4 
126 
We have c = — t when ¢s—WSt Ks, and (a) becomes, 
(.) tSp Kp=pStKp+Y/StKq+rStKr+s¢ StkKs. 
43. If w (p, q), @ (7, 8) be linear functions, we have 6. A. w(p, q) . ¢ (7, 8) 
=cSp’ Kp wherec—=Avy(1,i).A¢(j,k) +Av(j, kt). AG(1, 
i) + two similar terms found by advancing i, j, k. 
If » (p, g) =, (p K q), then 4 wv (1, 1) = AZ k= — y, 1, ete. 
te =v, (Kp q), then Ay (1, 1) =— A» (j, k) = y, 4, ete. 
pte == (pig)> ener BA. api (eL," 2) = OAania(eg) hh) an melee 
> = 2, (Sp. q) then Ap, 4= 4%, A) (1k) = 0, ete. 
We have thus the identities, 
(a.) A.W(p K q) ¢ (Kr s)=—0. 
(b.) A.W (pq)o(rs)=0. 
(c.) A.v(Sp-.g)¢(Sr.s)=—0. 
(d,)', 6. Ay pigiS (tr Kr) Seis Al Ors. 9,2 Spe 
In fact these methods may be employed to multiply formulas indefinitely. 
The above are interesting as giving the general relations between the six vector 
alternates of the same form that may be derived from the quaternions p, q, 7, s. 
44. We note the following geometrical interpretations: (p,, q,, ele. not 
affected by A). 
(a.) 2A.pS.4 Kk q, isa line in the plane (p, q) that is perpendicular to q, ; 
viz., tis A p K q . projection of q, on the plane (p, q). 
That it is a line perpendicular to q, in the plane (p, q) is seen by its form 
and the fact that the operator S. Ag, gives an alternate of a symmetric product 
which is zero by th 2’. 
Note.—We have, for the complete proof: 
A.p.Kq.q=3A(p. Kq.q—p.Kq-q+u- Kp.g 
2A. pS. 9K q=A(p-Kugs¢+ p+ Kya) SHAG eae | 
+9, Kq.p)=$4A(2p.Kq-n4+7-Kqi:¢—%:Kp.q), | 
so that A.p.Kq.q,-+2A.pS.qKq,=A pKq.q- In this result re- | 
solve q, into 4,’ -+-q,’” respectively parallel and perpendicular to the plane 
(p, q) and it becomes (since A. pp. Kq.q, =ApKq. 4”), 
2A Sg BiG ae pti 29 Qe SD: 
45, Operating on the last result by S. K p,, and remembering, since the planes 
(p,q) (Pis 91’) are perpendicular, that S.A p Kq. A, p, K q,” = o0,we find, 
(a) 2A.S(pKp,)S(qKq)=—S.ApKq.A4,p, KY 
= product of areas times cosine of angle between the planes of the parallelo- 
grams (p, 4), (pi, q,). It we drop the subscripts after expansion, we have the 
squared tensor of the area of ( p, q) viz, T? Ap Kq. 
