opposite, or the conjugate of the conjugate, of a quaternion is the quaternion itself:

K(n) = −n, K(noo) = (— n)c" = n➖c";

or, the conjugate of the versor of any quaternion is equal to versor of the conjugate (N). Hence,

whence, by multiplication and division,

qKq = T2q ; q : Kq = U2q .

46° (4) U q

=

(2)

(3)

(d). The conjugates of opposite quaternions are themselves opposite ; or,

K(q)=Kq;

an equation which is a particular case of a more general formula,

where x is any scalar.

This may be proved by supposing that the vectors OB, OB', fig. 21, are multiplied by any common scalar; or, that both are cut by any parallel to the line BB'.

(e). The conjugates of reciprocals are reciprocal; or,

For, suppose the two triangles AOB, AOB', fig. 21, to revolve inwards round O in the plane B'OB until the points B, B', coincide in D, a point in the line OA produced. Then FOD and EOD represent respectively the two triangles after the revolution. From B and B' draw lines parallel to ED

and FD, cutting OD produced in C; circumscribe a circle to the triangle ABC; and with O as centre and OB as radius describe the circle BDB'.

we can infer first, that a, ẞ, y are coinitial and coplanar; secondly, that Ty= Tß; and thirdly, that a bisects the angle between ẞ and y, or, that a (produced either way if necessary) bisects the join of the terms of ẞ and y at right angles; or, again, since the angles of incidence and reflexion of a ray of light are equal, that the ray y is the reflexion of the ray В (O being supposed to be a point on a plane mirror whose surface is perpendicular to a).

(g). Since OBA = OB'A = LODE = Z OCB, fig. 21, it follows-first, that AB and BC are antiparallel, or that the triangles AOB and COB are inversely similar (the triangles DOE and COB are directly similar); and, secondly, that OB is a tangent to the circle ABC at the point B. Hence, the circles BDB and ABC are orthogonal, because a tangent

to BDB' at B would be perpendicular to OB, which is a tangent to ABC.

(h). Again,

OA OB:: OB: OC.

Therefore, OB or OD is a mean proportional between OC and OA, and C and A are inverse points with respect to the circle BDB'. If, therefore,

an equation which expresses that AB and BC are antiparallel, or that the triangles AOB and COB are inversely similar, but expresses nothing more. Now, in order that this relation should hold good, it is only necessary either that (1) T. OB should be a geometric mean between T. OA and T. OC; or, that (2) T. OB = T. OD. If, then, O, A, D, C be fixed points, while B is a variable point and OB = OP = = P, it is evident that the locus of P is the surface of a sphere with centre O and radius T. OD = vTa. Equation (8) then

(i). If, then, we meet with an equation of this form, we can infer that the locus of P is the surface of a sphere with centre O and radius vTa. Further, if we take a point C such that OC = v2a, the sphere will be a common orthogonal to all the circles APC that pass through the fixed points A and C ; because every radius of the sphere is a tangent, at the variable point P, to the circle APC, AP and PC being antiparallel. (j). Since

and the first equation of (9) becomes

a

P

= Ko,

α

T2T UTK UK=TKU, (b) (1);

α

'

or,

α

α

AB = p — a, and T(BC)=T(— BC)=T(CB)=T(OB — OC). (1). Article (h) contains the solution of the problem of Apollonius of Perga: given any two points, C and A, in a plane, and a ratio of inequality, 1: v; to construct a circle BDB' in the plane such that the lengths of the two straight lines AB and CB, or AP and CP, which are inflected from the two given points to any common point, B or P, of its circumference, shall be to each other in the given ratio.

=

Cut AC externally at O in the duplicate of the given ratio of sides, so as to have OC: v2OA. Take OD, a geometric mean, to OA, OC; and, with O as centre and OD as radius, describe a circle. This is the locus of all points for which CPvAP.

Paragraphs (e) to (1) are chiefly from Sir W. R. Hamil

ton's Elements.'

CHAPTER VI

THE POWERS OF QUATERNIONS

50°. Let q = p, q= 0, and UVqe. Then, by 33° (2), we have Up == ', and

(Tp. U'p)" = (Tq. e1)": q"T"q (cos no + e sin no)

From this equation we have at once,

спв

Tq"T"q; Uq" = €¤no = (eco)n = U"q; ≤q" = n ≤ q;

=

Sq" Tq. cos no; Vq"T"q. sin no. e.

As this is the only power of a quaternion with which we will have to do in the following pages, it is desirable to inquire particularly into its nature.

The first question that arises is, has q3 two square roots

like an ordinary algebraic quantity?

Let OA, OB, fig. 22 (1) and (2), be the unit-vectors of any two vectors a and ß, inclined to each other at an acute angle in (1), and at an obtuse angle x in (2). and = x in (2); and Then, if

Χ

two square roots, either

T'q or U2q must have two square roots. But as Tq is always