T\int V \rho d\rho = TV\alpha\beta.\int dt$$
which will be easily understood later.]

Again,
$$\rho=\alpha t+\frac{\beta}{t}\textrm{ or }
\rho=\alpha\tan x + \beta\cot x$$
evidently represents a hyperbola referred to its asymptotes. [If
$t$ represent time, the sectorial area traced out is proportional to
$\log t$, taken between proper limits.]
Thus, also, the equation
$$\rho = \alpha(t + \sin t)+\beta\cos t$$
in which $\alpha$ and $\beta$ are of equal lengths, and at right angles to one
another, represents a cycloid. The origin is at the middle point of
the axis ($2\beta$) of the curve. [It may be added that, if t represent
{\sl time}, this equation shows the motion of the tracing point, provided
the generating circle rolls uniformly, revolving at the rate of a
radian per second.]

When the lengths of $\alpha$, $\beta$ are not equal, this equation gives the
cycloid distorted by elongation of its ordinates or abscissae : {\sl not} a
trochoid. The equation of a trochoid may be written
$$\rho = \alpha(et + \sin t)+\beta\cos t$$
$e$ being greater or less than 1 as the curve is prolate or curtate.
The lengths of $\alpha$ and $\beta$ are still taken as equal.

But, so far as we have yet gone with the explanation of the
calculus, as we are not prepared to determine the lengths or
inclinations of vectors, we can investigate only a very limited class of
the properties of curves, represented by such equations as those
above written.

(l) We may now, in extension of the statement in \S 29, make
the obvious remark that
$$\rho = \sum p\alpha$$
(where, as in \S 23, the number of vectors, $\alpha$, can always be reduced
to {\sl three}, at most) is the equation of a curve in space, if the
numbers $p_1$, $p_2$, \&c. are functions of one indeterminate. In such
a case the equation is sometimes written
$$\rho=\phi(t)$$
But, if $p_1$, $p_2$, \&c. be functions of {\sl two} indeterminates,
the locus of the extremity of $\rho$ is a {\sl surface};
whose equation is sometimes written
$$\rho = \phi(t,u)$$

[It may not be superfluous to call the reader's attention to the
fact that, in these equations, $\phi(t)$ or $\phi(t, u)$
is necessarily a vector expression, since it is equated to a vector, $\rho$.]

(m) Thus the equation
$$\rho = \alpha\cos t+\beta\sin t + \gamma t\eqno(1)$$
belongs to a helix,
\boxed{4.6in}{
\vskip 0.1cm
In Axiom we can draw this with the commands:
\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
tpdhere
}
while
$$\rho = \alpha\cos t+\beta\sin t + \gamma u\eqno(2)$$
represents a cylinder whose generating lines are parallel to $\gamma$,
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
tpdhere
}
and
whose base is the ellipse
$$\rho=\alpha\cos t + \beta\sin t$$
The helix above lies wholly on this cylinder.
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
tpdhere
}

Contrast with (2) the equation
$$\rho = u(\alpha\cos t + \beta\sin t + \gamma)\eqno(3)$$
which represents a cone of the second degree
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
tpdhere
}
made up, in fact,
of all lines drawn from the origin to the ellipse
$$\rho=\alpha\cos t + \beta\sin t + \gamma$$
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
tpdhere
}

If, however, we write
$$\rho = u(\alpha\cos t + \beta\sin t + \gamma t)$$
we form the equation of the transcendental cone whose vertex is
at the origin, and on which lies the helix (1).
\boxed{4.6in}{
\vskip 0.1cm
\spadcommand{draw(a*cos(t)+b*sin(t)+c*u,[t=0..1,u=0..1]}
tpdhere
}

In general
$$\rho=u\phi(t)$$
is the cone whose vertex is the origin, and on which lies the curve
$$\rho=\phi(t)$$
while\hbox{\hskip 4cm}$\rho=\phi(t)+u\alpha$\\
is a cylinder, with generating lines parallel to $\alpha$, standing on the
same curve as base.

Again,\hbox{\hskip 3cm}$\rho=p\alpha+q\beta+r\gamma$\\
with a condition of the form
$$ap^2+bq^2+cr^2=1$$
belongs to a central surface of the second order, of which
$\alpha$, $\beta$, $\gamma$
are the directions of conjugate diameters. If $a$, $b$, $c$ be all positive,
the surface is an ellipsoid.

{\bf 32}. In Example ($f$) above we performed an operation equivalent
to the differentiation of a vector with reference to a single
{\sl numerical} variable of which it was given as an explicit function.
As this process is of very great use, especially in quaternion
investigations connected with the motion of a particle or point; and as it
will afford us an opportunity of making a preliminary step towards
overcoming the novel difficulties which arise in quaternion differentiation;
we will devote a few sections to a more careful, though
very elementary, exposition of it.

{\bf 33}. It is a striking circumstance, when we consider the way
in which Newton's original methods in the Differential Calculus
have been decried, to find that Hamilton was {\sl obliged} to employ
them, and not the more modern forms, in order to overcome the
characteristic difficulties of quaternion differentiation. Such a thing
as {\sl a differential coefficient has absolutely no meaning in quaternions},
except in those special cases in which we are dealing with degraded
quaternions, such as numbers, Cartesian coordinates, \&c. But a
quaternion expression has always a {\sl differential}, which is, simply,
what Newton called a {\sl fluxion}.

As with the Laws of Motion, the basis of Dynamics, so with the
foundations of the Differential Calculus ; we are gradually coming
to the conclusion that Newton s system is the best after all.

{\bf 34}. Suppose $\rho$ to be the vector of a curve in space. Then,
generally, $\rho$ may be expressed as the sum of a number of terms,
each of which is a multiple of a constant vector by a function of some
{\sl one} indeterminate; or, as in \S 31 ($l$),
if $P$ be a point on the curve,
$$\overline{OP}=\rho=\phi(t)$$

And, similarly, if $Q$ be {\sl any other} point on the curve,
$$\overline{OQ}=\rho_1=\rho+\delta\rho=\phi(t_1)=\phi(t+\delta t)$$
where $\delta t$ is {\sl any number whatever}.

The vectorchord $\overline{PQ}$ is therefore, rigorously,
$$\delta p = \rho_1\rho = \phi(t+\delta t)\phi t$$

{\bf 35}. It is obvious that, in the present case, {\sl because the vectors
involved in $\phi$ are constant, and their numerical multipliers alone vary},
the expression $\phi(t+\delta t)$ is, by Taylor's Theorem, equivalent to
$$\phi(t)+\frac{d\phi(t)}{dt}\delta t+
\frac{d^2\phi(t)}{dt^2}\frac{(\delta t)^2}{1\textrm{ . }2}+\ldots$$

Hence,
$$\delta \rho=\frac{d\phi(t)}{dt}\delta t+
\frac{d^2\phi(t)}{dt^2}\frac{(\delta t)^2}{1\textrm{ . }2}+\textrm{\&c.}$$
And we are thus entitled to write, when $\delta t$ has been made
indefinitely small,
$$\textrm{Limit}\left(
\begin{array}{c}
\delta p\\
\delta t
\end{array}
\right)_{\delta t=0}
=\frac{d\rho}{dt}
=\frac{d\phi(t)}{dt}
=\phi^{'}(t)$$

In such a case as this, then, we are permitted to differentiate,
or to form the differential coefficient of, a vector, according to the
ordinary rules of the Differential Calculus. But great additional
insight into the process is gained by applying Newton's method.

{\bf 36}. Let $\overline{OP}$ be
$$\rho=\phi(t)$$
and $overline{OQ}_1$
$$\rho_1=\phi(t+dt)$$
where $dt$ is any number whatever.

\includegraphics{ps/quat5.ps}

The number $t$ may here be taken
as representing {\sl time}, i.e. we may
suppose a point to move along the
curve in such a way that the value
of $t$ for the vector of the point $P$ of
the curve denotes the interval which
has elapsed (since a fixed epoch) when the moving point has
reached the extremity of that vector. If, then, $dt$ represent any
interval, finite or not, we see that
$$\overline{OQ}_1=\phi(t+dt)$$
will be the vector of the point after the additional interval $dt$.

But this, in general, gives us little or no information as to the
velocity of the point at $P$. We shall get a better approximation
by halving the interval $dt$, and finding $Q_2$ ,
where $\overline{OQ}_2 = \phi(t + \frac{1}{2}dt)$,
as the position of the moving point at that time. Here the vector
virtually described in $\frac{1}{2}dt$ is
$\overline{PQ}_2$ . To find, on this supposition,
the vector described in $dt$, we must double
$\overline{PQ}_2$ , and we find, as a
second approximation to the vector which the moving point would
have described in time $dt$, if it had moved for that period in the
direction and with the velocity it had at $P$,
$$
\begin{array}{rcl}
\overline{Pq}_2=2\overline{PQ}_2 & = & 2(\overline{OQ}_2\overline{OP})\\
& = & 2\{\phi(t+\frac{1}{2}dt)\phi(t)\}
\end{array}
$$
The next approximation gives
$$
\begin{array}{rcl}
\overline{Pq}_3=3\overline{PQ}_3 & = & 3(\overline{OQ}_3\overline{OP})\\
& = & 3\{\phi(t+\frac{1}{3}dt)\phi(t)\}
\end{array}
$$
And so on, each step evidently leading us nearer the sought truth.
Hence, to find the vector which would have been described in time
$dt$ had the circumstances of the motion at $P$ remained undisturbed,
we must find the value of
$$d\rho=\overline{Pq}=L_{x=\infty}x\left\{\phi\left(t+\frac{1}{x}dt\right)
\phi(t)\right\}$$

We have seen that in this particular case we may use Taylor's
Theorem. We have, therefore,
$$
\begin{array}{rcl}
d\rho & = & L_{x=\infty}x
\left\{
\phi^{'}(t)\frac{1}{x}dt+
\phi^{''}(t)\frac{1}{x^2}\frac{(dt)^2}{1\textrm{ . }2}+
\textrm{\&c}
\right\}\\
&&\\
& = & \phi^{'}(t)dt
\end{array}
$$
And, if we choose, we may now write
$$\frac{d\rho}{dt}=\phi^{'}(t)$$

{\bf 37}. But it is to be most particularly remarked that in the
whole of this investigation no regard whatever has been paid to
the magnitude of $dt$. The question which we have now answered
may be put in the form  {\sl A point describes a given curve in a given
manner. At any point of its path its motion suddenly ceases to be
accelerated. What space will it describe in a definite interval?} As
Hamilton well observes, this is, for a planet or comet, the case
of a 'celestial Atwood's machine'.

{\bf 38}. If we suppose the variable, in terms of which $\rho$ is expressed,
to be the arc, $s$, of the curve measured from some fixed point, we
find as before
$$d\rho = \phi^{'}(x)ds$$
From the very nature of the question it is obvious that the length
of $dp$ must in this case be $ds$, so that $\phi^{'}(s)$
is necessarily a unitvector.
This remark is of importance, as we shall see later; and
it may therefore be useful to obtain afresh the above result without
any reference to time or velocity.

{\bf 39}. Following strictly the process of Newton s VIIth Lemma,
let us describe on $Pq_2$ an arc similar to $PQ_2$, and so on. Then
obviously, as the subdivision of $ds$ is carried farther, the new arc
(whose length is always $ds$) more and more nearly (and without
limit) coincides with the line which expresses the corresponding
approximation to $dp$.

{\bf 40}. As additional examples let us take some wellknown
{\sl plane} curves; and first the hyperbola (\S 31 ($k$))
$$\rho=\alpha t + \frac{\beta}{t}$$
Here
$$d\rho=\left(\alpha\frac{\beta}{t^2}\right)dt$$
This shows that the tangent is parallel to the vector
$$\alpha t  \frac{\beta}{t}$$
In words, {\sl if the vector (from the centre) of a point in a hyperbola
be one diagonal of a parallelogram, two of whose sides coincide with
the asymptotes, the other diagonal is parallel to the tangent at the
point, and cuts off a constant area from the space between the
asymptotes}. (For the sides of this triangular area are $t$ times the
length of $\alpha$, and $1/t$
times the length of $\beta$, respectively; the angle
between them being constant.)

Next, take the cycloid, as in \S 31 ($k$),
$$\rho=\alpha(t+\sin t)+\beta\cos t$$
We have
$$d\rho=\{\alpha(1+\cos t)\beta\sin t\}dt$$
At the vertex
$$t=0,\hbox{\hskip 0.5cm}
\cos t=1,\hbox{\hskip 0.5cm}
\sin t=0,\hbox{\hskip 0.5cm}
\textrm{ and }d\rho=2\alpha dt$$
At a cusp
$$t=\pi,\hbox{\hskip 0.5cm}
\cos t=1,\hbox{\hskip 0.5cm}
\sin t=0,\hbox{\hskip 0.5cm}
\textrm{ and }d\rho = 0$$

This indicates that, at the cusp, the tracing point is (
instantaneously) at rest. To find the direction of the tangent, and the
form of the curve in the vicinity of the cusp, put $t=\pi+\tau$,
where powers of $\tau$ above the second are omitted. We have
$$d\rho=\beta\tau dt + \frac{\alpha\tau^2}{2}dt$$
so that, at the cusp, the tangent is parallel to $\beta$. By making the
same substitution in the expression for $\rho$, we find that the part of
the curve near the cusp is a semicubical parabola,
$$\rho=\alpha(\pi+\tau^3/6)\beta(1\tau^2/2)$$
or, if the origin be shifted to the cusp ($\rho=\pi\alpha\beta$),
$$\rho=\alpha\tau^3/6+\beta\tau^2/2$$

{\bf 41}. Let us reverse the first of these questions, and {\sl seek the
envelope of a line which cuts off from two fixed axes a triangle of
constant area}.

If the axes be in the directions of $\alpha$ and $\beta$, the intercepts may
evidently be written $\alpha t$ and $\frac{\beta}{t}$.
Hence the equation of the line is (\S 30)
$$\rho=\alpha t + x\left(\frac{\beta}{t}\alpha t\right)$$

The condition of envelopment is, obviously, (see Chap. IX.)
$$d\rho = 0$$
This gives
$0 = \left\{\alphax\left(\frac{\beta}{t^2}+\alpha\right)\right\}dt+
\left(\frac{\beta}{t}\alpha t\right)dx$
%tpdhere  this should use an asterisk, a number looks like a superscript
\setcounter{footnote}{1}%% use a dagger because number looks like superscript
\footnote{
Here we have opportunity for a remark (very simple indeed, but)
of the utmost importance. {\sl We are not to equate separately to zero the
coefficients of dt and dx}; for we must remember that this equation is
of the form
$$0=p\alpha + q\beta$$
where $p$ and $q$ are numbers; and that, so long as $\alpha$ and $\beta$
are actual and nonparallel vectors, the existence of such an equation
requires (\S 24)}

%tpdhere this is a cheesy use of \leqno. figure out how to do it right.
\noindent
$$\leqno{\textrm{Hence}}\hbox{\hskip 4cm}(1x)dttdx=0$$
%tpdhere  this text comes out little.
$$\leqno{\textrm{and}}\hbox{\hskip 4cm}\frac{x}{t^2}dt+\frac{dx}{t}=0$$
From these, at once, $x = \frac{1}{2}$, since $dx$ and $dt$ are indeterminate.
Thus the equation of the envelope is
$$\begin{array}{rcl}
\rho & = & \alpha t + \frac{1}{2}\left(\frac{\beta}{t}\alpha t\right)\\
 & = & \frac{1}{2}\left(\alpha t + \frac{\beta}{t}\right)
\end{array}$$
the hyperbola as before; $\alpha$, $\beta$ being portions of its asymptotes.

{\bf 42}. It may assist the student to a thorough comprehension
of the above process, if we put it in a slightly different form.
Thus the equation of the enveloping line may be written
$$\rho=\alpha t(1x)+\beta\frac{x}{t}$$
which gives
$$d\rho = 0 = \alpha d\{t(1x)\}+\beta d\left(\frac{x}{t}\right)$$
Hence, as $\alpha$ is not parallel to $\beta$, we must have
$$d\{t(1x)\}=0,\hbox{\hskip 1cm}d\left(\frac{x}{t}\right)=0$$
and these are, when expanded, the equations we obtained in the
preceding section.

{\bf 43}. For farther illustration we give a solution not directly
employing the differential calculus. The equations of any two of
the enveloping lines are
$$\rho=\alpha t + x\left(\frac{\beta}{t}\alpha t\right)$$
$$\rho=\alpha t_1 + x_1\left(\frac{\beta}{t_1}\alpha t_1\right)$$
$t$ and $t_1$ being given, while $x$ and $x_1$ are indeterminate.

At the point of intersection of these lines we have (\S 26),
$$
\left.
\begin{array}{rcl}
t(1x) & = & t_1(1x_1)\\
\frac{x}{t} & = & \frac{x_1}{t_1}
\end{array}
\right\}$$
These give, by eliminating $x_1$
$$t(1x)=t_1\left(1\frac{t_1}{t}x\right)$$
$$\leqno{\textrm{or}}\hbox{\hskip 4cm}x=\frac{t}{t_1+t}$$
Hence the vector of the point of intersection is
$$\rho=\frac{\alpha tt_1+\beta}{t_1+t}$$
and thus, for the ultimate intersections, where $L\frac{t_1}{t}=1$,
$$\rho=\frac{1}{2}\left(\alpha t+\frac{\beta}{t}\right)
\textrm{ as before }$$
COR. If. instead of the {\sl ultimate} intersections, we consider
the intersections of pairs of these lines related by some law, we
obtain useful results. Thus let
$$tt_1 = 1$$
$$\rho=\frac{\alpha+\beta}{t+\frac{1}{t}}$$
or the intersection lies in the diagonal of the parallelogram on
$\alpha$, $\beta$.

If $t_1=mt$, where $m$ is constant,
$$\rho=\frac{mt\alpha+\frac{\beta}{t}}{m+1}$$
But we have also $x=\frac{1}{m+1}$

Hence {\sl the locus of a point which divides in a given ratio a line
cutting off a given area from two fixed axes, is a hyperbola of which
these axes are the asymptotes}.

If we take either
$$tt_1(t+t_1)=\textrm{constant, or }
\frac{t^2t_1^2}{t+t_1}=\textrm{constant}$$
the locus is a parabola; and so on.

It will be excellent practice for the student, at this stage, to
work out in detail a number of similar questions relating to the
envelope of, or the locus of the intersection of selected pairs from, a
series of lines drawn according to a given law. And the process
may easily be extended to planes. Thus, for instance, we may
form the general equation of planes which cut off constant tetrahedra
from the axes of coordinates. Their envelope is a surface of
the third degree whose equation may be written
$$\rho=x\alpha+y\beta+z\gamma$$
$$\leqno{\textrm{where}}\hbox{\hskip 4cm}xyz=\alpha^3$$

Again, find the locus of the point of intersection of three of
this group of planes, such that
the first intercepts on $\beta$ and $\gamma$,
the second on $\gamma$ and $\alpha$,
the third on $\alpha$ and $\beta$, lengths all equal to one
another, \&c. But we must not loiter with such simple matters as
these.

{\bf 44}. The reader who is fond of Anharmonic Ratios and Trans
versals will find in the early chapters of Hamilton's {\sl Elements of
Quaternions} an admirable application of the composition of vectors
to these subjects. The Theory of Geometrical Nets, in a plane,
and in space, is there very fully developed; and the method is
shown to include, as particular cases, the corresponding processes of
Grassmann's {\sl Ausdehnungslehre} and M\"obius' {\sl Barycentrische Calcul}.
Some very curious investigations connected with curves and surfaces
of the second and third degrees are also there founded upon the
composition of vectors.

\section{Examples To Chapter 1.}

1. The lines which join, towards the same parts, the extremities
of two equal and parallel lines are themselves equal and parallel.
({\sl Euclid}, I. xxxiii.)

2. Find the vector of the middle point of the line which joins
the middle points of the diagonals of any quadrilateral, plane or
gauche, the vectors of the corners being given; and so prove that
this point is the mean point of the quadrilateral.

If two opposite sides be divided proportionally, and two new
quadrilaterals be formed by joining the points of division, the mean
points of the three quadrilaterals lie in a straight line.

Show that the mean point may also be found by bisecting the
line joining the middle points of a pair of opposite sides.

3. Verify that the property of the coefficients of three vectors
whose extremities are in a line (\S 30) is not interfered with by
altering the origin.

4. If two triangles $ABC$, $abc$, be so situated in space that $Aa$,
$Bb$, $Cc$ meet in a point, the intersections of $AB$, $ab$, of $BC$, $bc$,
and of $CA$, $ca$, lie in a straight line.

5. Prove the converse of 4, i.e. if lines be drawn, one in each
of two planes, from any three points in the straight line in which
these planes meet, the two triangles thus formed are sections of a
common pyramid.

6. If five quadrilaterals be formed by omitting in succession
each of the sides of any pentagon, the lines bisecting the diagonals
of these quadrilaterals meet in a point. (H. Fox Talbot.)

7. Assuming, as in \S 7, that the operator
$$\cos\theta + \sqrt{1}\sin\theta$$
turns any radius of a given circle through an angle $\theta$ in the
positive direction of rotation, without altering its length, deduce
the ordinary formulae for $\cos(A+B)$, $\cos(AB)$, $\sin(A+B)$, and
$\sin(AB)$, in terms of sines and cosines of $A$ and $B$.

8. If two tangents be drawn to a hyperbola, the line joining
the centre with their point of intersection bisects the lines join
ing the points where the tangents meet the asymptotes : and the
secant through the points of contact bisects the intercepts on
the asymptotes.

9. Any two tangents, limited by the asymptotes, divide each
other proportionally.

10. If a chord of a hyperbola be one diagonal of a parallelogram
whose sides are parallel to the asymptotes, the other diagonal passes
through the centre.

11. Given two points $A$ and $B$, and a plane, $C$. Find the
locus of $P$, such that if $AP$ cut $C$ in $Q$, and $BP$ cut $C$ in $R$,
$\overline{QR}$ may be a given vector.

12. Show that\hbox{\hskip 1cm} $\rho = x^2\alpha+y^2\beta+(x+y)^2\gamma$\\
is the equation of a cone of the second degree, and that its section
by the plane
$$\rho=\frac{p\alpha+q\beta+r\gamma}{p+q+r}$$
is an ellipse which touches, at their middle points, the sides of the
triangle of whose corners $\alpha$, $\beta$, $\gamma$
are the vectors. (Hamilton, {\sl Elements}, p. 96.)

13. The lines which divide, proportionally, the pairs of opposite
sides of a gauche quadrilateral, are the generating lines of a
hyperbolic paraboloid. ({\sl Ibid}. p. 97.)

14. Show that\hbox{\hskip 2cm} $\rho=x^3\alpha+y^3\beta+z^3\gamma$\\
where\hbox{\hskip 4cm} $x+y+z=0$\\
represents a cone of the third order, and that its section by the plane
$$\rho=\frac{p\alpha+q\beta+r\gamma}{p+q+r}$$
is a cubic curve, of which the lines
$$\rho=\frac{p\alpha+q\beta}{p+q},\textrm{ \&c}$$
are the asymptotes and the three (real) tangents of inflection. Also
that the mean point of the triangle formed by these lines is a
conjugate point of the curve. Hence that the vector $\alpha+\beta+\gamma$
is a conjugate ray of the cone. ({\sl Ibid}. p. 96.)

\chapter{Products And Quotients of Vectors}

{\bf 45}. We now come to the consideration of questions in which
the Calculus of Quaternions differs entirely from any previous
mathematical method; and here we shall get an idea of what a
Quaternion is, and whence it derives its name. These questions
are fundamentally involved in the novel use of the symbols of
multiplication and division. And the simplest introduction to
the subject seems to be the consideration of the quotient, or ratio,
of two vectors.

{\bf 46}. If the given vectors be parallel to each other, we have
already seen (\S 22) that either may be expressed as a numerical
multiple of the other; the multiplier being simply the ratio of
their lengths, taken positively if they have similar currency,
negatively if they run opposite ways.

{\bf 47}. If they be not parallel, let $\overline{OA}$ and
$\overline{OB}$ be drawn parallel
and equal to them from any point $O$; and the question is reduced
to finding the value of the ratio of two vectors drawn from the
same point. Let us first find {\sl upon how many distinct numbers this
ratio depends}.

We may suppose $\overline{OA}$ to be changed into
$\overline{OB}$ by the following successive processes.

1st. Increase or diminish the length of $\overline{OA}$ till it becomes
equal to that of $\overline{OB}$. For this only one number is required, viz.
the ratio of the lengths of the two vectors. As Hamilton remarks,
this is a positive, or rather a {\sl signless}, number.

2nd. Turn $\overline{OA}$ about $O$, in the common plane of the two
vectors, until its direction coincides with that of $\overline{OB}$, and
(remembering the effect of the first operation) we see that the two vectors
now coincide or become identical. To specify this operation three
numbers are required, viz. two angles (such as node and inclination
in the case of a planet's orbit) to fix the plane in which the rotation
takes place, and {\sl one} angle for the amount of this rotation.

Thus it appears that the ratio of two vectors, or the multiplier
required to change one vector into another, in general depends upon
{\sl four} distinct numbers, whence the name QUATERNION.

A quaternion q is thus {\sl defined} as expressing a relation
$$\beta=q\alpha$$
between two vectors $\alpha$, $\beta$.
By what precedes, the vectors $\alpha$, $\beta$,
which serve for the definition of a given quaternion, must be in a
given plane, at a given inclination to each other, and with their
lengths in a given ratio ; but it is to be noticed that they may be
{\sl any} two such vectors. [{\sl Inclination} is understood to include sense,
or currency, of rotation from $\alpha$ to $\beta$.]

The particular case of perpendicularity of the two vectors, where
their quotient is a vector perpendicular to their plane, is fully
considered below; \S\S 64, 65, 72, \&c.

{\bf 48}. It is obvious that the operations just described may be
performed, with the same result, in the opposite order, being perfectly
independent of each other. Thus it appears that a quaternion,
considered as the factor or agent which changes one definite vector
into another, may itself be decomposed into two factors of which
the order is immaterial.

The {\sl stretching} factor, or that which performs the first operation
in \S 47, is called the TENSOR, and is denoted by prefixing $T$ to the
quaternion considered.

The {\sl turning factor}, or that corresponding to the second operation
in \S 47, is called the VERSOR, and is denoted by the letter $U$ prefixed
to the quaternion.

{\bf 49}. Thus, if $\overline{OA} = \alpha$,
$\overline{OB} = \beta$, and if $q$ be the quaternion
which changes $\alpha$ to $\beta$, we have
$$\beta = q\alpha$$
which we may write in the form
$$\frac{\beta}{\alpha} = q\textrm{, or }\beta\alpha^{1}=q$$
if we agree to {\sl define} that
$$\frac{\beta}{\alpha}\alpha = \beta\alpha^{1}\alpha = \beta$$
Here it is to be particularly noticed that we write $q$
{\sl before} $\alpha$ to
signify that $\alpha$ is multiplied by (or operated on by) $q$, not $q$
multiplied by $\alpha$.

This remark is of extreme importance in quaternions, for, as we
shall soon see, the Commutative Law does not generally apply to
the factors of a product.

We have also, by \S\S 47, 48,
$$q=TqUq=UqTq$$
where, as before, $Tq$ depends merely on the relative lengths of
$\alpha$ and $\beta$, and $Uq$ depends solely on their directions.

Thus, if $\alpha_1$ and $\beta_1$
be vectors of unit length parallel to $\alpha$ and $\beta$
respectively,
$$T\frac{\beta_1}{\alpha_1} = T\beta_1/T\alpha_1 = 1\textrm{, }
U\frac{\beta_1}{\alpha_1} = U\beta_1/U\alpha_1 = U\frac{\beta}{\alpha}$$
As will soon be shown, when $\alpha$ is perpendicular to $\beta$,
i.e. when the versor of the quotient is quadrantal, it is a unitvector.

{\bf 50}. We must now carefully notice that the quaternion which
is the quotient when $\beta$ is divided by $\alpha$ in no way depends upon
the {\sl absolute} lengths, or directions, of these vectors. Its value
will remain unchanged if we substitute for them any other pair
of vectors which

\noindent
\hbox{\hskip 2cm}(1) have their lengths in the same ratio,

\noindent
\hbox{\hskip 2cm}(2) have their common plane the same or parallel,

\noindent
and\hbox{\hskip 1.32cm} (3) make the same angle with each other.

Thus in the annexed figure

\includegraphics{ps/quat6.ps}

$$\frac{O_1B_1}{O_1A_1} = \frac{\overline{OB}}{\overline{OA}}$$
if, and only if,

\noindent
\hbox{\hskip 2cm}$(1)\textrm{ }\frac{O_1B_1}{O_1A_1} = \frac{OB}{OA}$

\noindent
\hbox{\hskip 2cm}$(2)\textrm{ plane }AOB\textrm{ parallel to plane }
A_1O_1B_1$

\noindent
\hbox{\hskip 2cm}$(3)\textrm{ }\angle{}AOB = \angle A_1O_1B_1$

[Equality of angles is understood to include
concurrency of rotation. Thus in the annexed
figure the rotation about an axis drawn upwards
from the plane is negative (or clock wise) from
$OA$ to $OB$, and also from $O_1A_1$ to $O_1B_1$.]

It thus appears that if
$$\beta = q\alpha\textrm{, }\delta = q\gamma$$
the vectors $\alpha$, $\beta$, $\gamma$, $\delta$
are parallel to one plane, and may be repre
sented (in a highly extended sense) as {\sl proportional} to one another,
thus: 
$$\beta : \alpha = \delta : \gamma$$

And it is clear from the previous part of this section that this
may be written not only in the form
$$\alpha : \beta = \gamma : \delta$$
but also in either of the following forms: 
$$\gamma : \alpha = \delta : \beta$$
$$\alpha : \gamma = \beta : \delta$$

While these proportions are true as equalities of ratios, they
do not usually imply equalities of products.

Thus, as the first of these was equivalent to the equation
$$\frac{\beta}{\alpha}=\frac{\delta}{\gamma}=q\textrm{, or }
\beta\alpha^{1}=\delta\gamma^{1}=q$$
the following three imply separately, (see next section)
$$\frac{\alpha}{\beta}=\frac{\gamma}{\delta}=q^{1}\textrm{, }
\frac{\gamma}{\alpha}=\frac{\delta}{\beta}=r\textrm{, }
\frac{\alpha}{\gamma}=\frac{\beta}{\delta}=r^{1}$$
or, if we please,
$$\alpha\beta^{1}=\gamma\delta^{1}=q^{1}\textrm{, }
\gamma\alpha^{1}=\delta\beta^{1}=r\textrm{, }
\alpha\gamma^{1}=\beta\delta^{1}=r^{1}$$
where $r$ is a {\sl new} quaternion, which has not necessarily anything
(except its plane), in common with $q$.

But here great caution is requisite, for we are {\sl not} entitled to
conclude from these that
$$\alpha\delta=\beta\gamma\textrm{, \&c.}$$

This point will be fully discussed at a later stage. Meanwhile
we may merely {\sl state} that from
$$\frac{\alpha}{\beta}=\frac{\gamma}{\delta}\textrm{, or }
\frac{\beta}{\alpha} = \frac{\delta}{\gamma}$$
we are entitled to deduce a number of equivalents such as
$$\alpha\beta^{1}\delta=\gamma\textrm{, or }
\alpha=\gamma\delta^{1}\beta\textrm{, or }
\beta^{1}\delta=\alpha^{1}\gamma\textrm{, \&c}$$

{\bf 51}. The {\sl Reciprocal} of a quaternion $q$ is defined by the
equation
$$\frac{1}{q}q=q^{1}=1=q\frac{1}{q}=qqe^{1}$$
Hence if
$$\frac{\beta}{\alpha}=q\textrm{, or}$$
$$\beta=q\alpha$$
we must have
$$\frac{\alpha}{\beta}=\frac{1}{q}=q^{1}$$
For this gives
$$\frac{\alpha}{\beta}\beta=q^{1}q\alpha$$
and each member of the equation is evidently equal to $\alpha$.
Or thus: 
$$\beta=q\alpha$$
Operate {\sl by} $q^{1}$
$$q^{1}\beta = \alpha$$
Operate {\sl on} $\beta^{1}$
$$q^{1} = \alpha\beta^{1} = \frac{\alpha}{\beta}$$

Or, we may reason thus:  since $q$ changes $\overline{OA}$ to
$\overline{OA}$, $q^{1}$ must
change $\overline{OB}$ to $\overline{OA}$,
and is therefore expressed by $\frac{\alpha}{\beta}$ (\S 49).

The tensor of the reciprocal of a quaternion is therefore the
reciprocal of the tensor; and the versor differs merely by the
{\sl reversal} of its representative angle. The versor, it must be
remembered, gives the plane and angle of the turning  it has
nothing to do with the extension.

[{\sl Remark}. In \S\S 4951, above, we had such expressions as
$\frac{\beta}{\alpha}=\beta\alpha^{1}$.
We have also met with $\alpha^{1}\beta$. Cayley suggests that this
also may be written in the ordinary fractional form by employing
the following distinctive notation: 
$$\frac{\beta}{\alpha}=\beta\alpha^{1}=\frac{\beta}{\alpha}\textrm{, }
\alpha^{1}\beta=\frac{\beta}{\alpha}$$

(It might, perhaps, be even simpler to use the {\sl solidus} as
recommended by Stokes, along with an obviously correlative
type: thus,
$$\frac{\beta}{\alpha}=\beta\alpha^{1}=\beta/\alpha\textrm{, }
\alpha^{1}\beta=\alpha\\ \beta$$

I have found such notations occasionally convenient for private
work, but I hesitate to introduce changes unless they are abso
lutely required. See remarks on this point towards the end of the
{\sl Preface to the Second Edition} reprinted above.]

{\bf 52}. The {\sl Conjugate} of a quaternion $q$, written $Kq$, has the
same tensor, plane, and angle, only the angle is taken the reverse
way; or the versor of the conjugate is the reciprocal of the versor
of the quaternion, or (what comes to the same thing) the versor of
the reciprocal.

\includegraphics{ps/quat7.ps}

Thus, if $OA$, $OB$, $OA^{'}$ , lie in one plane, and if
$OA^{'} = OA$, and $\angle A^{'}OB = \angle BOA$, we have
$$\frac{\overline{OB}}{\overline{OA}}=q$$,
and
$$\frac{\overline{OB}}{\overline{OA^{'}}}=\textrm{ congugate of }q=Kq$$

By last section we see that
$$Kq=(Tq)^2q^{1}$$
Hence\hbox{\hskip 4cm}$qKq=Kqq=(Tq)^2$

This proposition is obvious, if we recollect that
the tensors of $q$ and $Kq$ are equal, and that the
versors are such that either {\sl annuls} the effect of the other; while
the order of their application is indifferent. The joint effect of
these factors is therefore merely to multiply twice over by the
common tensor.

{\bf 53}. It is evident from the results of \S 50 that, if $\alpha$ and $\beta$
be of equal length, they may be treated as of unitlength so far as
their quaternion quotient is concerned. This quotient is therefore
a versor (the tensor being unity) and may be represented indifferently
by any one of an infinite number of concurrent arcs of
given length lying on the circumference of a circle, of which the
two vectors are radii. This is of considerable importance in the
proofs which follow.

\includegraphics{ps/quat8.ps}

Thus the versor
${\displaystyle \frac{\overline{OB}}{\overline{OA}}}$ may be represented
in magnitude, plane, and currency of rotation (\S 50)
by the arc $AB$, which may in this extended sense be written
${\stackrel{\frown}{AB}}$.

And, similarly, the versor
${\displaystyle \frac{\overline{OB_1}}{\overline{OA_1}}}$
may be represented by
${\stackrel{\frown}{A_1B_1}}$
which is equal to (and concurrent with)
${\stackrel{\frown}{AB}}$ if
$$\angle A_1OB_1 = \angle AOB$$
i.e. if the versors are {\sl equal}, in the quaternion meaning of the
word.

{\bf 54}. By the aid of this process, when a versor is represented as
an arc of a great circle on the unitsphere, we can easily prove
that {\sl quaternion multiplication is not generally commutative}.

\includegraphics{ps/quat9.ps}

Thus let $q$ be the versor ${\stackrel{\frown}{AB}}$ or
${\displaystyle \frac{\overline{OB}}{\overline{OA}}}$,
where $O$ is the centre of the sphere.

Take ${\stackrel{\frown}{BC}} = {\stackrel{\frown}{AB}}$,
(which, it must be remembered, makes the points $A$, $B$, $C$, lie
in one great circle), then $q$ may also be
represented by ${\displaystyle \frac{\overline{OC}}{\overline{OB}}}$.

In the same way any other versor $r$ may be represented by
${\stackrel{\frown}{DB}}$ or ${\stackrel{\frown}{BE}}$ and by
${\displaystyle \frac{\overline{OB}}{\overline{OD}}}$ or
${\displaystyle \frac{\overline{OE}}{\overline{OB}}}$.

[The line $OB$ in the figure is definite, and is given by the
intersection of the planes of the two versors.]

Now $r\overline{OD} = \overline{OB}$, and $q\overline{OB}=\overline{OC}$.

Hence $qr\overline{OD} = \overline{OC}$,

or $qr = {\displaystyle \frac{\overline{OC}}{\overline{OD}}}$,
and may therefore be represented by the arc ${\stackrel{\frown}{DC}}$ of
a great circle.

But $rq$ is easily seen to be represented by the arc
${\stackrel{\frown}{AE}}$.

For $q\overline{OA}=\overline{OB}$, and $r\overline{OB}=\overline{OE}$,

whence $rq\overline{OA} = \overline{OE}$. and
$rq = {\displaystyle \frac{\overline{OE}}{\overline{OA}}}$.

Thus the versors $rq$ and $qr$, though represented by arcs of equal
length, are not generally in the same plane and are therefore
unequal: unless the planes of $q$ and $r$ coincide.

Remark. We see that we have assumed, or defined, in the
above proof, that $q . r\alpha = qr . \alpha$.
and $r.q\alpha = rq.\alpha$ in the special case
when $q\alpha$, $r\alpha$, $q.r\alpha$ and $r.q\alpha$ are all {\sl vectors}.

{\bf 55}. Obviously ${\stackrel{\frown}{CB}}$ is $Kq$,
${\stackrel{\frown}{BD}}$ is $Kr$, and
${\stackrel{\frown}{CD}}$ is $K (qr)$. But
${\stackrel{\frown}{CD}} = {\stackrel{\frown}{BD}}.{\stackrel{\frown}{CB}}$
as we see by applying both to OC. This gives us
the very important theorem
$$K (qr) = Kr . Kq$$
i.e. {\sl the conjugate of the product of two versors is the product of their
conjugates in inverted order}. This will, of course, be extended to
any number of factors as soon as we have proved the associative
property of multiplication. (\S 58 below.)

{\bf 56}. The propositions just proved are, of course, true of quater
nions as well as of versors; for the former involve only an additional
numerical factor which has reference to the length merely, and not
the direction, of a vector (\S 48), and is therefore commutative with
all other factors.

{\bf 57}. Seeing thus that the commutative law does not in general
hold in the multiplication of quaternions, let us enquire whether
the Associative Law holds generally. That is if $p$, $q$, $r$ be three
quaternions, have we
$$p.qr = pq.r?$$

This is, of course, obviously true if $p$, $q$, $r$ be numerical quantities,
or even any of the imaginaries of algebra. But it cannot be con
sidered as a truism for symbols which do not in general give

$$pq = qp$$

We have assumed it, in definition, for the special case when $r$,
$qr$, and $pqr$ are all vectors. (\S 54.) But we are not entitled to
assume any more than is absolutely required to make our
definitions complete.

{\bf 58}. In the first place we remark that $p$, $q$, and $r$ may be
considered as versors only, and therefore represented by arcs of
great circles on the unit sphere, for their tensors may obviously
(\S 48) be divided out from both sides, being commutative with the
versors.

Let ${\stackrel{\frown}{AB}}=p$,
${\stackrel{\frown}{ED}} = {\stackrel{\frown}{CA}} = q$, and
${\stackrel{\frown}{FE}} = r$.

Join $BC$ and produce the great circle till it meets $EF$ in $H$, and
make ${\stackrel{\frown}{KH}}={\stackrel{\frown}{FE}} = r$,
and ${\stackrel{\frown}{HG}} = {\stackrel{\frown}{CB}} =pq$ (\S 54).

\includegraphics{ps/quat10.ps}

Join $GK$. Then
${\stackrel{\frown}{KG}} =
{\stackrel{\frown}{HG}} . {\stackrel{\frown}{KH}} = pq . r$.

Join $FD$ and produce it to meet $AB$ in $M$. Make
$${\stackrel{\frown}{LM}} = {\stackrel{\frown}{FD}}
\textrm{, and }{\stackrel{\frown}{MN}} = {\stackrel{\frown}{AB}}$$

and join $NL$. Then
$${\stackrel{\frown}{LN}}= {\stackrel{\frown}{MN}} .
{\stackrel{\frown}{LM}}=p.qr$$.

Hence to show that $p . qr = pq . r$

all that is requisite is to prove that $LN$, and $KG$, described as
above, are {\sl equal arcs of the same great circle}, since, by the figure,
they have evidently similar currency. This is perhaps most easily
effected by the help of the fundamental properties of the curves
known as {\sl Spherical Conics}. As they are not usually familiar to
students, we make a slight digression for the purpose of proving
these fundamental properties ; after Chasles, by whom and Magnus
they were discovered. An independent proof of the associative
principle will presently be indicated, and in Chapter VIII. we shall
employ quaternions to give an independent proof of the theorems
now to be established.

{\bf 59}.* DEF. {\sl A spherical conic is the curve of intersection of a
cone of the second degree with a sphere, the vertex of the cone being
the centre of the sphere}.

LEMMA. If a cone have one series of circular sections, it has
another series, and any two circles belonging to different series lie
on a sphere. This is easily proved as follows.

Describe a sphere, $A$, cutting the cone in one circular section,
$C$, and in any other point whatever, and let the side $OpP$ of the
cone meet $A$ in $p$, $P$ ; $P$ being a point in $C$. Then $PO.Op$ is
constant, and, therefore, since $P$ lies in a plane, $p$ lies on a sphere,
$a$, passing through $0$. Hence the locus, $c$, of $p$ is a circle, being
the intersection of the two spheres $A$ and $a$.

Let $OqQ$ be any other side of the cone, $q$ and $Q$ being points in
$c$, $C$ respectively. Then the quadrilateral $qQPp$ is inscribed in a
circle (that in which its plane cuts the sphere $A$) and the exterior

\includegraphics{ps/quat11.ps}

angle at $p$ is equal to the interior angle at $Q$. If $OL$, $OM$ be the
lines in which the plane $POQ$ cuts the {\sl cyclic planes} (planes through
$O$ parallel to the two series of circular sections) they are obviously
parallel to $pq$, $QP$, respectively; and therefore

$$\angle LOp = \angle Opq = \angle OQP = \angle MOQ$$

Let any third side, $OrR$, of the cone be drawn, and let the
plane $OPR$ cut the cyclic planes in $0l$, $Om$ respectively. Then,
evidently,
$$\angle lOL = \angle qpr$$
$$\angle MOm = \angle QPR$$

and these angles are independent of the position of the points $p$
and $P$, if $Q$ and $R$ be fixed points.

\includegraphics{ps/quat12.ps}

In the annexed section of the above spacediagram by a sphere
whose centre is $O$, $lL$, $Mm$ are the great circles which represent
the cyclic planes, $PQR$ is the spherical conic which represents the
cone. The point $P$ represents the line $OpP$, and so with the
others. The propositions above may now be stated thus,

$$\textrm{Arc } PL = \textrm{arc } MQ$$

and, if $Q$ and $R$ be fixed, $Mm$ and $lL$ are constant arcs whatever be
the position of $P$.

{\bf 60}. The application to \S 58 is now obvious. In the figure of
that article we have
$$
{\stackrel{\frown}{FE}}={\stackrel{\frown}{KH}}\textrm{, }
{\stackrel{\frown}{ED}}={\stackrel{\frown}{CA}}\textrm{, }
{\stackrel{\frown}{HG}}={\stackrel{\frown}{CB}}\textrm{, }
{\stackrel{\frown}{LM}}={\stackrel{\frown}{FD}}
$$

Hence $L$, $C$, $G$, $D$ are points of a spherical conic whose cyclic
planes are those of $AB$, $FE$. Hence also $KG$ passes through $L$,
and with $LM$ intercepts on $AB$ an arc equal to
${\stackrel{\frown}{AB}}$. That is, it
passes through $N$, or $KG$ and $LN$ are arcs of the same great circle :
and they are equal, for $G$ and $L$ are points in the spherical
conic.

Also, the associative principle holds for any number of
quaternion factors. For, obviously,

$$qr . st = qrs . t = \textrm{\&c., \&c.,}$$

since we may consider $qr$ as a single quaternion, and the above
proof applies directly.

{\bf 61}. That quaternion addition, and therefore also subtraction,
is commutative, it is easy to show.

\includegraphics{ps/quat13.ps}

For if the planes of two quaternions,
$q$ and $r$, intersect in the line $OA$, we
may take any vector $\overline{OA}$ in that line,
and at once find two others, $\overline{OB}$ and
$\overline{OC}$, such that

$$\overline{OB} = q\overline{OA}$$
and\hbox{\hskip 4cm}$\overline{CO} = r\overline{OA}$

And\hbox{\hskip 2cm}$(q + r)\overline{OA}
\overline{OB}+\overline{OC}=\overline{OC}+\overline{OB}=
(r + q) \overline{OA}$

since vector addition is commutative (\S 27).

Here it is obvious that $(q + r) \overline{OA}$, being the diagonal of the
parallelogram on $\overline{OB}$, $\overline{OC}$,
divides the angle between $OB$ and $OC$
in a ratio depending solely on the ratio of the lengths of these
lines, i.e. on the ratio of the tensors of $q$ and $r$. This will be useful
to us in the proof of the distributive law, to which we proceed.

{\bf 62}. Quaternion multiplication, and therefore division, is
distributive. One simple proof of this depends on the possibility,
shortly to be proved, of representing {\sl any} quaternion as a linear
function of three given rectangular unit vectors. And when the
proposition is thus established, the associative principle may readily
be deduced from it.

[But Hamilton seems not to have noticed that we may employ
for its proof the properties of Spherical Conies already employed

\includegraphics{ps/quat14.ps}

in demonstrating the truth of the associative principle. "For
continuity we give an outline of the proof by this process.

Let ${\stackrel{\frown}{BA}}$,
${\stackrel{\frown}{CA}}$
represent the versors of $q$ and $r$, and be the great
circle whose plane is that of $p$.

Then, if we take as operand the vector $\overline{OA}$, it is obvious that
$U (q + r)$ will be represented by some such arc as
${\stackrel{\frown}{DA}}$ where
$B$, $D$, $C$ are in one great circle;
for $(q + r) \overline{OA}$ is in the same plane
as $q\overline{OA}$ and $r\overline{OA}$,
and the relative magnitude of the arcs $BD$ and
$DC$ depends solely on the tensors of $q$ and $r$. Produce $BA$, $DA$,
$CA$ to meet be in $b$, $d$, $c$ respectively, and make

$${\stackrel{\frown}{Eb}} = {\stackrel{\frown}{BA}},
{\stackrel{\frown}{Fd}} = {\stackrel{\frown}{DA}},
{\stackrel{\frown}{Gc}} = {\stackrel{\frown}{CA}}$$

Also make
${\stackrel{\frown}{b\beta}} =
{\stackrel{\frown}{d\delta}} =
{\stackrel{\frown}{c\gamma}}=p$. Then $E$, $F$, $G$, $A$ lie on a spherical
conic of which $BC$ and $bc$ are the cyclic arcs. And, because
${\stackrel{\frown}{b\beta}} =
{\stackrel{\frown}{d\delta}} =
{\stackrel{\frown}{c\gamma}}$,
${\stackrel{\frown}{\beta E}}$,
${\stackrel{\frown}{\delta F}}$,
${\stackrel{\frown}{\gamma G}}$, when produced, meet in a point $H$
which is also on the spherical conic (\S 59*). Let these arcs meet $BC$
in $J$, $L$, $K$ respectively. Then we have
$${\stackrel{\frown}{JH}}={\stackrel{\frown}{E\beta}}=pUq$$
$${\stackrel{\frown}{LH}}={\stackrel{\frown}{F\delta}}=pU(q+r)$$
$${\stackrel{\frown}{KH}}={\stackrel{\frown}{G\gamma}}=pUr$$
Also\hbox{\hskip 4cm}${\stackrel{\frown}{LJ}}={\stackrel{\frown}{DB}}$\\
and\hbox{\hskip 4cm}${\stackrel{\frown}{KL}}={\stackrel{\frown}{CD}}$

And, on comparing the portions of the figure bounded respectively
by $HKJ$ and by $ACB$ we see that (when considered with reference
to their effects as factors multiplying
$\overline{OH}$ and $\overline{OA}$ respectively)

\hbox{\hskip 2cm}$pU(q4+r)$ bears the same relation to $pUq$ and $pUr$\\
that\hbox{\hskip 1cm}$U(q+r)$ bears to $Uq$ and $Ur$.\\
But\hbox{\hskip 1cm}$T(q+r)U(q+r)=q+r=TqUq+TrUr$. \\
Hence\hbox{\hskip 1cm}$T(q+r).pU(q+r)=Tq .pUq+Tr.pUr$;\\
or, since the tensors are mere numbers and commutative with all
other factors,
$$p(q+r)=pq+pr$$
In a similar manner it may be proved that
$$(q+ )p=qp+rp$$
And then it follows at once that
$$(p + q) (r + s) = pr + ps + qr + qs$$
where, by \S 61, the order of the partial products is immaterial.]

{\bf 63}. By similar processes to those of \S 53 we see that versors,
and therefore also quaternions, are subject to the indexlaw
$$q^m.q^n=q^{m+n}$$
at least so long as $m$ and $n$ are positive integers.

The extension of this property to negative and fractional
exponents must be deferred until we have defined a negative or
fractional power of a quaternion.

{\bf 64}. We now proceed to the special case of {\sl quadrantal} versors,
from whose properties it is easy to deduce all the foregoing
results of this chapter. It was, in fact, these properties whose
invention by Hamilton in 1843 led almost intuitively to the
establishment of the Quaternion Calculus. We shall content
ourselves at present with an assumption, which will be shown
to lead to consistent results ; but at the end of the chapter we
shall show that no other assumption is possible, following for this
purpose a very curious quasimetaphysical speculation of Hamilton.

{\bf 65}. Suppose we have a system of three mutually perpendicular
unitvectors, drawn from one point, which we may call for shortness
{\bf i}, {\bf j}, {\bf k}.
Suppose also that these are so situated that a positive
(i.e. {\sl lefthanded}) rotation through a right angle about {\bf i}
as an axis
brings {\bf j} to coincide with {\bf k}. Then it is obvious that positive
quadrantal rotation about {\bf j} will make {\bf k} coincide with {\bf i};
and, about {\bf k}, will make {\bf i} coincide with {\bf j}.

For defniteness we may suppose {\bf i} to be drawn {\sl eastwards}, {\bf j}
{\sl northwards}, and {\bf k} {\sl upwards}.
Then it is obvious that a positive
(lefthanded) rotation about the eastward line ({\bf i}) brings the northward
line ({\bf j}) into a vertically upward position ({\bf k}) ; and so of the
others.

{\bf 66}. Now the operator which turns {\bf j} into {\bf k} is a quadrantal
versor (\S 53) ; and, as its axis is the vector {\bf i},
we may call it {\sl i}.

Thus $$\frac{{\rm {\bf k}}}{{\rm {\bf j}}}=
i\textrm{, or }{\rm {\bf k}}=i{\rm {\bf j}}\eqno{(1)}$$

Similary we may put$$\frac{{\rm {\bf i}}}{{\rm {\bf k}}}=
j\textrm{, or }{\rm {\bf i}}=j{\rm {\bf k}}\eqno{(2)}$$

and $$\frac{{\rm {\bf j}}}{{\rm {\bf i}}}=
k\textrm{, or }{\rm {\bf j}}=k{\rm {\bf i}}\eqno{(3)}$$

[It may be here noticed, merely to show the symmetry of the
system we arc explaining, that if the three mutually perpendicular
vectors {\bf i}, {\bf j}, {\bf k}
be made to revolve about a line equally inclined to
all, so that {\bf i} is brought to coincide with {\bf j},
{\bf j} will then coincide
with {\bf k}, and {\bf k} with {\bf i}:
and the above equations will still hold good,
only (1) will become (2), (2) will become (3), and (3) will become (1).]

{\bf 67}. By the results of \S 50 we see that
$$\frac{{\rm {\bf j}}}{\rm {\bf k}}=\frac{{\rm {\bf k}}}{\rm {\bf j}}$$
i.e. a southward unit vector bears the same ratio to an upward
unitvector that the latter does to a northward one; and therefore
we have

Thus $$\frac{{\rm {\bf j}}}{{\rm {\bf k}}}=
i\textrm{, or }{\rm {\bf j}}=i{\rm {\bf k}}\eqno{(4)}$$

Similary t$$\frac{{\rm {\bf k}}}{{\rm {\bf i}}}=
j\textrm{, or }{\rm {\bf k}}=j{\rm {\bf i}}\eqno{(5)}$$

and $$\frac{{\rm {\bf i}}}{{\rm {\bf j}}}=
k\textrm{, or }{\rm {\bf i}}=k{\rm {\bf j}}\eqno{(6)}$$

{\bf 68}. By (4) and (1) we have

$$j = ik = i(ij)\textrm{ (by the assumption in \S 54) }= i^2j$$

Hence
$$i^2 =  1\eqno{(7)}$$

Arid in the same way, (5) and (2) give
$$j^2=1\eqno{(8)}$$
and (6) and (3)
$$k^2=1\eqno{(9)}$$

Thus, as the directions of {\bf i}, {\bf j}, {\bf k}
are perfectly arbitrary, we see that
{\sl the square of every quadrantal versor is negative unity}.

[Though the following proof is in principle exactly the same as
the foregoing, it may perhaps be of use to the student, in showing
him precisely the nature as well as the simplicity of the step we
have taken.

\includegraphics{ps/quat15.ps}

Let $ABA^{'}$ be a semicircle, whose centre
is $0$, and let $OB$ be perpendicular to $AOA^{'}$.

Then ${\displaystyle\frac{\overline{OB}}{\overline{OA^{'}}}}=q$
suppose, is a quadrantal versor, and is evidently equal to
${\displaystyle\frac{\overline{OA^{'}}}{\overline{OB}}}$ ;

\S\S 50, 53. Hence

$$q^2=\frac{\overline{OA^{'}}}{\overline{OB}}.
\frac{\overline{OB}}{\overline{OA}}=
\frac{\overline{OA^{'}}}{\overline{OA}}=1]$$

{\bf 69}.
Having thus found that the squares of {\sl i}, {\sl j}, {\sl k} are each
equal to negative unity ; it only remains that we find the values of
their products two and two. For, as we shall see, the result is such
as to show that the value of any other combination whatever of
{\sl i},{\sl j}, {\sl k}
(as factors of a product) may be deduced from the values of
these squares and products.

Now it is obvious that
$$\frac{{\rm {\bf k}}}{\rm {\bf i}}=
\frac{{\rm {\bf i}}}{\rm {\bf k}}=j$$
(i.e. the versor which turns a westward unitvector into an upward
one will turn the upward into an eastward unit) ; or
$${\rm {\bf k}}=j({\rm {\bf i}}) = j{\rm {\bf i}}\eqno{(10)}$$

Now let us operate on the two equal vectors in (10) by the
same versor, {\sl i}, and we have
$$i{\rm {\bf k}} = i(j{\rm {\bf i}}) = j{\rm {\bf i}}$$
But by (4) and (3)
$$i{\rm {\bf k}}={\rm {\bf j}}=k{\rm {\bf i}}$$

Comparing these equations, we have
$$ij{\rm {\bf i}}=k{\rm {\bf i}}$$
$$
\left.
\begin{array}{lr}
\textrm{or, \S 54 (end), } &ij=k\\
\textrm{and symmetry gives}&jk=i\\
 &ki=j\\
\end{array}
\right\}\eqno{(11)}
$$

The meaning of these important equations is very simple ; and
is, in fact, obvious from our construction in \S 54 for the multiplication
of versors ; as we see by the annexed figure, where we must
remember that {\sl i}, {\sl j}, {\sl k}
are quadrantal versors whose planes are at
right angles, so that the figure represents
a hemisphere divided into quadrantal
triangles. [The arrowheads indicate the
direction of each vector arc.]

\includegraphics{ps/quat16.ps}

Thus, to show that $ij = k$, we have,
$O$ being the centre of the sphere, $N$, $E$,
$S$, $W$ the north, east, south, and west,
and $Z$ the zenith (as in \S 65) ;

$$j\overline{OW}=\overline{OZ}$$
whence\hbox{\hskip 2cm}$ij\overline{OW}=i\overline{OZ}=
\overline{OS} = k\overline{OW}$

* The negative sign, being a mere numerical factor, is evidently commutative
with $j$ indeed we may, if necessary, easily assure ourselves of the fact
that to turn the negative (or reverse) of a vector through a right
(or indeed any) angle, is the same thing as to turn the vector through
that angle and then reverse it.

{\bf 70}. But, by the same figure,
$$i\overline{ON}=\overline{OZ}$$
whence\hbox{\hskip 1cm}$ji\overline{ON} =j\overline{OZ}
= \overline{OE} = \overline{OW} =  k\overline{ON}$.

{\bf 71}. From this it appears that
$$
\left.
\begin{array}{c}
ji = k\\
kj = i\\
ik = j\\
\end{array}
\right\}\eqno{(12)}
$$
and thus, by comparing (11),
$$
\left.
\begin{array}{c}
ij=ji=k\\
jk=kj=i\\
ki=ik=j\\
\end{array}
\right\}\eqno{(11),(12)}
$$

These equations, along with
$$i^2=j^2=k^2=1\eqno{((7),(8),(9))}$$
contain essentially the whole of Quaternions. But it is easy to see
that, for the first group, we may substitute the single equation
$$ijk=1\eqno{(13)}$$
since from it, by the help of the values of the squares of
{\sl i}, {\sl j}, {\sl k}, all
the other expressions may be deduced. We may consider it proved
in this way, or deduce it afresh from the figure above, thus
$$k\overline{ON}=\overline{OW}$$
$$jk\overline{ON}=j\overline{OW}=\overline{OZ}$$
$$ijk\overline{ON}=ij\overline{OW}=i\overline{OZ}=
\overline{OS}=\overline{ON}$$

{\bf 72}. One most important step remains to be made, to wit the
assumption referred to in \S 64. We have treated
{\sl i}, {\sl j}, {\sl k} simply as
quadrantal versors ; and
{\bf i}, {\bf j}, {\bf k} as unitvectors at right angles to
each other, and coinciding with the axes of rotation of these versors.
But if we collate and compare the equations just proved we have

\hbox{\hskip 4cm {\Huge \{}
\vbox{
\hbox{$i^2=1$\hbox{\hskip 5cm}(7)}
\hbox{${\rm {\bf i}}^2=1$\hbox{\hskip 5cm}(\S 9)}}}

\hbox{\hskip 4cm {\Huge \{}
\vbox{
\hbox{$ij=k$\hbox{\hskip 5.3cm}(11)}
\hbox{$i{\rm {\bf j}}={\rm {\bf k}}$\hbox{\hskip 5.3cm}(1)}}}

\hbox{\hskip 4cm {\Huge \{}
\vbox{
\hbox{$ji=k$\hbox{\hskip 5cm}(11)}
\hbox{$j{\rm {\bf i}}={\rm {\bf k}}$\hbox{\hskip 5cm}(1)}}}

with the other similar groups symmetrically derived from them.

Now the meanings we have assigned to {\sl i}, {\sl j}, {\sl k} are quite
independent of, and not inconsistent with, those assigned to
{\rm {\bf i}}, {\rm {\bf j}}, {\rm {\bf k}}.
And it is superfluous to use two sets of characters when one will
suffice. Hence it appears that {\sl i}, {\sl j}, {\sl k}
may be substituted for {\rm {\bf i}}, {\rm {\bf j}}, {\rm {\bf k}};
in other words, {\sl a unitvector when employed as a factor may be
considered as a quadrantal versor whose plane is perpendicular to the
vector}. (Of course it follows that every vector can be treated as the
product of a number and a quadrantal versor.) This is one of the
main elements of the singular simplicity of the quaternion calculus.

{\bf 73}. Thus {\sl the product, and therefore the quotient, of two
perpendicular vectors is a third vector perpendicular to both}.

Hence the reciprocal (\S 51) of a vector is a vector which has
the {\sl opposite} direction to that of the vector, arid its length is the
reciprocal of the length of the vector.

The conjugate (\S 52) of a vector is simply the vector reversed.

Hence, by \S 52, if $\alpha$ be a vector
$$(Ta)^2 = \alpha K\alpha = \alpha (  \alpha) = \alpha{}^2$$

{\bf 74}. We may now see that {\sl every versor may be represented by
a power of a unitvector}.

For, if $\alpha$ be any vector perpendicular to $i$ (which is
{\sl any} definite unitvector),
$i\alpha = \beta$ is a vector equal in length to $\alpha$,
but perpendicular to both $i$ and $\alpha$
$$
\begin{array}{ccl}
i^2\alpha & = & \alpha\\
\i^3\alpha & = & i\alpha = \beta\\
\i^4\alpha & = & i\beta = i^2\alpha = \alpha
\end{array}
$$
Thus, by successive applications of $i$, $\alpha$.
is turned round $i$ as an axis
through successive right angles. Hence it is natural to {\sl define}
$i^m$ {\sl as
a versor which turns any vector perpendicular to i through m right
angles in the positive direction of rotation about i as an axis}. Here
$m$ may have any real value whatever, whole or fractional, for it is
easily seen that analogy leads us to interpret a negative value of $m$
as corresponding to rotation in the negative direction.

{\bf 75}. From this again it follows that {\sl any quaternion may be
expressed as a power of a vector}. For the tensor and versor
elements of the vector may be so chosen that, when raised to the
same power, the one may be the tensor and the other the versor
of the given quaternion. The vector must be, of course, perpen
dicular to the plane of the quaternion.

{\bf 76}. And we now see, as an immediate result of the last two
sections, that the indexlaw holds with regard to powers of a
quaternion (\S 63).

{\bf 77}. So far as we have yet considered it, a quaternion has been
regarded as the {\sl product} of a tensor and a versor: we are now to
consider it as a {\sl sum}. The easiest method of so analysing it seems
to be the following.

\includegraphics{ps/quat17.ps}

Let ${\displaystyle\frac{\overline{OB}}{\overline{OA}}}$
represent any quaternion. Draw $BC$ perpendicular to $OA$,
produced if necessary.

Then, \S 19, $\overline{OB} = \overline{OC} + \overline{CB}$\\

But, \S 22, $\overline{OC}=x\overline{OA}$\\
where $x$ is a number, whose sign is the same
as that of the cosine of $\angle AOB$.

Also, \S 73, since $CB$ is perpendicular to $OA$,
$$\overline{CB}=\gamma\overline{OA}$$
where $\gamma$ is a vector perpendicular to $OA$ and $CB$, i.e. to the plane
of the quaternion; and, as the figure is drawn, directed {\sl towards} the
reader.

Hence
$$\frac{\overline{OB}}{\overline{OA}}=
\frac{x\overline{OA}+\gamma\overline{OA}}{\overline{OA}}=
x+\gamma$$

Thus a quaternion, in general, may be decomposed into the sum
of two parts, one numerical, the other a vector. Hamilton calls
them the SCALAR, and the VECTOR, and denotes them respectively
by the letters $S$ and $V$ prefixed to the expression for the
quaternion.

{\bf 78}. Hence $q = Sq+ Vq$, and if in the above example
$$\frac{\overline{OB}}{\overline{OA}}=q$$
then
$$\overline{OB}=\overline{OC}+\overline{CB}=
Sq.\overline{OA}+Vq.\overline{OA}
\setcounter{footnote}{1}%% use a dagger because number looks like superscript
\footnote{
The points are inserted to show that $S$ and $V$
apply only to $q$, and not to $q\overline{OA}$.}
$$

The equation above gives
$$\overline{OC}=Sq.\overline{OA}$$
$$\overline{CB}=Vq.\overline{OA}$$

{\bf 79}. If, in the last figure, we produce $BC$ to $D$, so as to double
its length, and join $OD$, we have, by \S 52,
$$\frac{\overline{OD}}{\overline{OA}}=Kq=SKq+VKq$$
so that\hbox{\hskip 1cm}
$\overline{OD}=\overline{OC}+\overline{CD}=
SKq.\overline{OA}+VKq.\overline{OA}$\\
Hence\hbox{\hskip 3.6cm}$\overline{OC}=SKq.\overline{OA}$\\
and\hbox{\hskip 4cm}$\overline{CD}=VKq.\overline{OA}$\\
Comparing this value of $\overline{OC}$ with that in last section, we find
$$SKq=Sq\eqno{(1)}$$
or {\sl the scalar of the conjugate of a quaternion is equal to the scalar of
the quaternion}.

Again, $\overline{CD} = \overline{CB}$
by the figure, and the substitution of their values gives
$$VKq=Vq\eqno{(2)}$$
or {\sl the vector of the conjugate of a quaternion is the vector of the
quaternion reversed}.

We may remark that the results of this section are simple con
sequences of the fact that the symbols $S$, $V$, $K$ are commutative
\setcounter{footnote}{1}%% use a dagger because number looks like superscript
\footnote{
It is curious to compare the properties of these quaternion symbols with those
of the Elective Symbols of Logic, as given in BOOLE'S
wonderful treatise on the {\sl Laws of Thought}; and to think that the
same grand science of mathematical analysis, by processes remarkably
similar to each other, reveals to us truths in the science of position
far beyond the powers of the geometer, and truths of deductive
reasoning to which unaided thought could never have led the logician. }.

Thus\hbox{\hskip 3cm}$SKq = KSq = Sq$, \\
since the conjugate of a number is the number itself; and
$$VKq=KVq=Vq (\S 73)$$

Again, it is obvious that,
$$\sum{Sq}=S\sum{q},\;\;\;\;\sum{Vq}=V\sum{q}$$
and thence\hbox{\hskip 3cm}$\sum{Kq}=K\sum{q}$

{\bf 80}. Since any vector whatever may be represented by
$$xi+yj+zk$$
where $x$, $y$, $z$ are numbers (or Scalars),
and $i$, $j$, $k$ may be any three
noncoplanar vectors, \S\S 23, 25 though they are usually understood
as representing a rectangular system of unitvectors and
since any scalar may be denoted by $w$; we may write, for any
quaternion $q$, the expression
$$q=w+xi+yj+zk (\S 78)$$

Here we have the essential dependence on four distinct numbers,
from which the quaternion derives its name, exhibited in the most
simple form.

And now we see at once that an equation such as
$$q^{'}=q$$
where\hbox{\hskip 3cm}$q^{'}=w^{'}+x^{'}i+y^{'}j+z^{'}k$\\
involves, of course, the {\sl four} equations
$$
w^{'}=w\textrm{, }
x^{'}=x\textrm{, }
y^{'}=y\textrm{, }
z^{'}=z
$$

{\bf 81}. We proceed to indicate another mode of proof of the distributive
law of multiplication.

We have already defined, or assumed (\S 61), that
$$\frac{\beta}{\alpha}+\frac{\gamma}{\alpha}=\frac{\beta+\gamma}{\alpha}$$
or\hbox{\hskip 3cm}$\beta\alpha^{1}+\gamma\alpha^{1}=
(\beta+\gamma)\alpha^{1}$\\
and have thus been able to understand what is meant by adding
two quaternions.

But, writing $\alpha$ for $\alpha^{1}$,
we see that this involves the equality
$$(\beta+\gamma)\alpha = \beta\alpha+\gamma\alpha$$
from which, by taking the conjugates of both sides, we derive
$$\alpha^{'}(\beta^{'}+\gamma^{'})=\alpha^{'}\beta^{'}+\alpha^{'}\gamma^{'}
(\S 55)$$
And a combination of these results (putting
$\beta+\gamma$ for $\alpha^{'}$ in the latter, for instance) gives
$$
\begin{array}{lcr}
(\beta+\gamma)(\beta^{'}+\gamma^{'}) & = &
(\beta+\gamma)\beta^{'}+(\beta+\gamma)\gamma^{'}\\
& = & \beta\beta^{'}+\gamma\beta^{'}+\beta\gamma^{'}+\gamma\gamma^{'}
\end{array}
$$
by the former.

Hence the {\sl distributive principle is true in the multiplication of
vectors}.

It only remains to show that it is true as to the scalar and
vector parts of a quaternion, and then we shall easily attain the
general proof.

Now, if $a$ be any scalar, $\alpha$ any vector, and $q$ any quaternion,
$$(a+\alpha)q=aq+\alpha q$$

For, if $\beta$ be the vector in which the plane of $q$ is intersected by
a plane perpendicular to $\alpha$, we can find other two vectors,
$\gamma$ and $\delta$ one in each of these planes such that
$$\alpha=\frac{\gamma}{\beta},\;\;\;\;\;q=\frac{\beta}{\delta}$$
And, of course, $a$ may be written
${\displaystyle\frac{a\beta}{\beta}}$; so that
$$
\begin{array}{ccl}
(a+\alpha)q & = & \frac{a\beta+\gamma}{\beta}.\frac{\beta}{\delta}
=\frac{a\beta+\gamma}{\delta}\\
& & \\
& = & a\frac{\beta}{\delta}+\frac{\gamma}{\delta}=
a\frac{\beta}{\delta}+\frac{\gamma}{\beta}.\frac{\beta}{\delta}\\
& & \\
& = & aq + \alpha q
\end{array}
$$
And the conjugate may be written
$$q^{'}(a^{'}+\alpha^{'})=q^{'}a^{'}+q^{'}\alpha^{'} (\S 55)$$
Hence, generally,
$$(a+\alpha)(b+\beta)=ab+a\beta+b\alpha+\alpha\beta$$
or, breaking up $a$ and $b$ each into the sum of two scalars, and
$\alpha$, $\beta$ each into the sum of two vectors, \\
$(a_1+a_2+\alpha_1+\alpha_2)(b_1+b_2+\beta_1+\beta_2)$
$$=(a_1+a_2)(b_1+b_2)
+(a_1+a_2)(\beta_1+\beta_2)
+(b_1+b_2)(\alpha_1+\alpha_2)
+(\alpha_1+\alpha_2)(\beta_1+\beta_2)
$$
(by what precedes, all the factors on the right are distributive, so
that we may easily put it in the form)
$$=(a_1+\alpha_1)(b_1+\beta_1)
+(a_1+\alpha_1)(b_2+\beta_2)
+(a_2+\alpha_2)(b_1+\beta_1)
+(a_2+\alpha_2)(b_2+\beta_2)
$$

Putting $a_1+\alpha_1=p,\;\;\;$
$a_2+\alpha_2=q,\;\;\;$
$b_1+\beta_1=r,\;\;\;$ $b_2+\beta_2=s$,\\
we have $(p+q)(r+s)=pr+ps+qr+qs$

{\bf 82}. Cayley suggests that the laws of quaternion multiplication
may be derived more directly from those of vector multiplication,
supposed to be already established. Thus, let $\alpha$ be the unit vector
perpendicular to the vector parts of $q$ and of $q^{'}$. Then let
$$\rho=q.\alpha,\;\;\;\sigma=\alpha .q^{'}$$
as is evidently permissible, and we have
$$p\alpha=q.\alpha\alpha=q;\;\;\;\alpha\sigma=\alpha\alpha.q^{'}=q^{'}$$
so that\hbox{\hskip 4cm}$q.q^{'}=\rho\alpha.\alpha\sigma=\rho.\sigma$

The student may easily extend this process.

For variety, we shall now for a time forsake the geometrical
mode of proof we have hitherto adopted, and deduce some of our
next steps from the analytical expression for a quaternion given in
\S 80, and the properties of a rectangular system of unitvectors as
in \S 71.

We will commence by proving the result of \S 77 anew.

{\bf 83}. Let
$$\alpha=xi+yj+zk$$
$$\beta=x^{'}i+y^{'}j+z^{'}k$$
Then, because by \S 71 every product or quotient of $i$, $j$, $k$ is reducible
to one of them or to a number, we are entitled to assume
$$q=\frac{\beta}{\alpha}=\omega+\xi i+\eta j +\zeta k$$
where $\omega$, $\xi$, $\eta$, $\zeta$ are numbers.
This is the proposition of \S 80.

[Of course, with this expression for a quaternion, there is no
necessity for a formal proof of such equations as
$$p + (q+r) = (p + q) + r$$
where the various sums are to be interpreted as in \S 61.

All such things become obvious in view of the properties of $i$, $j$ ,$k$.]

{\bf 84}. But it may be interesting to find $\omega$, $\xi$, $\eta$, $\zeta$
in terms of $x$, $y$, $z$, $x^{'}$, $y^{'}$ , $z^{'}$ .

We have
$$\beta=q\alpha$$
or
$$x^{'}i+y^{'}j+z^{'}k=(\omega+\xi i+\eta j+\zeta k)(xi+yj+zk)$$
$$=(\xi x+\eta y+\zeta z)
+(\omega x+\eta z\zeta y)i
+(\omega y+\zeta x\xi z)j
+(\omega z+\xi y\eta x)k
$$
as we easily see by the expressions for the powers and products of
$i$, $j$, $k$ given in \S 71. But the student must pay particular attention
to the {\sl order} of the factors, else he is certain to make mistakes.

This (\S 80) resolves itself into the four equations
$$
\begin{array}{lllllllll}
0 & = & & & \xi x & + & \eta y & + & \zeta z\\
x^{'} & = & \omega x & & & + & \eta z &  & \zeta y\\
y^{'} & = & \omega y &  & \xi z & & & + & \zeta x\\
z^{'} & = & \omega z & + & \xi y &  & \eta x\\
\end{array}
$$
The three last equations give
$$xx^{'}+yy^{'}+zz^{'}=\omega(x^2+y^2+z^2)$$
which determines $\omega$.

Also we have, from the same three, by the help of the first,
$$\xi x^{'}+\eta y^{'}+\zeta z^{'} = 0$$
which, combined with the first, gives
$$\frac{\xi}{yz^{'}zy^{'}}
=\frac{\eta}{zx^{'}xz^{'}}
=\frac{\zeta}{xy^{'}yx^{'}}
$$
and the common value of these three fractions is then easily seen
to be
$$\frac{1}{x^2+y^2+z^2}$$

It is easy enough to interpret these expressions by means of
ordinary coordinate geometry : but a much simpler process will
be furnished by quaternions themselves in the next chapter, and, in
giving it, we shall refer back to this section.

{\bf 85}. The associative law of multiplication is now to be proved
by means of the distributive (\S 81). We leave the proof to the
student. He has merely to multiply together the factors
$$w+xi+yj+zk,\;\;\;\;
w+x^{'}i+y^{'}j+z^{'}k,\;\;\;\;\textrm{ and }
w^{''} + x^{''}i + y^{''}j + z^{''}k$$

as follows :

First, multiply the third factor by the second, and then multiply
the product by the first; next, multiply the second factor by the
first and employ the product to multiply the third: always remembering
that the multiplier in any product is placed {\sl before} the
multiplicand. He will find the scalar parts and the coefficients of
$i$, $j$, $k$, in these products, respectively equal, each to each.

{\bf 86}.
With the same expressions for $\alpha$, $\beta$, as in section 83, we
have
$$\alpha\beta=(xi+yj+zk)(x^{'}i+y^{'}j+z^{'}k)$$
$$\;\;=(xx^{'}+yy^{'}+zz^{'})
+(yz^{'}zy^{'})i
+(zx^{'}xz^{'})j
+(xy^{'}yx^{'})k
$$

But we have also
$$\beta\alpha=
(xx^{'}+yy^{'}+zz^{'})
(yz^{'}zy^{'})i
(zx^{'}xz^{'})j
(xy^{'}yx^{'})k
$$

The only difference is in the sign of the vector parts. Hence
$$S\alpha\beta=S\beta\alpha\eqno{(1)}$$
$$V\alpha\beta=V\beta\alpha\eqno{(2)}$$
$$\alpha\beta+\beta\alpha=2S\alpha\beta\eqno{(3)}$$
$$\alpha\beta\beta\alpha=2V\alpha\beta\eqno{(4)}$$
$$\alpha\beta=K.\beta\alpha\eqno{(5)}$$

{\bf 87}. If $\alpha=\beta$ we have of course (\S 25)
$$x=x^{'},\;\;\;\;y=y^{'},\;\;\;\;z=z^{'}$$
and the formulae of last section become
$$\alpha\beta=\beta\alpha=\alpha^2=(x^2+y^2+z^2)$$
which was anticipated in \S 73, where we proved the formula
$$(T\alpha)^2=\alpha^2$$
and also, to a certain extent, in \S 25.

{\bf 88}. Now let $q$ and $r$ be any quaternions, then
$$
\begin{array}{rcl}
S.qr & = & S.(Sq+Vq)(Sr+Vr)\\
 & = & S.(SqSr+Sr.Vq+Sq.Vr+VqVr)\\
 & = & SqSr+S.VqVr
\end{array}
$$
since the two middle terms are vectors.
Similarly,
$$S.rq=SrSq+S.VrVq$$
Hence, since by (1) of \S 86 we have
$$S.VqVr=S.VrVq$$
we see that
$$S.qr=S.rq\eqno{(1)}$$
a formula of considerable importance.

It may easily be extended to any number of quaternions,
because, $r$ being arbitrary, we may put for it $rs$. Thus we have
$$
\begin{array}{rcl}
S.qrs & = & S.rsq\\
& = & S.sqr
\end{array}
$$
by a second application of the process. In words, we have the
theorem {\sl the scalar of the product of any number of given
quaternions depends only upon the cyclical order in which they are
arranged}.

{\bf 89}. An important case is that of three factors, each a vector.
The formula then becomes
$$S.\alpha\beta\gamma=S.\beta\gamma\alpha=S.\gamma\alpha\beta$$
But
$$\begin{array}{rcll}
S.\alpha\beta\gamma & = & S\alpha(S\beta\gamma+V\beta\gamma) &\\
 & = & S\alpha V\beta\gamma & \textrm{since }\alpha S\beta\gamma
\textrm{ is a vector}\\
 & = & S\alpha V\gamma\beta & \textrm{by (2) of \S 86}\\
 & = & S\alpha(S\gamma\beta+V\gamma\beta) &\\
 & = & S.\alpha\gamma\beta
\end{array}
$$
Hence {\sl the scalar of the product of three vectors changes sign when
the cyclical order is altered.}

By the results of \S\S 55, 73, 79 we see that, for any number
of vectors, we have
$$K.\alpha\beta\gamma\ldots\phi\chi=
\pm\chi\phi\ldots\gamma\beta\alpha$$
(the positive sign belonging to the product of an even number of
vectors) so that
$$S.\alpha\beta\ldots\phi\chi=\pm S.\chi\phi\ldots\beta\alpha$$

Similarly
$$V.\alpha\beta\ldots\phi\chi=\mp V.\chi\phi\ldots\beta\alpha$$
Thus we may generalize (3) and (4) of \S 86 into
$$2S.\alpha\beta\ldots\phi\chi=
\alpha\beta\ldots\chi\phi\pm\phi\chi\ldots\beta\alpha$$
$$2V.\alpha\beta\ldots\phi\chi=
\alpha\beta\ldots\chi\phi\mp\phi\chi\ldots\beta\alpha$$
the upper sign still being used when the number of factors is
even.

Other curious propositions connected with this will be given
later (some, indeed, will be found in the Examples appended to
this chapter), as we wish to develop the really fundamental
formulae in as compact a form as possible.

{\bf 90}. By (4) of \S 86,
$$2V\beta\gamma=\beta\gamma\gamma\beta$$
Hence
$$2V.\alpha V\beta\gamma=V.\alpha(\beta\gamma\gamma\beta)$$
(by multiplying both by $\alpha$, and taking the vector parts of each side)
$$=V(\alpha\beta\gamma+\beta\alpha\gamma\beta\alpha\gamma\alpha\gamma\beta)$$
(by introducing the null term $\beta\alpha\gamma\beta\alpha\gamma$).

\noindent
That is
$$2V.\alpha V\beta\gamma=V.(\alpha\beta+\beta\alpha)\gamma
V(\beta S\alpha\gamma+\beta V\alpha\gamma+S\alpha\gamma .\beta+
V\alpha\gamma .\beta$$
$$=V.(2S\alpha\beta)\gamma2V\beta S\alpha\gamma$$
(if we notice that $V(V\alpha\gamma .\beta)=V.\beta V\alpha\gamma$
by (2) of \S 86).
Hence
$$V.\alpha V\beta\gamma=\gamma S\alpha\beta\beta S\gamma\alpha\eqno{(1)}$$
a formula of constant occurrence.

Adding $\alpha S\beta\gamma$ to both sides, we get another most valuable
formula
$$V.\alpha\beta\gamma
=\alpha S\beta\gamma
\beta S\gamma\alpha
+\gamma S\alpha\beta\eqno{(2)}
$$
and the form of this shows that we may interchange $\gamma$ and $\alpha$
without altering the righthand member. This gives
$$V.\alpha\beta\gamma = V.\gamma\beta\alpha$$
a formula which may be greatly extended. (See \S89, above.)

Another simple mode of establishing (2) is as follows :
$$
\begin{array}{rcl}
K.\alpha\beta\gamma & = & \gamma\beta\alpha\\
\therefore 2V.\alpha\beta\gamma & = &
\alpha\beta\gammaK.\alpha\beta\gamma\textrm{ (by \S 79(2))}\\
& = & \alpha\beta\gamma + \gamma\beta\alpha\\
& = & \alpha(\beta\gamma+\gamma\beta)
(\alpha\gamma+\gamma\alpha)\beta
+\gamma(\alpha\beta+\beta\alpha)\\
& = & 2\alpha S\beta\gamma2\beta S\alpha\gamma+2\gamma S\alpha\beta
\end{array}
$$

{\bf 91}. We have also
$$VV\alpha\beta V\gamma\delta = VV\gamma\delta V\alpha\beta\;\;\;\;
\textrm{ by (2) of \S 86}$$
$$=\delta S\gamma V\alpha\beta\gamma S\delta V\alpha\beta
=\delta S.\alpha\beta\gamma\gamma S.\alpha\beta\delta$$
$$=\beta S\alpha V\gamma\delta+\alpha S\beta V\gamma\delta
=\beta S.\alpha\gamma\delta+\alpha S.\beta\gamma\delta$$
all of these being arrived at by the help of \S 90 (1) and of \S 89;
and by treating alternately $V\alpha\beta$ and
$V\gamma\delta$ as {\sl simple} vectors.

Equating two of these values, we have
$$\delta S.\alpha\beta\gamma
=\alpha S.\beta\gamma\delta
+\beta S.\gamma\alpha\delta
+\gamma S.\alpha\beta\delta\eqno{(3}
$$
a very useful formula, expressing any vector whatever in terms
of three given vectors. [This, of course, presupposes that
$\alpha$, $\beta$, $\gamma$
are not coplanar, \S 23. In fact, if they be coplanar, the factor
$S.\alpha\beta\gamma$ vanishes,
and thus (3) does not give an expression for $\delta$.
This will be shown in \S 101 below.]

{\bf 92}. That such an expression as (3) is possible we knew already
by \S 23. For variety we may seek another expression of a similar
character, by a process which differs entirely from that employed
in last section.

$\alpha$, $\beta$, $\gamma$
being any three noncoplanar vectors, we may derive
from them three others $V\alpha\beta$, $V\beta\gamma$, $V\gamma\alpha$
and, as these will not be
coplanar, any other vector $\delta$ may be expressed as the sum of the
three, each multiplied by some scalar. It is required to find this
expression for $\delta$.

Let
$$\delta=xV\alpha\beta+yV\beta\gamma+zV\gamma\alpha$$
Then
$$S\gamma\delta=xS.\gamma\alpha\beta =xS.\alpha\beta\gamma$$
the terms in y and z going out, because
$$S\gamma V\beta\gamma = S.\gamma\beta\gamma=S\beta\gamma^2
=\gamma^2 S\beta=0$$
for $\gamma^2$ is (\S 73) a number.

Similarly
$$S\beta\delta=zS.\beta\gamma\alpha=zS.\alpha\beta\gamma$$
and
$$S\alpha\delta=qS.\alpha\beta\gamma$$
Thus
$$\delta S.\alpha\beta\gamma=V\alpha\beta S\gamma\delta
+V\beta\gamma S\alpha\delta
+V\gamma\alpha S\beta\delta\eqno{(4)}
$$

{\bf 93}. We conclude the chapter by showing (as promised in \S 64)
that the assumption that the product of two parallel vectors is
a number, and the product of two perpendicular vectors a third
vector perpendicular to both, is not only useful and convenient,
but absolutely inevitable, if our system is to deal indifferently with
all directions in space. We abridge Hamilton s reasoning.

Suppose that there is no direction in space preeminent, and
that the product of two vectors is something which has quantity,
so as to vary in amount if the factors are changed, and to have its
sign changed if that of one of them is reversed ; if the vectors be
parallel, their product cannot be, in whole or in part, a vector
{\sl inclined} to them, for there is nothing to determine the direction in
which it must lie. It cannot be a vector {\sl parallel} to them; for by
changing the signs of both factors the product is unchanged,
whereas, as the whole system has been reversed, the product
vector ought to have been reversed. Hence it must be a number.
Again, the product of two perpendicular vectors cannot be wholly
or partly a number, because on inverting one of them the sign of
that number ought to change; but inverting one of them is simply
equivalent to a rotation through two right angles about the other,
and (from the symmetry of space) ought to leave the number
unchanged. Hence the product of two perpendicular vectors must
be a vector, and a simple extension of the same reasoning shows
that it must be perpendicular to each of the factors. It is easy to
carry this farther, but enough has been said to show the character
of the reasoning.

\section{Examples To Chapter 2.}

{\bf 1}. It is obvious from the properties of polar triangles that any
mode of representing versors by the {\sl sides} of a spherical triangle
must have an equivalent statement in which they are represented
by {\sl angles} in the polar triangle.

Show directly that the product of two versors represented
by two angles of a spherical triangle is a third versor represented
by the {\sl supplement} of the remaining angle of the triangle ; and
determine the rule which connects the {\sl directions} in which these
angles are to be measured.

{\bf 2}. Hence derive another proof that we have not generally
$$pq=qp$$

{\bf 3}. Hence show that the proof of the associative principle,
\S 57, may be made to depend upon the fact that if from any point
of the sphere tangent arcs be drawn to a spherical conic, and also
arcs to the foci, the inclination of either tangent arc to one of the
focal arcs is equal to that of the other tangent arc to the other
focal arc.

{\bf 4}. Prove the formulae
$$2S.\alpha\beta\gamma = \alpha\beta\gamma\gamma\beta\alpha$$
$$2V.\alpha\beta\gamma = \alpha\beta\gamma+\gamma\beta\alpha$$

{\bf 5}. Show that, whatever odd number of vectors be represented
by $\alpha$, $\beta$, $\gamma$ \&c., we have always
$$
V.\alpha\beta\gamma\delta\epsilon=V.\epsilon\delta\gamma\beta\alpha
$$
$$
V.\alpha\beta\gamma\delta\epsilon\zeta\eta
=V.\eta\zeta\epsilon\delta\gamma\beta\alpha,\textrm{ \&c.}
$$

{\bf 6}. Show that
$$
S.V\alpha\beta V\beta\gamma V\gamma\alpha=(S.\alpha\beta\gamma)^2
$$
$$
V.V\alpha\beta V\beta\gamma V\gamma\alpha=
V\alpha\beta(\gamma^2S\alpha\betaS\beta\gamma S\gamma\alpha)+\ldots
$$
and
$$
V(V\alpha\beta V.V\beta\gamma V\gamma\alpha)
=(\beta S\alpha\gamma\alpha S\beta\gamma)S.\alpha\beta\gamma
$$

{\bf 7}. If $\alpha$, $\beta$, $\gamma$
be any vectors at right angles to each other, show that
$$
(\alpha^3+\beta^3+\gamma^3)S.\alpha\beta\gamma
=\alpha^4V\beta\gamma
+\beta^4V\gamma\alpha
+\gamma^4V\alpha\beta
$$
$$
(\alpha^{2n1}+\beta^{2n1}+\gamma^{2n1})S.\alpha\beta\gamma
=\alpha^{2n}V\beta\gamma
+\beta^{2n}V\gamma\alpha
+\gamma^{2n}V\alpha\beta
$$

{\bf 8}. If $\alpha$, $\beta$, $\gamma$
be noncoplanar vectors, find the relations among
the six scalars, $x$, $y$, $z$ and $\xi$, $\eta$, $\zeta$
which are implied in the
equation
$$
x\alpha+y\beta+z\gamma
=\xi V\beta\gamma+\eta V\gamma\alpha+\zeta V\alpha\beta
$$

{\bf 9}. If $\alpha$, $\beta$, $\gamma$
be any three noncoplanar vectors, express any
fourth vector, $\delta$, as a linear function of each of the following sets of
three derived vectors.
$$
V.\gamma\alpha\beta,\;\;\;\;V.\alpha\beta\gamma,\;\;\;\;
V.\beta\gamma\alpha
$$
and
$$
V.V\alpha\beta V\beta\gamma V\gamma\alpha,\;\;\;\;
V.V\beta\gamma V\gamma\alpha V\alpha\beta,\;\;\;\;
V.V\gamma\alpha V\alpha\beta V\beta\gamma
$$

{\bf 10}. Eliminate $\rho$ from the equations
$$
S\alpha\rho=a,\;\;\;\;
S\beta\rho=b,\;\;\;\;
S\gamma\rho=c,\;\;\;\;
S\delta\rho=d
$$
where $\alpha$, $\beta$, $\gamma$, $\delta$
are vectors, and $a$, $b$, $c$, $d$ scalars.

{\bf 11}. In any quadrilateral, plane or gauche, the sum of the
squares of the diagonals is double the sum of the squares of the
lines joining the middle points of opposite sides.

\chapter{Interpretations And Transformations}

{\bf 94}. Among the most useful characteristics of the Calculus of
Quaternions, the ease of interpreting its formulae geometrically,
and the extraordinary variety of transformations of which the
simplest expressions are susceptible, deserve a prominent place.
We devote this Chapter to some of the more simple of these,
together with a few of somewhat more complex character but of
constant occurrence in geometrical and physical investigations.
Others will appear in every succeeding Chapter. It is here,
perhaps, that the student is likely to feel most strongly the
peculiar difficulties of the new Calculus. But on that very account
he should endeavour to master them, for the variety of forms
which any one formula may assume, though puzzling to the
beginner, is of the utmost advantage to the advanced student, not
alone as aiding him in the solution of complex questions, but
as affording an invaluable mental discipline.

{\bf 95}. If we refer again to the figure of \S 77 we see that
$$OC=OB\cos AOB$$
$$CB=OB \sin AOB$$
Hence if
$$\overline{AB}=\alpha,\;\;\;\;
\overline{OB}=\beta,\;\;\;\;\textrm{ and }
\angle AOB=\theta
$$
we have
$$OB=T\beta,\;\;\;\;OA=T\alpha$$
$$OC=T\beta\cos\theta,\;\;\;\;CB=T\beta\sin\theta
$$
Hence
$$S\frac{\beta}{\alpha}=
\frac{OC}{OA}=
\frac{T\beta}{T\alpha}\cos\theta
$$
Similarly,
$$
TV\frac{\beta}{\alpha}=\frac{CB}{OA}=\frac{T\beta}{T\alpha}\sin\theta
$$

Hence, if $\eta$ be a unitvector perpendicular to
$\alpha$ and $\beta$, and such
that positive rotation about it, through the angle $\theta$, turns $\alpha$
towards $\beta$ or
$$
\eta=
\frac{U\overline{CB}}{U\overline{OA}}=
U\frac{\overline{CB}}{\overline{OA}}=
UV\frac{\beta}{\alpha}
$$
we have
$$
V\frac{\beta}{\alpha}=
\frac{T\beta}{T\alpha}\sin\theta .\eta\;\;\;\;\;\textrm{ (See, again, \S 84)}
$$

{\bf 96}. In the same way, or by putting
$$
\begin{array}{rcl}
\alpha\beta & = & S\alpha\beta+V\alpha\beta\\
 & = & S\beta\alpha  V\beta\alpha\\
 & = & \alpha^2\left(S\frac{\beta}{\alpha}V\frac{\beta}{\alpha}\right)\\
 & = & T\alpha^2\left(S\frac{\beta}{\alpha}+V\frac{\beta}{\alpha}\right)
\end{array}
$$
we may show that
$$S\alpha\beta=T\alpha T\beta\cos\theta$$
$$TV\alpha\beta = T\alpha T\beta\sin\theta$$
and
$$V\alpha\beta=T\alpha T\beta\sin\theta . \eta$$
where
$$\eta=UV\alpha\beta = U(V\beta\alpha)=UV\frac{\beta}{\alpha}$$

Thus {\sl the scalar of the product of two vectors is the continued
product of their tensors and of the cosine of the supplement of the
contained angle}.

{\sl The tensor of the vector of the product of two vectors is the con
tinued product of their tensors and the sine of the contained angle ;
and the versor of the same is a unitvector perpendicular to both,
and such that the rotation about it from the first vector (i. e. the
multiplier) to the second is lefthanded or positive}.

{\sl Hence also $TV\alpha\beta$
is double the area of the triangle two of whose
sides are $\alpha$, $\beta$.}

{\bf 97}. (a) In any plane triangle $ABC$ we have
$$\overline{AC}=\overline{AB}+\overline{BC}$$
Hence,
$$
\overline{AC}^2=S.\overline{AC}\overline{AC}=
S.\overline{AC}(\overline{AB}+\overline{BC})
$$

With the usual notation for a plane triangle the interpretation
of this formula is
$$b^2 = bc\cos Aab\cos C$$
or
$$b=c\cos C+c\cos A$$

(b) Again we have, obviously,
$$
\begin{array}{rcl}
V.\overline{AB}\;\overline{AC}&=&V.\overline{AB}(\overline{AB}+\overline{BC})\\
&=&V.\overline{AB}\;\overline{BC}
\end{array}
$$
or
$$cb\sin A = ca\sin B$$
whence
$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$

These are truths, but not truisms, as we might have been led
to fancy from the excessive simplicity of the process employed.

{\bf 98}.
From \S 96 it follows that, if $\alpha$ and $\beta$ be both actual (i. e.
real and nonevanescent) vectors, the equation
$$S\alpha\beta = 0$$


shows that $\cos\theta=0$, or that
$\alpha$ is {\sl perpendicular} to $\beta$. And, in fact,
we know already that the product of two perpendicular vectors is
a vector.

Again if
$$V\alpha\beta=0$$
we must have $\sin\theta=0$, or
$\alpha$ is {\sl parallel} to $\beta$. We know already
that the product of two parallel vectors is a scalar.

Hence we see that
$$S\alpha\beta=0$$
is equivalent to
$$\alpha=V\gamma\beta$$
where $\gamma$ is an undetermined vector; and that
$$V\alpha\beta=0$$
is equivalent to
$$\alpha=x\beta$$
where $x$ is an undetermined scalar.

{\bf 99}. If we write, as in \S\S 83, 84,
$$\alpha=ix+jy+kz$$
$$\beta=ix^{'}+jy^{'}+kz^{'}$$
we have, at once, by \S 86,
$$\begin{array}{rcl}
S\alpha\beta&=&xx^{'}yy^{'}zz^{'}\\
&=&rr^{'}\left(
\frac{x}{r}\frac{x^{'}}{r^{'}}+
\frac{y}{r}\frac{y^{'}}{r^{'}}+
\frac{z}{r}\frac{z^{'}}{r^{'}}
\right)
\end{array}
$$
where
$$
r=\sqrt{x^2+y^2+z^2},\;\;\;\;
r^{'}=\sqrt{x^{'2}+y^{'2}+z^{'2}}
$$
Also
$$
V\alpha\beta=rr^{'}\left\{
\frac{yz^{'}zy^{'}}{rr^{'}}i+
\frac{zx^{'}xz^{'}}{rr^{'}}j+
\frac{xy^{'}=yx^{'}}{rr^{'}}k
\right\}
$$

These express in Cartesian coordinates the propositions we have
just proved. In commencing the subject it may perhaps assist
the student to see these more familiar forms for the quaternion
expressions ; and he will doubtless be induced by their appearance
to prosecute the subject, since he cannot fail even at this stage to
see how much more simple the quaternion expressions are than
those to which he has been accustomed.

{\bf 100}. The expression
$$S.\alpha\beta\gamma$$
may be written
$$SV(\alpha\beta)\gamma$$
because the quaternion $\alpha\beta\gamma$ may be broken up into
$$S(\alpha\beta)\gamma+V(\alpha\beta)\gamma$$
of which the first term is a vector.

But, by \S 96,
$$SV(\alpha\beta)\gamma=T\alpha T\beta\sin\theta S\eta\gamma$$
Here $T\eta=1$, let $\phi$ be the angle between $\eta$ and $\gamma$,
then finally
$$S.\alpha\beta\gamma = T\alpha T\beta T\gamma\sin\theta\cos\phi$$

But as $\eta$ is perpendicular to $\alpha$ and $\beta$,
$T\gamma\cos\phi$ is the length of the
perpendicular from the extremity of $\gamma$
upon the plane of $\alpha$, $\beta$. And
as the product of the other three factors is (\S 96) the area of the
parallelogram two of whose sides are $\alpha$, $\beta$, we see that the
magnitude of $S.\alpha\beta\gamma$,
independent of its sign, is {\sl the volume of the
parallelepiped of which three coordinate edges
are $\alpha$, $\beta$, $\gamma$};
or six times the volume of the pyramid which has
$\alpha$, $\beta$, $\gamma$ for edges.

{\bf 101}. Hence the equation
$$S.\alpha\beta\gamma=0$$
if we suppose $\alpha\beta\gamma$ to be actual vectors, shows either that
$$\sin\theta=0$$
or
$$\cos\phi=0$$
i. e. {\sl two of the three vectors are parallel},
or {\sl all three are parallel to one plane}.

This is consistent with previous results, for if $\gamma=p\beta$ we have
$$S.\alpha\beta\gamma=pS.\alpha\beta^2=0$$
and, if $\gamma$ be coplanar with $\alpha$,$\beta$, we have
$\gamma=p\alpha+q\beta$ and
$$S.\alpha\beta\gamma=S.\alpha\beta(p\alpha+q\beta)=0$$

{\bf 102}.
This property of the expression $S.\alpha\beta\gamma$ prepares us to
find that it is a determinant. And, in fact, if we take $\alpha$,$\beta$ as in
\S 83, and in addition
$$\gamma=ix^{''}+jy^{''}+kz^{''}$$
we have at once
$$S.\alpha\beta\gamma=x^{''}(yz^{'}zy^{'})
y^{''}(zx^{'}xz^{'})
z^{''}(xy^{'}yx^{'})$$
$$
=\left\vert
\begin{array}{ccc}
x & y & z\\
x^{'} & y^{'} & z^{'}\\
x^{''}&y^{''}&z^{''}
\end{array}
\right\vert
$$
The determinant changes sign if we make any two rows change
places. This is the proposition we met with before (\S 89) in the
form
$$S.\alpha\beta\gamma=S.\beta\alpha\gamma=S.\beta\gamma\alpha
\textrm{, \&c}$$

If we take three new vectors
$$\alpha_1=ix+jx^{'}+kx^{''}$$
$$\beta_1 =iy+jy^{'}+ky^{''}$$
$$\gamma_1=iz+jz^{'}+kz^{''}$$
we thus see that they are coplanar if $\alpha$, $\beta$, $\gamma$ are so.
That is, if
$$S.\alpha\beta\gamma=0$$
then
$$S.\alpha_1\beta_1\gamma_1=0$$

{\bf 103}. We have, by \S 52,
$$
\begin{array}{rcl}
(Tq)^2 &=&qKq = (Sq+Vq)(SqVq)\;\;\;\;\textrm{(\S 79)}\\
&=&(Sq)^2(Vq)^2\;\;\;\;\;\textrm{by algebra}\\
&=&(Sq)^2+(TVq)^2\;\;\;\;\textrm{(\S 73)}\\
\end{array}
$$
If $q=\alpha\beta$, we have $Kq = \beta\alpha$, and the formula becomes
$$\alpha\beta . \beta\alpha = \alpha^2\beta^2=
(S\alpha\beta)^2(V\alpha\beta)^2
$$

In Cartesian coordinates this is\\
\vskip 0.1cm
$(x^2+y^2+z^z)(x^{'2}+y^{'2}+z^{'2})$
$$
=(xx^{'}+yy^{'}+zz^{'})^2+(yz^{'}zy^{'})^2+
(zx^{'}xz^{'})^2+(xy^{'}yx^{'})^2
$$
More generally we have
$$
\begin{array}{rcl}
(T(qr))^2&=&(Tq)^2(Tr)^2\\
&=&(S.qr)^2(V.qr)^2
\end{array}
$$
If we write
$$q=w+\alpha=w+ix+jy+kz$$
$$r=w^{'}+\beta=w^{'}+ix^{'}+jy^{'}+kz^{'}$$
this becomes
$$(w^2+x^2+y^2+z^2)(w^{'2}+x^{'2}+y^{'2}+z^{'2})$$
$$=(ww^{'}xx^{'}yy^{'}zz^{'})^2+(wx^{'}+w^{'}x+yz^{'}zy^{'})^2$$
$$=(xy^{'}+w^{'}y+zx^{'}xz^{'})^2+(wz^{'}+w^{'}z+xy^{'}yx^{'})^2$$
a formula of algebra due to Euler.

{\bf 104}. We have, of course, by multiplication,
$$
(\alpha+\beta)^2=
\alpha^2+\alpha\beta+\beta\alpha+\beta^2=
\alpha^2+2S\alpha\beta+\beta^2\;\;\;\;\;\textrm{(\S 86 (3))}
$$
Translating into the usual notation of plane trigonometry, this
becomes
$$c^2=a^22ab\cos C+b^2$$
the common formula.

Again,
$$
V.(\alpha+\beta)(\alpha\beta)=
V\alpha\beta+V\beta\alpha=
2V\alpha\beta\;\;\;\;\;\textrm{(\S 86 (2)}
$$
Taking tensors of both sides we have the theorem, {\sl the parallelogram
whose sides are parallel and equal to the diagonals of a
given parallelogram, has double its area} (\S 96).

Also
$$S(\alpha+\beta)(\alpha\beta)=\alpha^2\beta^2$$
and vanishes only when $\alpha^2=\beta^2$,
or $T\alpha=T\beta$; that is, {\sl the diagonals
of a parallelogram are at right angles to one another, when, and
only when, it is a rhombus}.

Later it will be shown that this contains a proof that the angle
in a semicircle is a right angle.

{\bf 105}. The expression\hbox{\hskip 1cm}$\rho=\alpha\beta\alpha^{1}$\\
obviously denotes a vector whose tensor is equal to that of $\beta$.

But we have\hbox{\hskip 2cm}$S.\beta\alpha\rho=0$\\
so that $\rho$ is in the plane of $\alpha$, $\beta$

Also we have\hbox{\hskip 2cm}$S\alpha\rho=S\alpha\beta$\\
so that $\beta$ and $\rho$ make equal angles with $\alpha$,
evidently on opposite
sides of it. Thus if $\alpha$ be the perpendicular to a reflecting surface
and $\beta$ the path of an incident ray, $\rho$ will be the path of the
reflected ray.

Another mode of obtaining these results is to expand the above
expression, thus, \S 90 (2),
$$
\begin{array}{rcl}
\rho&=&2\alpha^{1}S\alpha\beta\beta\\
&=&2\alpha^{1}S\alpha\beta\alpha^{1}(S\alpha\beta+V\alpha\beta)\\
&=&\alpha^{1}(S\alpha\betaV\alpha\beta)
\end{array}
$$
so that in the figure of \S 77 we see that if $\overline{OA}=\alpha$,
and $\overline{OB}=\beta$, we
have $\overline{OD} = \rho = \alpha\beta\alpha^{1}$

Or, again, we may get the result at once by transforming the
equation to $\frac{\rho}{\alpha}=K(\alpha^{1}\rho)=K\frac{\beta}{\alpha}$

{\bf 106}. For any three coplanar vectors the expression
$$\rho=\alpha\beta\gamma$$
is (\S 101) a vector. It is interesting to determine what this vector
is. The reader will easily see that if a circle be described about
the triangle, two of whose sides are (in order) $\alpha$ and $\beta$,
and if from
the extremity of $\beta$ a line parallel to $\gamma$ be drawn,
again cutting the
circle, the vector joining the point of intersection with the origin
of $\alpha$ is the direction of the vector $\alpha\beta\gamma$.
For we may write it in the form
$$
\rho=\alpha\beta^2\beta^{1}\gamma=
(T\beta)^2\alpha\beta^{1}\gamma=
(T\beta)^2\frac{\alpha}{\beta}\gamma
$$
which shows that the versor $\displaystyle\left(\frac{\alpha}{\beta}\right)$
which turns $\beta$ into a direction
parallel to $\alpha$, turns $\gamma$ into a direction parallel to $\rho$.
And this expresses the longknown property of opposite angles of a
quadrilateral inscribed in a circle.

Hence if $\alpha$, $\beta$, $\gamma$
be the sides of a triangle taken in order, the
tangents to the circumscribing circle at the angles of the triangle
are parallel respectively to
$$
\alpha\beta\gamma,\;\;\;\;
\beta\gamma\alpha,\;\;\;\;\textrm{ and }
\gamma\alpha\beta
$$

Suppose two of these to be parallel, i. e. let
$$\alpha\beta\gamma=x\beta\gamma\alpha=x\alpha\gamma\beta\;\;\;\;(\S 90)$$
since the expression is a vector. Hence
$$\beta\gamma=x\gamma\beta$$
which requires either
$$x=1,\;\;\;\;V\gamma\beta=0\;\;\;\;\textrm{ or }\gamma \vert\vert \beta$$
a case not contemplated in the problem; or
$$x=1,\;\;\;\;S\beta\gamma=0$$
i. e. the triangle is rightangled. And geometry shows us at once
that this is correct.

Again, if the triangle be isosceles, the tangent at the vertex is
parallel to the base. Here we have
$$x\beta=\alpha\beta\gamma$$
or
$$x(\alpha+\gamma)=\alpha(\alpha+\gamma)\gamma$$
whence $x=\gamma^2=\alpha^2$, or $T\gamma=T\alpha$, as required.

As an elegant extension of this proposition the reader may
prove that the vector of the continued product $\alpha\beta\gamma\delta$
of the vectorsides of any quadrilateral inscribed in a sphere
is parallel to the radius drawn to the corner ($\alpha$, $\delta$).
[For, if $\epsilon$ be the vector from $\delta$,
$\alpha$ to $\beta$, $\gamma$, $\alpha\beta\epsilon$ and
$\epsilon\gamma\delta$ are (by what precedes) vectors {\sl touching} the
sphere at $\alpha$, $\delta$. And their product (whose vector part must be
parallel to the radius at $\alpha$, $\delta$) is
$$\alpha\beta\epsilon . \epsilon\gamma\delta=\epsilon^2 .
\alpha\beta\gamma\delta]$$

{\bf 107}. To exemplify the variety of possible transformations
even of simple expressions, we will take cases which are of
frequent occurrence in applications to geometry.

Thus $$T(\rho+\alpha)=T(\rho\alpha)$$
[which expresses that if
$$
\overline{OA}=\alpha\;\;\;\;
\overline{OA^{'}}=\alpha\;\;\;\;\textrm{ and }\;\;\;\;
\overline{OP}=\rho
$$
we have\hbox{\hskip 4cm}$AP=A^{'}P$\\
and thus that $P$ is any point equidistant from two fixed points,]
may be written $$(\rho+\alpha)^2=(\rho\alpha)^2$$
or\hbox{\hskip 3cm}$\rho^2+2S\alpha\rho+\alpha^2=
\rho^22S\alpha\rho+\alpha^2\;\;\;\;\textrm{(\S 104)}$\\
whence\hbox{\hskip 4cm}$S\alpha\rho=0$\\
This may be changed to
$$\alpha\rho+\rho\alpha=0$$
or
$$\alpha\rho+K\alpha\rho=0$$
$$SU\frac{\rho}{\alpha}=0$$
or finally,
$$TVU\frac{\rho}{\alpha}=1$$
all of which express properties of a plane.

Again,\hbox{\hskip 4cm}$T\rho=T\alpha$\\
may be written\hbox{\hskip 3.2cm}$\displaystyle T\frac{\rho}{\alpha}=1$
$$\left(S\frac{\rho}{\alpha}\right)^2\left(V\frac{\rho}{\alpha}\right)^2=1$$
$$(\rho+\alpha)^22S\alpha(\rho+\alpha)=0$$
$$\rho=(\rho+\alpha)^{1}\alpha(\rho+\alpha)$$
$$S(\rho+\alpha)(\rho\alpha)=0$$
or finally,
$$T.(\rho+\alpha)(\rho\alpha)=2TV\alpha\rho$$

All of these express properties of a sphere. They will be
interpreted when we come to geometrical applications.

{\bf 108}. {\sl To find the space relation among five points.}

A system of five points, so far as its internal relations are
concerned, is fully given by the vectors from one to the other four.
If three of these be called $\alpha$, $\beta$, $\gamma$, the fourth,
$\delta$, is necessarily expressible as
$x\alpha+y\beta+z\gamma$. Hence the relation required must be
independent of x, y, z.

But
$$
\left.
\begin{array}{rlll}
S\alpha\delta &=\;x\alpha^2 &+\;yS\alpha\beta &+\;zS\alpha\gamma\\
S\beta\delta &=\;xS\beta\alpha &+\;y\beta^2 &+\;zS\beta\gamma\\
S\gamma\delta &=\;xS\gamma\alpha &+\;yS\gamma\beta &+\;z\gamma^2\\
S\delta\delta=\delta^2 &=\;xS\delta\alpha &+\;yS\delta\beta &+\;zS\delta\gamma
\end{array}
\right\}\eqno{(1)}
$$
The elimination of $x$, $y$, $z$ gives a determinant of the fourth order,
which may be written
$$
\left\vert
\begin{array}{cccc}
S\alpha\alpha & S\alpha\beta & S\alpha\gamma & S\alpha\delta\\
S\beta\alpha & S\beta\beta & S\beta\gamma & S\beta\delta\\
S\gamma\alpha & S\gamma\beta & S\gamma\gamma & S\gamma\delta\\
S\delta\alpha & S\delta\beta & S\delta\gamma & S\delta\delta
\end{array}
\right\vert=0
$$
Now each term may be put in either of two forms, thus
$$S\beta\gamma=\frac{1}{2}\left\{\beta^2+\gamma^2(\beta\gamma)^2\right\}=
T\beta T\gamma\cos\widehat{\beta\gamma}$$


If the former be taken we have the expression connecting the
distances, two and two, of five points in the form given by Muir
(Proc. R. S. E. 1889) ; if we use the latter, the tensors divide out
(some in rows, some in columns), and we have the relation among
the cosines of the sides and diagonals of a spherical quadrilateral.

We may easily show (as an exercise in quaternion manipulation
merely) that this is the {\sl only} condition, by showing that from it
we can get the condition when any other of the points is taken as
origin. Thus, let the origin be at $\alpha$, the vectors are
$\alpha$, $\beta\alpha$, $\gamma\alpha$, $\delta\alpha$.
But, by changing the signs of the first row, and first
column, of the determinant above, and then adding their values
term by term to the other rows and columns, it becomes
$$
\left\vert
\begin{array}{cccc}
S(\;\;\;\alpha)(\alpha) & S(\;\;\;\alpha)(\beta\alpha)
& S(\;\;\;\alpha)(\gamma\alpha) & S(\;\;\;\alpha)(\delta\alpha)\\
S(\beta\alpha)(\alpha) & S(\beta\alpha)(\beta\alpha)
& S(\beta\alpha)(\gamma\alpha) & S(\beta\alpha)(\delta\alpha)\\
S(\gamma\alpha)(\alpha) & S(\gamma\alpha)(\beta\alpha)
& S(\gamma\alpha)(\gamma\alpha) & S(\gamma\alpha)(\delta\alpha)\\
S(\delta\alpha)(\alpha) & S(\delta\alpha)(\beta\alpha)
& S(\delta\alpha)(\gamma\alpha) & S(\delta\alpha)(\delta\alpha)
\end{array}
\right\vert
$$
which, when equated to zero, gives the same relation as before.
[See Ex. 10 at the end of this Chapter.]

An additional point, with $\epsilon=x^{'}\alpha+y^{'}\beta+z^{'}\gamma$
gives six additional equations like (1) ; i. e.
$$
\begin{array}{rlll}
S\alpha\epsilon&=x^{'}\alpha^2&+y^{'}S\alpha\beta&+z^{'}S\alpha\gamma\\
S\beta\epsilon&=x^{'}S\beta\alpha&+y^{'}\beta^2&+z^{'}S\beta\gamma\\
S\gamma\epsilon&=x^{'}S\gamma\alpha&+y^{'}S\gamma\beta&+z^{'}\gamma^2\\
S\delta\epsilon&=x^{'}S\delta\alpha&+y^{'}S\delta\beta&+z^{'}S\delta\gamma\\
&=xS\epsilon\alpha&+yS\epsilon\beta&+zS\epsilon\gamma\\
\epsilon^2&=x^{'}S\alpha\epsilon&+y^{'}S\beta\epsilon&+z^{'}S\gamma\epsilon
\end{array}
$$
from which corresponding conclusions may be drawn.

Another mode of solving the problem at the head of this
section is to write the {\sl identity}
$$
\sum m(\alpha\theta)^2=\sum m\alpha^2sS.\theta\sum m\alpha+\theta^2\sum m
$$
where the $m$s are undetermined scalars, and the $\alpha$s are given
vectors, while $\theta$ is any vector whatever.

Now, {\sl provided that the number of given vectors exceeds four}, we
do not completely determine the ms by imposing the conditions
$$\sum m=0,\;\;\;\;\sum m\alpha=0$$
Thus we may write the above identity, for each of five vectors
successively, as
$$
\begin{array}{rcl}
\sum m(\alpha\alpha_1)^2 &=& \sum m\alpha^2\\
\sum m(\alpha\alpha_2)^2 &=& \sum m\alpha^2\\
\ldots\ldots &=& \ldots\\
\sum m(\alpha\alpha_n)^2 &=& \sum m\alpha^2\\
\end{array}
$$
Take, with these,\hbox{\hskip 3cm}$\sum m = 0$\\
and we have six linear equations from which to eliminate the $m$s.
The resulting determinant is
$$
\left\vert
\begin{array}{cccccc}
\overline{\alpha_1\alpha_1^2} & \overline{\alpha_1\alpha_s^2} &
\overline{\alpha_1\alpha_3^2} & . &
\overline{\alpha_1\alpha_5^2} & 1\\
\overline{\alpha_2\alpha_1^2} & \overline{\alpha_2\alpha_s^2} &
\overline{\alpha_2\alpha_3^2} & . &
\overline{\alpha_2\alpha_5^2} & 1\\
. & . & . & & . & \\
. & . & . & & . & \\
\overline{\alpha_5\alpha_1^2} & \overline{\alpha_5\alpha_s^2} &
\overline{\alpha_5\alpha_3^2} & . &
\overline{\alpha_5\alpha_5^2} & 1\\
1 & 1 & . & . & 1 & 0\\
\end{array}
\right\vert
\sum m\alpha^2=0
$$

This is equivalent to the form in which Cayley gave the
relation among the mutual distances of five points. (Camb. Math.
Journ. 1841.)

{\bf 109}. We have seen in \S 95 that a quaternion may be divided
into its scalar and vector parts as follows:
$$
\frac{\beta}{\alpha}=S\frac{\beta}{\alpha}+V\frac{\beta}{\alpha}=
\frac{T\beta}{T\alpha}(\cos\theta+\epsilon\sin\theta)
$$
where $\theta$ is the angle between the directions of
$\alpha$ and $\beta$ and $\displaystyle \epsilon=UV\frac{\beta}{\alpha}$
is the unitvector perpendicular to the plane of $\alpha$
and $\beta$ so situated
that positive (i.e. lefthanded) rotation about it turns
$\alpha$ towards $\beta$

Similarly we have (\S 96)
$$
\begin{array}{rl}
\alpha\beta&=S\alpha\beta + V\alpha\beta\\
&=T\alpha T\beta(\cos\theta +\epsilon\sin\theta)
\end{array}
$$
$\theta$ and $\epsilon$ having the same signification as before.

{\bf 110}. Hence, considering the versor parts alone, we have
$$U\frac{\beta}{\alpha}=\cos\theta+\epsilon\sin\theta$$
Similarly
$$U\frac{\gamma}{\beta}=\cos\phi+\epsilon\sin\phi$$
$\phi$ being the positive angle between the directions of
$\gamma$ and $\beta$, and $\epsilon$
the same vector as before, if $\alpha$, $\beta$, $\gamma$ be coplanar.

Also we have
$$U\frac{\gamma}{\alpha}=\cos(\theta+\phi)+\epsilon\sin(\theta+\phi)$$
But we have always
$$\frac{\gamma}{\beta}.\frac{\beta}{\alpha}=\frac{\gamma}{\alpha}$$
and therefore
$$U\frac{\gamma}{\beta}.U\frac{\beta}{\alpha}=U\frac{\gamma}{\alpha}$$
or
$$
\cos(\phi+\theta)+\epsilon\sin(\phi+\theta)=
(\cos\phi+\epsilon\sin\phi)(\cos\theta+\epsilon\sin\theta)
$$
$$
=\cos\phi\cos\theta\sin\phi\sin\theta+
\epsilon(\sin\phi\cos\theta+\cos\phi\sin\theta)
$$
from which we have at once the fundamental formulae for the
cosine and sine of the sum of two arcs, by equating separately the
scalar and vector parts of these quaternions.

And we see, as an immediate consequence of the expressions
above, that
$$\cos m\theta+\epsilon\sin m\theta=(\cos\theta+\epsilon\sin\theta)^m$$
if $m$ be a positive whole number. For the lefthand side is a versor
which turns through the angle $m\theta$ at once, while the righthand
side is a versor which effects the same object by $m$ successive turn
ings each through an angle $\theta$. See \S\S 8, 9.

{\bf 111}. To extend this proposition to fractional indices we have
only to write $\displaystyle \frac{\theta}{n}$ for $\theta$,
when we obtain the results as in ordinary trigonometry.

From De Moivre's Theorem, thus proved, we may of course
deduce the rest of Analytical Trigonometry. And as we have
already deduced, as interpretations of selfevident quaternion
transformations (\S\S 97, 104), the fundamental formulae for the solution
of plane triangles, we will now pass to the consideration of spherical
trigonometry, a subject specially adapted for treatment by qua
ternions; but to which we cannot afford more than a very few
sections. (More on this subject will be found in Chap. XI in
connexion with the Kinematics of rotation.) The reader is referred to
Hamilton s works for the treatment of this subject by quaternion
exponentials.

{\bf 112}. Let $\alpha$, $\beta$, $\gamma$
be unitvectors drawn from the centr to the
corners $A$, $B$, $C$ of a triangle on the unitsphere. Then it is evident
that, with the usual notation, we have (\S 96),
$$
S\alpha\beta=\cos c,\;\;\;\;
S\beta\gamma=\cos a,\;\;\;\;
S\gamma\alpha=\cos b
$$
$$
TV\alpha\beta=\sin c,\;\;\;\;
TV\beta\gamma=\sin a,\;\;\;\;
TV\gamma\alpha=\sin b
$$
Also $UV\alpha\beta$, $UV\beta\gamma$, $UV\gamma\alpha$
are evidently the vectors of the corners of the polar triangle.

Hence
$$S.UV\alpha\beta UV\beta\gamma=\cos B\textrm{, \&c.}$$
$$TV.UV\alpha\beta UV\beta\gamma=\sin B\textrm{, \&c.}$$

Now (\S 90 (1)) we have
$$
\begin{array}{rcl}
SV\alpha\beta V\beta\gamma&=&S.\alpha V(\beta V\beta\gamma)\\
&=&S\alpha\beta S\beta\gamma + \beta^2S\alpha\gamma
\end{array}
$$
Remembering that we have
$$
SV\alpha\beta V\beta\gamma=
TV\alpha\beta TV\beta\gamma S.UV\alpha\beta UV\beta\gamma
$$
we see that the formula just written is equivalent to
$$\sin a \sin c \cos B = \cos a \cos c + \cos b$$
or
$$\cos b = \cos a \cos c + \sin a \sin c \cos B$$

{\bf 113}. Again,
$$V.V\alpha\beta V\beta\gamma=\beta S\alpha\beta\gamma$$
which gives
$$
TV.V\alpha\beta V\beta\gamma=
TS.\alpha\beta\gamma=
TS.\alpha V\beta\gamma=
TS.\beta V\gamma\alpha=
TS.\gamma V\alpha\beta
$$
or
$$\sin a \sin c \sin B = \sin a \sin p_a = \sin b \sin p_b = \sin c \sin p_c$$
where $p_a$ is the arc drawn from $A$ perpendicular to $BC$, \&c.
Hence
$$\sin p_a = \sin c \sin B$$
$$\sin p_b = \frac{\sin a \sin c}{\sin b} \sin B$$
$$\sin p_c = \sin a \sin B$$

{\bf 114}. Combining the results of the last two sections, we have
$$V\alpha\beta .V\beta\gamma=\sin a \sin c \cos B\beta \sin a \sin c \sin B$$
$$= \sin a \sin c (\cos B  \beta \sin B)$$
$$
\left.
\begin{array}{lcl}
\textrm{Hence} & \hbox{\hskip 1cm}
& U.V\alpha\beta V\beta\gamma=(\cos B\beta\sin B)\\
\textrm{and} & \hbox{\hskip 1.1cm}
& U.V\gamma\beta V\beta\alpha=(\cos B+\beta\sin B)
\end{array}
\right\}
$$
These are therefore versors which turn all vectors perpendicular to
$OB$ negatively or positively about $OB$ through the angle $B$.

[It will be shown later (\S 119) that, in the combination
$$(\cos B+\beta\sin B)(\;\;\;\;)(\cos B \beta\sin B)$$
the system operated on is made to rotate, as if rigid, round the
vector axis $\beta$ through an angle $2B$.]

As another instance, we have
$$
\begin{array}{rl}
\tan B &=\displaystyle \frac{\sin B}{\cos B}\\
&\\
&=\displaystyle
\frac{TV.V\alpha\beta V\beta\gamma}{S.V\alpha\beta V\beta\gamma}\\
&\\
&=\displaystyle
\beta^{1}\frac{V.V\alpha\beta V\beta\gamma}{S.V\alpha\beta V\beta\gamma}\\
&\\
&=\displaystyle
\frac{S.\alpha\beta\gamma}{S\alpha\gamma+S\alpha\beta S\beta\gamma}
=\textrm{ \&c}
\end{array}
\eqno{(1)}
$$

The interpretation of each of these forms gives a different theorem
in spherical trigonometry.

{\bf 115}. Again, let us square the equal quantities
$$
V.\alpha\beta\gamma\;\;\;\;\textrm{ and }\;\;\;\;
\alpha S\beta\gamma\beta S\alpha\gamma+\gamma S\alpha\beta$$
supposing $\alpha$, $\beta$, $\gamma$ to be any unitvectors whatever. We have
$$
(V.\alpha\beta\gamma)^2=
S^2\beta\gamma + S^2\gamma\alpha +
S^2\alpha\beta + 2S\beta\gamma S\gamma\alpha S\alpha\beta
$$
But the lefthand member may be written as
$$T^2.\alpha\beta\gammaS^2.\alpha\beta\gamma$$
whence
$$
1S^2.\alpha\beta\gamma=S^2\beta\gamma +S^2\gamma\alpha +
S^2\alpha\beta+2S\beta\gamma S\gamma\alpha S\alpha\beta
$$
or
$$
1\cos^2 a  \cos^2 b  \cos^2 c + 2 \cos a \cos b \cos c$$
$$= \sin^2 a \sin^2 p_a = \textrm{ \&c.}$$
$$= \sin^2 a \sin^2 b \sin^2 C = \textrm{ \&c.}$$
all of which are wellknown formulae.

{\bf 116}. Again, for any quaternion,
$$q=Sq+Vq$$
so that, if $n$ be a positive integer,
$$q^n=(Sq)^n+n(Sq)^{n1}Vq+
\frac{n.\overline{n1}}{1.2}(Sq)^{n2}(Vq)^2+\ldots$$
From this at once
$$S.q^n=(Sq)^n\frac{n.\overline{n1}}{1.2}(Sq)^{n2}T^2Vq$$
$$+\frac{n.\overline{n1}.\overline{n2}.\overline{n3}}{1.2.3.4}
(Sq)^{n4}T^4(Vq)\textrm{\&c.,}$$
$$V.q^n=Vq\left[n(Sq)^{n1}
\frac{n.\overline{n1}.\overline{n2}}{1.2.3}
(Sq)^{n3}T^2Vq+\textrm{\&c.,}\right]$$
If $q$ be a versor we have
$$q=\cos u +\theta\sin u$$
so that
$$
\begin{array}{rl}
S.q^n & =\displaystyle
(\cos u)^n\frac{n.\overline{n1}}{1.2}(\cos u)^{n2}(\sin u)^2+\ldots\\
&\\
&=\displaystyle\cos nu;\\
&\\
V.q^n & =\displaystyle
\theta\sin u\left[n(\cos u)^{n1}
\frac{n.\overline{n1}.\overline{n2}}{1.2.3}
(\cos u)^{n3}(\sin u)^2+\ldots\right]\\
&\\
&=\displaystyle\theta\sin nu;
\end{array}
$$
as we might at once have concluded from \S 110.

Such results may be multiplied indefinitely by any one who has
mastered the elements of quaternions.

{\bf 117}. A curious proposition, due to Hamilton, gives us a
quaternion expression for the {\sl spherical excess} in any triangle.
The following proof, which is very nearly the same as one of his,
though by no means the simplest that can be given, is chosen here
because it incidentally gives a good deal of other information.
We leave the quaternion proof as an exercise.

Let the unitvectors drawn from the centre of the sphere to
$A$, $B$, $C$, respectively, be $\alpha$, $\beta$, $\gamma$.
It is required to express, as an
arc and as an angle on the sphere, the quaternion
$$\beta\alpha^{1}\gamma$$

\includegraphics{ps/quat18.ps}

The figure represents an orthographic projection made on a
plane perpendicular to $\gamma$. Hence $C$ is the centre of the circle $DEe$.
Let the great circle through $A$, $B$ meet $DEe$ in $E$, $e$, and let $DE$ be
a quadrant. Thus
${\stackrel{\frown}{DE}}$ represents $\gamma$ (\S 72). Also make
${\stackrel{\frown}{EF}} = {\stackrel{\frown}{AB}}$
$=\beta\alpha^{1}$ Then, evidently,
$${\stackrel{\frown}{DF}}=\beta\alpha^{1}\gamma$$
which gives the arcual representation required.

Let $DF$ cut $Ee$ in $G$. Make $Ca = EG$, and join $D$, $a$, and $a$, $F$.
Obviously, as $D$ is the pole of $Ee$, $Da$ is a quadrant ; and since
$EG = Ca$, $Ga = EG$, a quadrant also. Hence $a$ is the pole of $DG$,
and therefore the quaternion may be represented by the angle
$DaF$.

Make $Cb = Ca$, and draw the arcs $Pa\beta$, $Pb\alpha$ from $P$, the pole of
$AB$. Comparing the triangles $Eb\alpha$ and $ea\beta$,
we see that $E\alpha = e\beta$.
But, since $P$ is the pole of $AB$, $F\beta a$ is a right angle: and therefore
as $Fa$ is a quadrant, so is $F\beta$. Thus $AB$ is the complement of $E\alpha$
or $\beta e$, and therefore
$$\alpha\beta=2AB$$

Join $bA$. and produce it to $c$ so that $Ac = bA$; join $c$, $P$, cutting
$AB$ in $o$. Also join $c$, $B$, and $B$, $a$.

Since $P$ is the pole of $AB$, the angles at $o$ are right angles;
and therefore, by the equal triangles $b\alpha A$, $coA$, we have
$$\alpha A = Ao$$
But
$$\alpha\beta = 2AB$$
whence
$$oB=B\beta$$
and therefore the triangles $coB$ and $Ba\beta$ are equal, and $c$, $B$, $a$
lie on the same great circle.

Produce $cA$ and $cB$ to meet in $H$ (on the opposite side of the
sphere). $H$ and $c$ are diametrically opposite, and therefore $cP$,
produced, passes through $H$.

Now $Pa = Pb = PH$, for they differ from quadrants by the
equal arcs $a\beta$, $b\alpha$, $oc$. Hence these arcs divide the
triangle $Hab$ into three isosceles triangles.

But
$$\angle PHb + \angle PHA = \angle aHb = \angle bca$$
Also
$$\angle Pab = \pi  \angle cab  \angle PaH$$
$$\angle Pba = \angle Pab = \pi  \angle cba  \angle PbH$$
Adding,
$$2\angle Pab = 2\pi  \angle cab  \angle cba  \angle bca$$
$$= \pi  (\textrm{spherical excess of }abc)$$
But, as $\angle Fa\beta$ and $\angle Dae$ are right angles, we have
$$
\textrm{angle of }\beta\alpha^{1}\gamma =
\angle FaD = \beta ae = \angle Pab
$$
$$=\frac{\pi}{2}  \frac{1}{2}(\textrm{spherical excess of }abc)$$

[Numerous singular geometrical theorems, easily proved {\sl ab
initio} by quaternions, follow from this: e.g. The arc $AB$, which
bisects two sides of a spherical triangle $abc$, intersects the base at
the distance of a quadrant from its middle point. All spherical
triangles, with a common side, and having their other sides
bisected by the same great circle (i.e. having their vertices in a
small circle parallel to this great circle) have equal areas, \&c. ]

{\bf 118}. Let $\overline{Oa}=\alpha^{'}$, $\overline{Ob}=\beta^{'}$,
$\overline{Oc}=\gamma^{'}$, and we have
$$
\begin{array}{rcl}
\left(\frac{\alpha^{'}}{\beta^{'}}\right)^{\frac{1}{2}}
\left(\frac{\beta^{'}}{\gamma^{'}}\right)^{\frac{1}{2}}
\left(\frac{\gamma^{'}}{\alpha^{'}}\right)^{\frac{1}{2}}&=&
{\stackrel{\frown}{Ca}}.{\stackrel{\frown}{cA}}.{\stackrel{\frown}{Bc}}\\
&=&{\stackrel{\frown}{Ca}}.{\stackrel{\frown}{BA}}\\
&=&{\stackrel{\frown}{EG}}.{\stackrel{\frown}{FE}}=
{\stackrel{\frown}{FG}}
\end{array}
$$

But $FG$ is the complement of $DF$. Hence the {\sl angle of the
quaternion}
$$
\left(\frac{\alpha^{'}}{\beta^{'}}\right)^{\frac{1}{2}}
\left(\frac{\beta^{'}}{\gamma^{'}}\right)^{\frac{1}{2}}
\left(\frac{\gamma^{'}}{\alpha^{'}}\right)^{\frac{1}{2}}
$$
{\sl is half the spherical excess of the triangle whose angular points are
at the extremities of the unitvectors} $\alpha^{'}$, $\beta^{'}$, and
$\gamma^{'}$.

[In seeking a purely quaternion proof of the preceding proposi
tions, the student may commence by showing that for any three
unitvectors we have
$$
\frac{\beta}{\alpha}\frac{\gamma}{\beta}\frac{\alpha}{\gamma}=
(\beta\alpha^{1}\gamma)^2
$$

The angle of the first of these quaternions can be easily assigned;
and the equation shows how to find that of $\beta\alpha^{1}\gamma$.

Another easy method is to commence afresh by forming from
the vectors of the corners of a spherical triangle three new vectors
thus:
$$
\alpha^{'}=\left(\frac{\beta+\gamma}{\alpha}^{2}\right)^2 .\;\alpha,\;\;\;\;\;
\textrm{\&c.}
$$

Then the angle between the planes of $\alpha$, $\beta^{'}$ and
$\gamma^{'}$, $\alpha$; or of $\beta$, $\gamma^{'}$ and $\alpha^{'}$,
$\beta$; or of $\gamma$, $\alpha^{'}$ and $\beta^{'}$, $\gamma$
is obviously the spherical excess.

But a still simpler method of proof is easily derived from the
composition of rotations.]

{\bf 119}. It may be well to introduce here, though it belongs
rather to Kinematics than to Geometry, the interpretation of the
operator
$$q(\;\;\;)q^{1}$$

By a rotation, about the axis of $q$, through double the angle of $q$,
the quaternion $r$ becomes the quaternion $qrq^{1}$ . Its tensor and
angle remain unchanged, its plane or axis alone varies.

\includegraphics{ps/quat19.ps}

A glance at the figure is sufficient for
the proof, if we note that of course
$T . qrq^{1} = Tr$, and therefore that we need
consider the {\sl versor} parts only. Let $Q$
be the pole of $q$.
$$
{\stackrel{\frown}{AB}}=q,\;\;\;\;
{\stackrel{\frown}{AB^{1}}}=q^{1},\;\;\;\;
{\stackrel{\frown}{B^{'}C^{'}}}=r
$$
Join $C^{'}A$, and make
${\stackrel{\frown}{AC}}={\stackrel{\frown}{C^{'}A}}$. Join $CB$.

Then ${\stackrel{\frown}{CB}}$ is $qrq^{1}$,
its arc $CB$ is evidently equal in length to that
of $r$, $B^{'}C^{'}$; and its plane (making the same angle with
$B^{'}B$ that that of
$B^{'}C^{'}$ does) has evidently been made to revolve about $Q$, the
pole of $q$, through double the angle of $q$.

It is obvious, from the nature of the above proof, that this
operation is distributive; i.e. that
$$q(r+s)q^{1}=qrq^{1}+qsq^{1}$$

If $r$ be a vector, $=\rho$, then $q\rho q^{1}$
(which is also a vector) is the
result of a rotation through double the angle of $q$ about the axis
of $q$. Hence, as Hamilton has expressed it, if $B$ represent a rigid
system, or assemblage of vectors,
$$qBq^{1}$$
is its new position after rotating through double the angle of $q$
about the axis of $q$.

{\bf 120}. To compound such rotations, we have
$$r . qBq^{1} . r^{1} = rq . B . (rq)^{1}$$

To cause rotation through an angle $t$fold the double of the angle
of $q$ we write
$$q^{t}Bq^{t}$$

To reverse the direction of this rotation write
$$q^{t}Bq^{t}$$

To {\sl translate} the body $B$ without rotation, each point of it moving
through the vector $\alpha$, we write $\alpha + B$.

To produce rotation of the translated body about the same axis,
and through the same angle, as before,
$$q(\alpha+B)q^{1}$$

Had we rotated first, and then translated, we should have had
$$\alpha+qBq^{1}$$

From the point of view of those who do not believe in the
Moon s rotation, the former of these expressions ought to be
$$q\alpha q^{1}+B$$
instead of
$$q\alpha q^{1}+qBq^{1}$$
But to such men quaternions are unintelligible.

{\bf 121}. The operator above explained finds, of course, some
of its most direct applications in the ordinary questions of
Astronomy, connected with the apparent diurnal rotation of the
stars. If $\lambda$ be a unitvector parallel to the polar axis, and $h$ the
hour angle from the meridian, the operator is
$$
\left(\cos\frac{h}{2}\lambda\sin\frac{h}{2}\right)
\left(\;\;\;\;\right)
\left(\cos\frac{h}{2}+\lambda\sin\frac{h}{2}\right)
$$
or
$$L^{1}\left(\;\;\;\;\right)L$$

the inverse going first, because the {\sl apparent} rotation is negative
(clockwise).

If the upward line be $i$, and the southward $j$, we have
$$\lambda = i\sin lj\cos l$$
where $l$ is the latitude of the observer. The meridian equatorial
unit vector is
$$\mu = i\cos l+j\sin l$$
and $\lambda$, $\mu$, $k$ of course form a rectangular unit system.

The meridian unitvector of a heavenly body is
$$\delta=i\cos(ld)+j\sin(ld)$$
$$=\lambda\sin d+\mu\cos d$$
where $d$ is its declination.

Hence when its hourangle is $h$, its vector is
$$\delta^{'}=L^{1}\delta L$$

The vertical plane containing it intersects the horizon in
$$iVi\delta^{'}=jSj\delta^{'}+kSk\delta^{'}$$
so that
$$\tan(azimuth)=\frac{Sk\delta^{'}}{Sj\delta^{'}}\eqno{(1)}$$

[This may also be obtained directly from the last formula (1)
of \S 114.]

To find its Amplitude, i.e. its azimuth at rising or setting,
the hourangle must be obtained from the condition
$$Si\delta^{'}=0\eqno{(2)}$$

These relations, with others immediately deducible from them,
enable us (at once and for ever) to dispense with the hideous
formulae of Spherical Trigonometry.

{\bf 122}. To show how readily they can be applied, let us
translate the expressions above into the ordinary notation. This
is effected at once by means of the expressions for $\lambda$, $\mu$, $L$,
and $\delta$ above, which give by inspection
$$\delta^{'}=\lambda\sin d+(\mu\cos hk\sin h)\cos d$$
= x sin d + (fjb cos h k sin h) cos d,
and we have from (1) and (2) of last section respectively
$$
\tan(azimuth)=
\frac{\sin h\cos d}{\cos l\sin d\sin l\cos d\cos h}\eqno{(1)}
$$
$$
\cos h+\tan l \tan d=0\eqno{(2)}
$$

In Capt. Weir s ingenious {\sl Azimuth Diagram}, these equations
are represented graphically by the rectangular coordinates of a
system of confocal conics: viz.
$$
\left.
\begin{array}{c}
x = \sin h \sec l \\
y = \cos h \tan l
\end{array}
\right\}\eqno{(3)}
$$

The ellipses of this system depend upon $l$ alone, the hyperbolas
upon $h$. Since (1) can, by means of (3), be written as
$$\tan(azimuth)=\frac{x}{\tan dy}$$
we see that the azimuth can be constructed at once by joining
with the point $0$, $\tan d$, the intersection of the proper ellipse and
hyperbola.

Equation (2) puts these expressions for the coordinates in the
form
$$
\left.
\begin{array}{c}
x=\sec l\sqrt{1\tan^{2} l\tan^{2} d}\\
y=\tan^{2} l \tan d
\end{array}
\right\}
$$

The elimination of $d$ gives the ellipse as before, but that of $l$
gives, instead of the hyperbolas, the circles
$$x^{2}+y^{2}y(\tan d\cot d)=1$$

The radius is
$$\frac{1}{2}(\tan d+ \cot d)$$
and the coordinates of the centre are
$$0,\;\;\;\frac{1}{2}(\tan d \cot d)$$

123. A scalar equation in $\rho$, the vector of an undetermined
point, is generally the equation of a {\sl surface}; since we may use
in it the expression
$$\rho=x\alpha$$
where $x$ is an unknown scalar, and $\alpha$ any assumed unitvector.
The result is an equation to determine $x$. Thus one or more
points are found on the vector $x\alpha$, whose coordinates satisfy the
equation; and the locus is a surface whose degree is determined
by that of the equation which gives the values of $x$.

But a {\sl vector} equation in $\rho$, as we have seen, generally leads to
three scalar equations, from which the three rectangular or other
components of the sought vector are to be derived. Such a vector
equation, then, usually belongs to a definite number of {\sl points} in
space. But in certain cases these may form a {\sl line}, and even a
{\sl surface}, the vector equation losing as it were one or two of the
three scalar equations to which it is usually equivalent.

Thus while the equation
$$\alpha\rho=\beta$$
gives at once
$$\rho=\alpha^{1}\beta$$
which is the vector of a definite point, since by making $\rho$ a {\sl vector}
we have evidently assumed
$$S\alpha\beta=0$$
the closely allied equation
$$V\alpha\rho=\beta$$
is easily seen to involve
$$S\alpha\beta=0$$
and to be satisfied by
$$\rho=\alpha^{1}\beta+x\alpha$$
whatever be $x$. Hence the vector of any point whatever in the
line drawn parallel to $\alpha$ from the extremity of $\alpha^{1}\beta$
satisfies the given equation. [The difference between the results depends
upon the fact that $S\alpha\rho$ is indeterminate in the second form, but
definite (= 0) in the first.]

{\bf 124}. Again,
$$V\alpha\rho . V\rho\beta=(V\alpha\beta)^{2}$$
is equivalent to but two scalar equations. For it shows that $V\alpha\rho$
and $V\beta\rho$ are parallel, i.e. $\rho$
lies in the same plane as $\alpha$ and $\beta$, and
can therefore be written (\S 24)
$$\rho=x\alpha+y\beta$$
where $x$ and $y$ are scalars as yet undetermined.

We have now
$$V\alpha\rho=yV\alpha\beta$$
$$V\rho\beta=xV\alpha\beta$$
which, by the given equation, lead to
$$xy=1,\;\;\;\textrm{or}\;\;\;y=\frac{1}{x}$$
or finally
$$\rho=x\alpha+\frac{1}{x}\beta$$
which (\S 40) is the equation of a hyperbola whose asymptotes are
in the directions of $\alpha$ and $\beta$.

{\bf 125}. Again, the equation
$$V . V\alpha\beta V\alpha\rho=0$$
though apparently equivalent to three scalar equations, is really
equivalent to one only. In fact we see by \S 91 that it may be
written
$$\alpha S.\alpha\beta\rho=0$$
whence, if $\alpha$ be not zero, we have
$$S . \alpha\beta\rho=0$$
and thus (\S 101) the only condition is that $\rho$ is coplanar with
$\alpha$, $\beta$.
Hence the equation represents the plane in which
$\alpha$ and $\beta$ lie.

{\bf 126}. Some very curious results are obtained when we extend
these processes of interpretation to functions of a {\sl quaternion}
$$q=w+\rho$$
instead of functions of a mere {\sl vector} $\rho$.

A scalar equation containing such a quaternion, along with
quaternion constants, gives, as in last section, the equation of a
surface, if we assign a definite value to $w$. Hence for successive
values of $w$, we have successive surfaces belonging to a system ;
and thus when $w$ is indeterminate the equation represents not a
{\sl surface}, as before, but a {\sl volume},
in the sense that the vector of any
point within that volume satisfies the equation.

Thus the equation
$$(Tq)^2=a^2$$
or
$$w^2\rho^2=a^2$$
or
$$(TP)^2=a^2w^2$$
represents, for any assigned value of $w$, not greater than $a$, a sphere
whose radius is $\sqrt{a^2w^2}$. Hence the equation is satisfied by the
vector of any point whatever in the {\sl volume} of a sphere of radius $a$,
whose centre is origin.

Again, by the same kind of investigation,
$$(T (q\beta))^2=a^2$$
where $q=w+\rho$, is easily seen to represent the volume of a sphere
of radius $a$ described about the extremity of $\beta$ as centre.

Also $S(q^2) = a^2$ is the equation of infinite space less the space
contained in a sphere of radius $a$ about the origin.

Similar consequences as to the interpretation of vector
equations in quaternions may be readily deduced by the reader.

{\bf 127}. The following transformation is enuntiated without proof
by Hamilton ({\sl Lectures}, p. 587, and {\sl Elements}, p. 299).
$$r^{1}(r^2q^2)^{\frac{1}{2}}q^{1}=U(rq+KrKq)$$
To prove it, let
$$r^{1}(r^2q^2)^{\frac{1}{2}}q^{1}=t$$
then
$$Tt=1$$
and therefore
$$Kt=t^{1}$$
But
$$(r^2q^2)^{\frac{1}{2}}=rtq$$
or
$$r^2q^2=rtqrtq$$
or
$$rq=tqrt$$
Hence
$$KqKr=t^{1}KrKqt^{1}$$
or
$$KrKq=tKqKrt$$
Thus we have
$$U(rq\pm KrKq)=tU(qr\pm KqKr)t$$
or, if we put
$$s=U(qr\pm KqKr)$$
$$Ks=\pm tst$$
Hence
$$sKs=(Ts)^2=1=\pm stst$$
which, if we take the positive sign, requires
$$st=\pm 1$$
or
$$t=\pm s^{1}=\pm UKs$$
which is the required transformation.

[It is to be noticed that there are other results which might
have been arrived at by using the negative sign above ; some
involving an arbitrary unitvector, others involving the imaginary
of ordinary algebra.]

{\bf 128}. As a final example, we take a transformation of Hamilton's,
of great importance in the theory of surfaces of the second order.

Transform the expression
$$(S\alpha\rho)^2+(S\beta\rho)^2+(S\gamma\rho)^2$$
in which $\alpha$, $\beta$, $\gamma$
are any three mutually rectangular vectors, into the form
$$\left(\frac{T(\iota\rho+\rho\kappa)}{\kappa^2\iota^2}\right)^2$$
which involves only two vectorconstants, $\iota$, $\kappa$.

[The student should remark here that $\iota$, $\kappa$, two undetermined
vectors, involve six disposable constants : and that $\alpha$, $\beta$,
$\gamma$, being a {\sl rectangular} system, involve also only six constants.]
$$
\begin{array}{rcl}
\{T(\iota\rho+\rho\kappa)\}^2
&=& (\iota\rho+\rho\kappa)(\rho\iota+\kappa\rho)\;\;\;\;(\S\S 52,55)\\
&=& (\iota^2+\kappa^2)\rho^2+(\iota\rho\kappa\rho+\rho\kappa\rho\iota)\\
&=& (\iota^2+\kappa^2)\rho^2+2S.\iota\rho\kappa\rho\\
&=& (\iota\kappa)^2\rho^2+4S\iota\rho S\kappa\rho
\end{array}
$$
Hence
$$
(S\alpha\rho)^2+(S\beta\rho)^2+(S\gamma\rho)^2=
\frac{(\iota\kappa)^2}{(\kappa^2\iota^2)^2}\rho^2+
4\frac{S\iota\rho S\kappa\rho}{(\kappa^2\iota^2)^2}
$$
But
$$
\alpha^{2}(S\alpha\rho)^2+
\beta^{2}(S\beta\rho)^2+
\gamma^{2}(S\gamma\rho)^2=
\rho^2\;\;\;\;(\S\S 25,73).
$$
Multiply by $\beta^2$ and subtract, we get
$$
\left(1\frac{\beta^2}{\alpha^2}\right)(S\alpha\rho)^2
\left(\frac{\beta^2}{\gamma^2}1\right)(S\gamma\rho)^2=
\left\{\frac{(\iota\kappa)^2}{(\kappa^2\iota^2)^2}\beta^2\right\}\rho^2+
4\frac{S\iota\rho S\kappa\rho}{(\kappa^2\iota^2)^2}
$$

The left side breaks up into two real factors if $\beta^2$ be intermediate
in value to $\alpha^2$ and $\gamma^2$:
and that the right side may do so the term
in $\rho^2$ must vanish. This condition gives
$$\beta^2=\frac{(\iota\kappa)^2}{(\kappa^2\iota^2)^2}$$
and the identity becomes
$$
S\left\{
\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}+
\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
\right\}
\rho S\left\{
\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}
\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
\right\}\rho=
4\frac{S\iota\rho S\kappa\rho}{(\kappa^2\iota^2)^2}
$$
Hence we must have
$$
\frac{2\iota}{\kappa^2\iota^2}=
p\left\{
\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}+
\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
\right\}
$$
$$
\frac{2\kappa}{\kappa^2\iota^2}=
\frac{1}{p}\left\{
\alpha\sqrt{\left(1\frac{\beta^2}{\alpha^2}\right)}
\gamma\sqrt{\left(\frac{\beta^2}{\gamma^2}1\right)}
\right\}
$$
where $\rho$ is an undetermined scalar.

To determine $\rho$, substitute in the expression for $\beta^2$, and we find
$$
\begin{array}{rcl}
4\beta^2=\frac{4(\iota\kappa)^2}{(\kappa^2\iota^2)^2}
&=&\left(p\frac{1}{p}\right)^2(\alpha^2\beta^2)+
\left(p+\frac{1}{p}\right)^2(\beta^2\gamma^2)\\
&=&\left(p^2+\frac{1}{p^2}\right)(\alpha^2\gamma^2)
2(\alpha^2+\gamma^2)+4\beta^2
\end{array}
$$

Thus the transformation succeeds if
$$p^2+\frac{1}{p^2}=\frac{2(\alpha^2+\gamma^2)}{\alpha^2\gamma^2}$$
which gives
$$p+\frac{1}{p}=\pm 2\sqrt{\frac{\alpha^2}{\alpha^2\gamma^2}}$$
$$p\frac{1}{p}=\pm 2\sqrt{\frac{\gamma^2}{\alpha^2\gamma^2}}$$
Hence
$$
\frac{4(\kappa^2\iota^2)}{(\kappa^2\iota^2)^2}=
\left(\frac{1}{p^2}p^2\right)(\alpha^2\gamma^2)=
\pm 4\sqrt{\alpha^2\gamma^2}
$$
$$
(\kappa^2\iota^2)^{1}=\pm T\alpha T\gamma$$

Again
$$
p=\frac{T\alpha +T\gamma}{\sqrt{\gamma^2\alpha^2}},\;\;\;\;
\frac{1}{p}=\frac{T\alpha T\gamma}{\sqrt{\gamma^2\alpha^2}}
$$
and therefore
$$
2\iota=\frac{T\alpha +T\gamma}{T\alpha T\gamma}
\left(
\sqrt{\frac{\beta^2\alpha^2}{\gamma^2\alpha^2}}U\alpha+
\sqrt{\frac{\gamma^2\beta^2}{\gamma^2\alpha^2}}U\gamma
\right)
$$
$$
2\kappa=\frac{T\alpha T\gamma}{T\alpha T\gamma}
\left(
\sqrt{\frac{\beta^2\alpha^2}{\gamma^2\alpha^2}}U\alpha
\sqrt{\frac{\gamma^2\beta^2}{\gamma^2\alpha^2}}U\gamma
\right)
$$
Thus we have proved the possibility of the transformation, and
determined the transforming vectors $\iota$, $\kappa$.

{\bf 129}. By differentiating the equation
$$
(S\alpha\rho)^2+(S\beta\rho)^2+(S\gamma\rho)^2=
\left(\frac{T(\iota\rho+\rho\kappa)}{(\kappa^2\iota^2)}\right)^2
$$
we obtain, as will be seen in Chapter IV, the following,
$$
S\alpha\rho S\alpha\rho^{'}+
S\beta\rho S\beta\rho^{'}+
S\gamma\rho S\gamma\rho^{'}=
\frac{S.(\iota\rho+\rho\kappa)(\kappa\rho^{'}+\rho^{'}\iota)}
{(\kappa^2\iota^2)^2}
$$
where $\rho$ also may be any vector whatever.

This is another very important formula of transformation ; and
it will be a good exercise for the student to prove its truth by
processes analogous to those in last section. We may merely
observe, what indeed is obvious, that by putting $\rho^{'}=\rho$ it becomes
the formula of last section. And we see that we may write, with
the recent values of $\iota$ and $\kappa$ in terms of
$\alpha$, $\beta$, $\gamma$, the identity
$$
\begin{array}{rcl}
\alpha S\alpha\rho+\beta S\beta\rho+\gamma S\gamma\rho
&=&\displaystyle
\frac{(\iota^2+\kappa^2)\rho+2V.\iota\rho\kappa}{(\kappa^2\iota^2)^2}\\
&&\\
&=&\displaystyle
\frac{(\iota\kappa)^2\rho+2(\iota S\kappa\rho+\kappa S\iota\rho)}
{(\kappa^2\iota^2)^2}
\end{array}
$$

{\bf 130}. In various quaternion investigations, especially in such
as involve {\sl imaginary} intersections of curves and surfaces, the old
imaginary of algebra of course appears. But it is to be particularly
noticed that this expression is analogous to a scalar and not to a
vector, and that like real scalars it is commutative in
multiplication with all other factors. Thus it appears, by the same proof
as in algebra, that any quaternion expression which contains this
imaginary can always be broken up into the sum of two parts, one
real, the other multiplied by the first power of $\sqrt{1}$. Such an
expression, viz.
$$q=q^{'}+\sqrt{1}q^{''}$$
where $q^{'}$ and $q^{''}$ are real quaternions, is called by Hamilton a
BIQUATERNION. [The student should be warned that the term
Biquaternion has since been employed by other writers in the
sense sometimes of a ``set'' of 8 elements, analogous to the
Quaternion 4 ; sometimes for an expression $q^{'} + \theta q^{''}$
where $\theta$ is not
the algebraic imaginary. By them Hamilton s Biquaternion is
called simply a quaternion with nonreal constituents.] Some
little care is requisite in the management of these expressions, but
there is no new difficulty. The points to be observed are: first,
that any biquaternion can be divided into a real and an imaginary
part, the latter being the product of $\sqrt{1}$ by a real quaternion;
second, that this $\sqrt{1}$ is commutative with all other quantities in
multiplication; third, that if two biquaternions be equal, as
$$q^{'}+\sqrt{1}\;q^{''}=r^{'}+\sqrt{1}\;r^{''}$$
we have, as in algebra,
$$q^{'}=r^{'},\;\;\;\;q^{''}=r^{''}$$
so that an equation between biquaternions involves in general
{\sl eight} equations between scalars. Compare \S 80.

{\bf 131}. We have obviously, since $\sqrt{1}$ is a scalar,
$$S(q^{'}+\sqrt{1}\;q^{''})=Sq^{'}+\sqrt{1}\;Sq^{''}$$
$$V(q^{'}+\sqrt{1}\;q^{''})=Vq^{'}+\sqrt{1}\;Vq^{''}$$
Hence (\S 103)
$$\{T(q^{'}+\sqrt{1}\;q^{''})\}^2$$
$$
=(Sq^{'}+\sqrt{1}\;Sq^{''}+Vq^{'}+\sqrt{1}\;Vq^{''})
(Sq^{'}+\sqrt{1}\;Sq^{''}Vq^{'}\sqrt{1}\;Vq^{''})
$$
$$=(Sq^{'}+\sqrt{1}\;Sq^{''})^2(Vq^{'}+\sqrt{1}\;Vq^{''})^2$$
$$=(Tq^{'})^2(Tq^{''})^2+2\sqrt{1}\;S.q^{'}Kq^{''}$$

The only remark which need be made on such formulae is this, that
{\sl the tensor of a biquaternion may vanish while both of the component
quaternions are finite}.

Thus, if
$$Tq^{'}=Tq^{''}$$
and
$$S.q^{'}Kq^{''}=0$$
the above formula gives
$$T(q^{'}+\sqrt{1}\;q^{''})=0$$
The condition
$$S.q^{'}Kq^{''}=0$$
may be written
$$
Kq^{''}=q^{'1}\alpha,\;\;\;\textrm{ or }\;\;\;
q^{''}=\alpha Kq^{'1}=\frac{\alpha q^{'}}{(Tq^{'})^2}
$$
where $\alpha$ is any vector whatever.

Hence
$$Tq^{'}=Tq^{''}=TKq^{''}=\frac{T\alpha}{Tq^{''}}$$
and therefore
$$
Tq^{'}(Uq^{'}\sqrt{1}\;U\alpha . Uq^{'})=
(1\sqrt{1}\;U\alpha)q^{'}
$$
is the general form of a biquaternion whose tensor is zero.

{\bf 132}. More generally we have, $q$, $r$, $q^{'}$, $r^{'}$
being any four real and nonevanescent quaternions,
$$
(q+\sqrt{1}\;q^{'})(r+\sqrt{1}\;r^{'})=
qrq^{'}r^{'}+\sqrt{1}\;(qr^{'}+q^{'}r)
$$
That this product may vanish we must have
$$qr=q^{'}r^{'}$$
and
$$qr^{'}=q^{'}r$$
Eliminating $r^{'}$ we have
$$qq^{'1}qr=q^{'}r$$
which gives
$$(q^{'1}q)^2=1$$
i.e.
$$q=q^{'}\alpha$$
where $\alpha$ is some unitvector.

And the two equations now agree in giving
$$r=\alpha r^{'}$$
so that we have the biquaternion factors in the form
$$q^{'}(\alpha +\sqrt{1})\;\;\;\textrm{ and }\;\;\;(\alpha\sqrt{1})r^{'}$$
and their product is
$$q^{'}(\alpha +\sqrt{1})(\alpha \sqrt{1})r^{'}$$
which, of course, vanishes.

[A somewhat simpler investigation of the same proposition
may be obtained by writing the biquaternions as
$$
q^{'}(q^{'1}q+\sqrt{1})\;\;\;\textrm{ and }\;\;\;
(rr^{'1}+\sqrt{1})r^{'}
$$
or
$$
q^{'}(q^{''}+\sqrt{1})\;\;\;\textrm{ and }\;\;\;
(r^{''}+\sqrt{1})r^{'}
$$
and showing that
$$q^{''}=r^{''}=\alpha \;\;\;\textrm{ where }\;T\alpha=1]$$

From this it appears that if the product of two {\sl bivectors}
$$
\rho+\sigma\sqrt{1}\;\;\;\textrm{ and }\;\;\;
\rho^{'}+\sigma^{'}\sqrt{1}
$$
is zero, we must have
$$\sigma^{1}\rho=\rho^{'}\sigma^{'1}=U\alpha$$
where $\alpha$ may be any vector whatever. But this result is still more
easily obtained by means of a direct process.

{\bf 133}. It may be well to observe here (as we intend to avail our
selves of them in the succeeding Chapters) that certain abbreviated
forms of expression may be used when they are not liable to confuse,
or lead to error. Thus we may write
$$T^2q\;\;\;\textrm{for}\;\;\;(Tq)^2$$
just as we write
$$\cos^2\theta\;\;\;\textrm{for}\;\;\;(\cos\theta)^2$$
although the true meanings of these expressions are
$$T(Tq)\;\;\;\textrm{and}\;\;\;\cos(\cos\theta)$$

The former is justifiable, as $T(Tq) = Tq$, and therefore $T^2q$ is not
required to signify the second tensor (or tensor of the tensor) of $q$.
But the trigonometrical usage is defensible only on the score of
convenience, and is habitually violated by the employment of
$cos^{1}x$ in its natural and proper sense.
Similarly we may write
$$S^2q\;\;\;\textrm{for}\;\;\;(Sq)^2,\;\;\;\textrm{\&c.}$$
but it may be advisable not to use
$$Sq^2$$
as the equivalent of either of those just written; inasmuch as it
might be confounded with the (generally) different quantity
$$S.q^2\;\;\;\textrm{or}\;\;\;S(q^2)$$
although this is rarely written without the point or the brackets.

The question of the use of points or brackets is one on which
no very definite rules can be laid down. A beginner ought to use
them freely, and he will soon learn by trial which of them are
absolutely necessary to prevent ambiguity.

In the present work this course has been adopted: the
earlier examples in each part of the subject being treated with
a free use of points and brackets, while in the later examples
superfluous marks of the kind are gradually got rid of.

It may be well to indicate some general principles which
regulate the omission of these marks. Thus in $S.\alpha\beta$ or
$V.\alpha\beta$
the point is obviously unnecessary: because $S\alpha=0$, and
$V\alpha=\alpha$
so that the $S$ would annihilate the term if it applied to $\alpha$ alone,
while in the same case the $V$ would be superfluous. But in $S.qr$
and $V.qr$, the point (or an equivalent) is indispensable, for $Sq.r$,
and $Vq.r$ are usually quite different from the first written
quantities. In the case of $K$, and of $d$ (used for scalar differentiation),
the {\sl omission} of the point indicates that the operator acts
{\sl only} on the nearest factor: thus
$$Kqr=(Kq)r=Kq.r,\;\;\;dqr=(dq)r=dq.r$$
Kqr = (Kq) r = Kq . r, dqr = (dq) r=dq.r;
while, if its action extend farther, we write
$$K.qr=K(qr),\;\;\;d.qr=d(qr)\;\;\;\textrm{\&c.}$$

In more complex cases we must be ruled by the general
principle of dropping nothing which is essential. Thus, for
instance
$$V(pK(dq)V(Vq.r))$$
may be written without ambiguity as
$$V(pK(dq)V(Vq.r))$$
but nothing more can be dropped without altering its value.

Another peculiarity of notation, which will occasionally be
required, shows {\sl which portions} of a complex product are affected
by an operator. Thus we write
$$\nabla S\sigma\tau$$
if $\nabla$ operates on $\sigma$ and also on $\tau$, but
$$\nabla_1S\sigma\tau_1$$
if it operates on $\tau$ alone. See, in this connection, the last Example
at the end of Chap. IV. below.

{\bf 134}. The beginner may expect to be at first a little puzzled
with this aspect of the notation; but, as he learns more of the
subject, he will soon see clearly the distinction between such an
expression as
$$S.V\alpha\beta V\beta\gamma$$
where we may omit at pleasure either the point or the first V
without altering the value, and the very different one
$$S\alpha\beta .V\beta\gamma$$
which admits of no such changes, without alteration of its value.

All these simplifications of notation are, in fact, merely examples
of the transformations of quaternion expressions to which part of
this Chapter has been devoted. Thus, to take a very simple ex
ample, we easily see that
$$
\begin{array}{rcl}
S.V\alpha\beta V\beta\gamma
&=&SV\alpha\beta V\beta\gamma
=S.\alpha\beta V\beta\gamma
=S\alpha V.\beta V\beta\gamma
=S\alpha V.(V\beta\gamma)\beta\\
&=&S\alpha V.(V\gamma\beta)\beta
=S.\alpha V(\gamma\beta)\beta
=S.V(\gamma\beta)\beta\alpha
=SV\gamma\beta V\beta\alpha\\
&=&S.\gamma\beta V\beta\alpha
=S.K(\beta\gamma)V\beta\alpha
=S.\beta\gamma KV\beta\alpha
=S.\beta\gamma V\beta\alpha\\
&=&S.V\gamma\beta V\beta\alpha,\textrm{\&c., \&c.}
\end{array}
$$

The above group does not nearly exhaust the list of even the simpler
ways of expressing the given quantity. We recommend it to the
careful study of the reader. He will find it advisable, at first, to
use stops and brackets pretty freely; but will gradually learn to
dispense with those which are not absolutely necessary to prevent
ambiguity.

There is, however, one additional point of notation to which
the reader s attention should be most carefully directed. A very
simple instance will suffice. Take the expressions
$$
\frac{\beta}{\gamma}.\frac{\gamma}{\alpha}\;\;\;\;\textrm{and}\;\;\;\;
\frac{\beta\gamma}{\gamma\alpha}
$$

The first of these is
$$\beta\gamma^{1}.\gamma\alpha^{1}=\beta\alpha^{1}$$
and presents no difficulty. But the second, though at first sight
it closely resembles the first, is in general totally different in
value, being in fact equal to
$$\beta\gamma\alpha^{1}\gamma^{1}$$

For the denominator must be treated as {\sl one quaternion}. If,
then, we write
$$\frac{\beta\gamma}{\gamma\alpha}=q$$
we have
$$\beta\gamma=q\gamma\alpha$$
so that, as stated above,
$$q=\beta\gamma\alpha^{1}\gamma^{1}$$
We see therefore that
$$
\frac{\beta}{\gamma}.\frac{\gamma}{\alpha}=
\frac{\beta}{\alpha}=
\frac{\beta\gamma}{\alpha\gamma};\;\;\;\textrm{but {\sl not}}\;\;=
\frac{\beta\gamma}{\gamma\alpha}
$$

\section{Examples to Chapter 3}

{\bf 1}. Investigate, by quaternions, the requisite formulae for
changing from any one set of coordinate axes to another ; and
derive from your general result, and also from special investiga
tions, the usual expressions for the following cases:

\begin{itemize}
\item[(a)] Rectangular axes turned about z through any angle.

\item[(b)] Rectangular axes turned into any new position by rota
tion about a line equally inclined to the three.

\item[(c)] Rectangular turned to oblique, one of the new axes
lying in each of the former coordinate planes.
\end{itemize}

{\bf 2}. Point out the distinction between
$$
\left(\frac{\alpha+\beta}{\alpha}\right)^2\;\;\;\textrm{and}\;\;\;
\frac{(\alpha+\beta)^2}{\alpha^2}
$$
and find the value of their difference.

If
$$
T\beta/\alpha=1\;\;\;\textrm{and}\;\;\;
U\frac{\alpha+\beta}{\alpha}=\left(\frac{\beta}{\alpha}\right)^{\frac{1}{2}}
$$

Show also that
$$
\frac{\alpha+\beta}{\alpha\beta}=
\frac{V\alpha\beta}{1+S\alpha\beta^{'}}
$$
and
$$
\frac{\alpha\beta}{\alpha+\beta}=
\frac{V\alpha\beta}{1S\alpha\beta^{'}}
$$
provided $\alpha$ and $\beta$ be unitvectors. If these conditions are not
fulfilled, what are the true values ?

{\bf 3}. Show that, whatever quaternion $r$ may be, the expression
$$\alpha r+r\beta$$
in which $\alpha$ and $\beta$ are any two unit vectors, is reducible to the
form
$$l(\alpha+\beta)+m(\alpha\beta1)$$
where $l$ and $m$ are scalars.

{\bf 4}. If $Tp=T\alpha=T\beta=1$, and $S.\alpha\beta\rho=0$
show by direct transformations that
$$S.U(\rho\alpha)U(\rho\beta)=\pm\sqrt{\frac{1}{2}(1S\alpha\beta)}$$
Interpret this theorem geometrically.

{\bf 5}. If $S\alpha\beta=0$, $T\alpha=T\beta=1$, show that
$$
(1+\alpha^{m})\beta=
2\cos\frac{m\pi}{4}\alpha^{\frac{m}{2}}\beta=
2S\alpha^{\frac{m}{2}}.\alpha^{\frac{m}{2}}\beta
$$

{\bf 6}. Put in its simplest form the equation
$$
\rho S.V\alpha\beta V\beta\gamma V\gamma\alpha=
aV.V\gamma\alpha V\alpha\beta+
bV.V\alpha\beta V\beta\gamma+
cV.V\beta\gamma V\gamma\alpha
$$
and show that
$$a=S.\beta\gamma\rho,\;\;\;\textrm{\&c.}$$

{\bf 7}. Show that any quaternion may in general, in one way only,
be expressed as a homogeneous linear function of four given
quaternions. Point out the nature of the exceptional cases. Also
find the simplest form in which any quaternion may generally be
expressed in terms of two given quaternions.

{\bf 8}. Prove the following theorems, and exhibit them as properties
of determinants :

\begin{itemize}
\item[(a)] $S.(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=
2S.\alpha\beta\gamma$
\item[(b)]$S.V\alpha\beta V\beta\gamma V\gamma\alpha=
(S.\alpha\beta\gamma)^2$
\item[(c)]$S.V(\alpha+\beta)(\beta+\gamma)V(\beta+\gamma)(\gamma+\alpha)
V(\gamma+\alpha)(\alpha+\beta)=4(S.\alpha\beta\gamma)^2$
\item[(d)]$S.V(V\alpha\beta V\beta\gamma)V(V\beta\gamma V\gamma\alpha)
V(V\gamma\alpha V\alpha\beta)=(S.\alpha\beta\gamma)^4$
\item[(e)]$S.\delta\epsilon\zeta=16(S.\alpha\beta\gamma)^4$\\
where
$$\delta=V(V(\alpha+\beta)(\beta+\gamma)V(\beta+\gamma)(\gamma+\alpha))$$
$$\epsilon=V(V(\beta+\gamma)(\gamma+\alpha)V(\gamma+\alpha)(\alpha+\beta))$$
$$\zeta=V(V(\gamma+\alpha)(\alpha+\beta)V(\alpha+\beta)(\beta+\gamma))$$
\end{itemize}

{\bf 9}. Prove the common formula for the product of two determinants
of the third order in the form
$$
S.\alpha\beta\gamma S.\alpha_1\beta_1\gamma_1=
\left
\begin{array}{ccc}
S\alpha\alpha_1 & S\beta\alpha_1 & S\gamma\alpha_1\\
S\alpha\beta_1 & S\beta\beta_1 & S\gamma\beta_1\\
S\alpha\gamma_1 & S\beta\gamma_1 & S\gamma\gamma_1
\end{array}
\right
$$

{\bf 10}. Show that, whatever be the eight vectors involved,
$$
\left
\begin{array}{cccc}
S\alpha\alpha_1 & S\alpha\beta_1 & S\alpha\gamma_1 & S\alpha\delta_1\\
S\beta\alpha_1 & S\beta\beta_1 & S\beta\gamma_1 & S\beta\delta_1\\
S\gamma\alpha_1 & S\gamma\beta_1 & S\gamma\gamma_1 & S\gamma\delta_1\\
S\delta\alpha_1 & S\delta\beta_1 & S\delta\gamma_1 & S\delta\delta_1
\end{array}
\right
=S.\alpha\beta\gamma S. \beta_1\gamma_1\delta_1S\alpha_1(\delta\delta)=0
$$

If the single term $S\alpha\alpha_1$, be changed to $S\alpha_0\alpha_1$,
the value of the determinant is
$$S.\beta\gamma\delta S.\beta_1\gamma_1\delta_1 S\alpha_1(\alpha_0\alpha)$$

State these as propositions in spherical trigonometry.

Form the corresponding null determinant for any two groups
of five quaternions : and give its geometrical interpretation.

{\bf 11}. If, in \S 102, $\alpha$, $\beta$, $\gamma$ be three mutually
perpendicular vectors, can anything be predicated as to $\alpha_1$,
$\beta_1$, $\gamma_1$? If $\alpha$, $\beta$, $\gamma$ be rectangular
unitvectors, what of $\alpha_1$, $\beta_1$, $\gamma_1$?

{\bf 12}. If $\alpha$, $\beta$, $\gamma$, $\alpha^{'}$, $\beta^{'}$,
$\gamma^{'}$ be two sets of rectangular unitvectors, show that
$$
S\alpha\alpha^{'}=
S\gamma\beta^{'}S\beta\gamma^{'}=
S\beta\beta^{'}S\gamma\gamma^{'}\;\;\;\textrm{\&c. \&c.}
$$

{\bf 13}. The lines bisecting pairs of opposite sides of a quadrilateral
(plane or gauche) are perpendicular to each other when the
diagonals of the quadrilateral are equal.

{\bf 14}. Show that
\begin{itemize}
\item [(a)]$S.q^2=2S^2qT^2q$
\item [(b)]$S.q^3=S^3q3SqT^2Vq$
\item [(c)]$\alpha^2\beta^2\gamma^2+S^2.\alpha\beta\gamma=
V^2.\alpha\beta\gamma$
\item [(d)]$S(V.\alpha\beta\gamma V.\beta\gamma\alpha V.\gamma\alpha\beta)=
4S\alpha\beta S\beta\gamma S\gamma\alpha S.\alpha\beta\gamma$
\item [(e)]$V.q^3=(2S^2qT^2Vq)Vq$
\item [(f)]$qUVq^{1}=Sq.UVq+TVq$
\end{itemize}

and interpret each as a formula in plane or spherical trigonometry.

{\bf 15}. If $q$ be an undetermined quaternion, what loci are represented by
\begin{itemize}
\item[(a)]$(q\alpha^{1})^2=a^2$
\item[(b)]$(q\alpha^{1})^4=a^4$
\item[(c)]$S.(q\alpha)^2=a^2$
\end{itemize}
where $a$ is any given scalar and $\alpha$ any given vector ?

{\bf 16}. If $q$ be any quaternion, show that the equation
$$Q^2=q^2$$
is satisfied, not alone by $Q = \pm q$, but also by
$$Q=\pm \sqrt{1}(Sq.UVqTVq)$$

\begin{flushright}
(Hamilton, {\sl Lectures}, p. 673.)
\end{flushright}

{\bf 17}. Wherein consists the difference between the two equations
$$
T^2\frac{\rho}{\alpha}=1\;\;\;\textrm{and}\;\;\;
\left(\frac{\rho}{\alpha}\right)^2=1
$$

What is the full interpretation of each, $\alpha$ being a given, and p an
undetermined, vector?

{\bf 18}. Find the {\sl full} consequences of each of the following
groups of equations, as regards both the unknown vector $\rho$ and
the given vectors $\alpha$, $\beta$, $\gamma$:
$$
\begin{array}{crcrcr}
 & S.\alpha\beta\rho=0 & & S\alpha\rho=0 & & S\alpha\rho=0\\
(a) & & (b) & S.\alpha\beta\rho=0 & (c) & S.\alpha\beta\rho=0\\
 & S.\beta\gamma\rho=0 & & S\beta\rho=0 & & S.\alpha\beta\gamma\rho=0
\end{array}
$$

{\bf 19}. From \S\S 74, 110, show that, if $\epsilon$
be any unitvector, and $m$ any scalar,
$$\epsilon^{m}=\cos\frac{m\pi}{2}+\epsilon\sin\frac{m\pi}{2}$$
Hence show that if $\alpha$, $\beta$, $\gamma$
be radii drawn to the corners of a triangle on the unitsphere,
whose spherical excess is $m$ right angles,
$$
\frac{\alpha+\beta}{\beta+\gamma}.
\frac{\gamma+\alpha}{\alpha+\beta}.
\frac{\beta+\gamma}{\gamma+\alpha}=
\alpha^m
$$
Also that, if $A$, $B$, $C$ be the angles of the triangle, we have
$$
\gamma^{\frac{2C}{\pi}}
\beta^{\frac{2B}{\pi}}
\alpha^{\frac{2A}{\pi}}
=1
$$

{\bf 20}. Show that for any three vectors $\alpha$, $\beta$, $\gamma$ we have
$$
(U\alpha\beta)^2+(U\beta\gamma)^2+(U\alpha\gamma)^2+(U.\alpha\beta\gamma)^2+
4U\alpha\gamma .SU\alpha\beta SU\beta\gamma=2
$$

\begin{flushright}
(Hamilton, {\sl Elements}, p. 388.)
\end{flushright}

{\bf 21}. If $a_1$, $a_2$, $a_3$, $x$ be any four scalars,
and $\rho_1$, $\rho_2$, $\rho_3$ any three vectors, show that
$$
(S.\rho_1\rho_2\rho_3)^2+
(\sum.a_1V\rho_2\rho_3)^2+
x^2(\sum V\rho_1\rho_2)^2
$$
$$
x^2(\sum.a_1(\rho_2\rho_3))^2
+2\prod(x^2+S\rho_1\rho_2+a_1a_2)
$$
$$
=2\prod(x^2+\rho^2)+
2\prod a^2+
$$
$$
\sum\{(x^2+a_1^2+\rho_1^2)((V\rho_2\rho_3)^2+
2a_2a_3(x^2+S\rho_2\rho_3)x^2(\rho_2\rho_3)^2)\}
$$
where $\displaystyle \prod a^2=a_1^2a_2^2a_3^2$

Verify this formula by a simple process in the particular case
$$a_1=a_2=a_3=x=0$$

\begin{flushright}
({\sl Ibid})
\end{flushright}

{\bf 22}. Eliminate $p$ from the equations
$$V.\beta\rho\alpha\rho=0,\;\;\;S\gamma\rho=0$$
and state the problem and its solution in a geometrical form.

{\bf 23}. If $p$, $q$, $r$, $s$ be four versors, such that
$$qp=sr=\alpha$$
$$rq=ps=\beta$$
where $\alpha$ and $\beta$ are unitvectors; show that
$$S(V.VsVqV.VrVp)=0$$
Interpret this as a property of a spherical quadrilateral.

{\bf 24}. Show that, if $pq$, $rs$, $pr$, and $qs$ be vectors, we have
$$S(V.VpVsV.VqVr)=0$$

{\bf 25}. If $\alpha$, $\beta$, $\gamma$ be unitvectors,
$$
V\beta\gamma S.\alpha\beta\gamma=
\alpha(1S^2\beta\gamma)
\beta(S\alpha\gamma S\beta r + S\alpha\beta)
\gamma(S\alpha\beta S\beta\gamma+S\alpha\gamma)
$$

{\bf 26}. If $i$, $j$, $k$, $i^{'}$, $j^{'}$, $k^{'}$,
be two sets of rectangular unitvectors, show that
$$
\begin{array}{rcl}
S.Vii^{'}Vjj^{'}Vkk^{'}&=&(Sij^{'})^2(Sji^{'})^2\\
 &=&(Sjk^{'})^2(Skj^{'})^2=\textrm{\&c.}
\end{array}
$$
and find the values of the vector of the same product.

{\bf 27}. If $\alpha$, $\beta$, $\gamma$
be a rectangular unitvector system, show that,
whatever be $\lambda$, $\mu$, $\nu$
$$\lambda S^2i\alpha +\mu S^2j\gamma +\nu S^2k\beta$$
$$\lambda S^2k\gamma +\mu S^2i\beta +\nu S^2j\alpha$$
and
$$\lambda S^2j\beta +\mu S^2k\alpha +\nu S^2i\gamma$$
are coplanar vectors. What is the connection between this and
the result of the preceding example ?

\vfill
\newpage
\chapter{Axiom Algebra Code}

\section{package QUATCT2 QuaternionCategoryFunctions2}
<>=
)abbrev package QUATCT2 QuaternionCategoryFunctions2
++ Author: Robert S. Sutor
++ Date Created: 23 May 1990
++ Change History:
++ 23 May 1990
++ Basic Operations: map
++ Related Constructors: QuaternionCategory, Quaternion
++ Also See:
++ AMS Classifications: 11R52
++ Keywords: quaternions, division ring, map
++ Description:
++ \spadtype{QuaternionCategoryFunctions2} implements functions between
++ two quaternion domains. The function \spadfun{map} is used by
++ the system interpreter to coerce between quaternion types.

QuaternionCategoryFunctions2(QR,R,QS,S) : Exports ==
 Implementation where
 R : CommutativeRing
 S : CommutativeRing
 QR : QuaternionCategory R
 QS : QuaternionCategory S
 Exports == with
 map: (R > S, QR) > QS
 ++ map(f,u) maps f onto the component parts of the quaternion
 ++ u.
 Implementation == add
 map(fn : R > S, u : QR): QS ==
 quatern(fn real u, fn imagI u, fn imagJ u, fn imagK u)$QS

@
\chapter{Further Notes}
\section{Axiom Examples}
The basic operation for creating quaternions is {\bf quatern}.
This is a quaternion over the rational numbers.
\spadcommand{q:=quatern(2/11,8,3/4,1)}
$$
{2 \over {11}} {8 \ i}+{{3 \over 4} \ j}+k
$$
\returnType{Type: Quaternion Fraction Integer}

This is a quaternion over the integers.
\spadcommand{r:=quatern(1,2,3,4)}
$$
1+{2 \ i}+{3 \ j}+{4 \ k}
$$
\returnType{Type: Quaternion Integer}

We can also construct quaternions with complex components.
First we construct a complex number.
\spadcommand{b:=complex(3,4)}
$$
3+{4 \ i}
$$
\returnType{Type: Complex Integer}
and then we use it as a component in a quaternion.
\spadcommand{s:=quatern(3,1/7,b,2)}
$$
3+{{1 \over 7} \ i}+{{\left( 3+{4 \ i} \right)}\ j}+{2 \ k}
$$
\returnType{Type: Quaternion Complex Fraction Integer}
Notice that the $i$ component of the complex number has no
relation to the $i$ component of the quaternion even though
they use the same symbol by convention.

The four parts of a quaternion are the real part, the $i$ imaginary
part, the $j$ imaginary part, and the $k$ imaginary part. The
{\bf real} function returns the real part.
\spadcommand{real q}
$$
2 \over {11}
$$
\returnType{Type: Fraction Integer}

The {\bf imagI} function returns the $i$ imaginary part.
\spadcommand{imagI q}
$$
8
$$
\returnType{Type: Fraction Integer}

The {\bf imagJ} function returns the $j$ imaginary part.
\spadcommand{imagJ q}
$$
3 \over 4
$$
\returnType{Type: Fraction Integer}

The {\bf imagK} function returns the $k$ imaginary part.
\spadcommand{imagK q}
$$
1
$$
\returnType{Type: Fraction Integer}

Quaternions satisfy a very fundamental relationship between the parts,
namely that
$$i^2 = j^2 = k^2 = ijk = 1$$. This is similar to the requirement
in complex numbers of the form $a+bi$ that $i^2 = 1$.

The set of quaternions is denoted by $\mathbb{H}$, whereas the integers
are denoted by $\mathbb{Z}$ and the complex numbers by $\mathbb{C}$.

Quaternions are not commutative which means that in general
$$AB \ne BA$$
for any two quaternions, A and B. So, for instance,
\spadcommand{q*r}
$$
{{437} \over {44}} {{{84} \over {11}} \ i}+{{{1553} \over {44}} \ j}
{{{523} \over {22}} \ k}
$$
\returnType{Type: Quaternion Fraction Integer}
\spadcommand{r*q}
$$
{{437} \over {44}} {{{84} \over {11}} \ i} {{{1439} \over {44}} \
j}+{{{599} \over {22}} \ k}
$$
\returnType{Type: Quaternion Fraction Integer}
and these are clearly not equal.

Complex $2\times2$ matrices form an alternate, equivalent
representation of quaternions. These matrices have the form:
$$
\left[
\begin{array}{cc}
u & v \\
\overline{v} & \overline{u}
\end{array}
\right]
$$
=
$$
\left[
\begin{array}{cc}
a+bi & c+di \\
c+di & abi
\end{array}
\right]
$$
where $u$ and $v$ are complex, $\overline{u}$ is complex conjugate
of $u$, $\overline{z}$ is the complex conjugate of $z$, and a,b,c,
and d are real.

Within the quaternion each component operator represents a basis
element in $\mathbb{R}^4$ thus:
$$
1 =
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{array}
\right]
$$

$$
i =
\left[
\begin{array}{cccc}
0 & 1 & 0 & 0\\
1 & 0 & 0 & 1\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0
\end{array}
\right]
$$


\chapter{Licenses}
<>=
A Primer of Quaternions by Arthur S. Hathaway is a public domain book.

portions Copyright (c) 19912002, The Numerical ALgorithms Group Ltd.
All rights reserved.

Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are
met:

  Redistributions of source code must retain the above copyright
 notice, this list of conditions and the following disclaimer.

  Redistributions in binary form must reproduce the above copyright
 notice, this list of conditions and the following disclaimer in
 the documentation and/or other materials provided with the
 distribution.

  Neither the name of The Numerical ALgorithms Group Ltd. nor the
 names of its contributors may be used to endorse or promote products
 derived from this software without specific prior written permission.

THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS
IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED
TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A
PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER
OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL,
EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO,
PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF
LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING
NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS
SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
@
<<*>>=
<>

<>
@
\vfill
\newpage
<>=
Qauternions
@
\vfill
\newpage
\addcontentsline{toc}{chapter}{Biblography}
\begin{thebibliography}{99}
\bibitem{1} Lamport, Leslie,
{\it LaTeX: A Document Preparation System,} \\
Reading, Massachusetts,
AddisonWesley Publishing Company, Inc.,
1986. ISBN 020115790X
\bibitem{2} Knuth, Donald, {\it The \TeX{}book} \\
Reading, Massachusetts,
AddisonWesley Publishing Company, Inc.,
1984. ISBN 0201134489
\bibitem{3} Jenks, Richard D. and Sutor, Robert S.,\\
{\it Axiom, The Scientific Computation System} \\
SpringerVerlag, New York, NY 1992 ISBN 0387978550
\bibitem{4} Hathway, Arthur S., "A Primer Of Quaternions" (1896)
\bibitem{5} Conway, John H. and Smith, Derek, A.,
"On Quaternions and Octonions", A.K Peters, Natick, MA. (2003)
ISBN 1568811349
\bibitem{6} http://mathworld.wolfram.com/Quaternion.html
\bibitem{7} Tait, P.G.,
{\it An Elementary Treatise on Quaternions} \\
C.J. Clay and Sons, Cambridge University Press Warehouse,
Ave Maria Lane 1890
\end{thebibliography}
\end{document}
diff git a/src/axiomwebsite/patches.html b/src/axiomwebsite/patches.html
index 5b5d90f..e3a61a8 100644
 a/src/axiomwebsite/patches.html
+++ b/src/axiomwebsite/patches.html
@@ 1691,5 +1691,7 @@ bookvol10.3 fix = in Any, add regression and help files
Makefile, src/algebra/Makefile parallel make help files
20090706.02.tpd.patch
src/axiomwebsite/patches.html fixup patch order
+20090709.01.tpd.patch
+quat.spad move QUATCT2 to bookvol10.4