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 EC, 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 Paß, Pba from P, the pole of AB. Comparing the triangles Eba and eaß, we see that Ea eß. But, since P is the pole of AB, Fßa is a right angle: and therefore as Fa is a quadrant, so is FB. Thus AB is the complement of Ea or Be, and therefore

Join b 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 baA, coA, we have

and therefore the triangles coB and Baß 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 aß, ba, oc. Hence these arcs divide the triangle Hab into three isosceles triangles.

But, as Faß and ▲ Dae are right angles, we have

angle of Bay L FaD Lẞae Pab

= = = L

[Numerous singular geometrical theorems, easily proved 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., &c.]

115. Let бa a', Ob = B, Oc = y', and we have

=

But FG is the complement of DF. Hence the angle of the quaternion

is half the spherical excess of the triangle whose angular points are at the extremities of the unit-vectors a', B', y.

[In seeking a purely quaternion proof of the preceding propositions, the student may commence by showing that for any three unit-vectors we have

The angle of the first of these quaternions can be easily assigned; and the equation shows how to find that of Ba-1y. But a still simpler method of proof is easily derived from the composition of rotations.]

116. A scalar equation in p, the vector of an undetermined point, is generally the equation of a surface; since we may substitute for p the expression

p = xa,

where a is an unknown scalar, and a any assumed unit-vector. The result is an equation to determine x. Thus one or more points are found on the vector xa whose cöordinates 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 vector equation in p, 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 points in space. But in certain cases these may form a line, and even a 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

which is the vector of a definite point (since we have evidently

Saẞ=0);

the closely allied equation

Tap=B

is easily seen to involve

Saẞ=0,

and to be satisfied by

p=a ̄1B+xα,

whatever be x. Hence the vector of any point whatever in the line drawn parallel to a from the extremity of a 1ẞ satisfies the given equation.

is equivalent to but two scalar equations. For it shews that

Vap and Vẞp are parallel, i. e. p lies in the same plane as a and ẞ, and can therefore be written (§ 24)

p=xa+yẞ,

where x and y are scalars as yet undetermined.

which (§ 40) is the equation of a hyperbola whose asymptotes are in the directions of a and ß.

118. Again, the equation

V.VaẞVap=0,

though apparently equivalent to three scalar equations, is really equivalent to one only. In fact we see by § 91 that it may be written

- aS.aẞp=0,

whence, if a be not zero, we have

S.aßp=0,

and thus (§ 101) the only condition is that p is coplanar with a, B. Hence the equation represents the plane in which a and ẞ lie.

119. Some very curious results are obtained when we extend these processes of interpretation to functions of a quaternion

q=w+p

instead of functions of a mere vector p.

A scalar equation containing such a quaternion, along with quaternion constants, gives, as in last section, the equation of

Hence for succes

a surface, if we assign a definite value to w. sive values of w, we have successive surfaces belonging to a system; and thus when @ is indeterminate the equation represents not a surface, as before, but a volume, in the sense that the vector of any point within that volume satisfies the equation. Thus the equation

or

or

(Tq)2=a2,

w2-p2=a2,

(Tp)2= a2-w2,

represents, for any assigned value of w, not greater than a, a sphere whose radius is √a1-w2. Hence the equation is satisfied by the vector of any point whatever in the volume of a sphere of radius a, whose centre is origin.

Again, by the same kind of investigation,

(T(q—B))2=a2,

where q=w+p is easily seen to represent the volume of a sphere of radius a described about the extremity of ẞ as centre.

Also S(q)=-a 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.

120. The following transformation is enunciated by Hamilton (Lectures, p. 587, and Elements, p. 299).