125 
of the parallelopiped in the relative direction of 1 tot, j,k. WehaveA.p.Kq.1 
=—8Apq. AlsoA. Kp.q. Kr is the quaternion volume of the parallelopiped 
(Kp, Kq, Kr.) 
38. The alternates of fourth order are: 
(a) A.Vp.Vq.Vr.Vs.=O, since it is its own conjugate and is 
in form the vector A. Vp.S.Vq.Vr. Vs. 
(b). 44A.Sp.VqVrVs=4A.Sp.8S.Vq.Vr.Vs=—  D, a scalar. 
D is the coefficient of © (1, 7%, j, &) in any linear alternate 
©(p,4q,7, 8). All our alternates of fourth order are scalars, 
zero when they can be shown formally as vectors. 
39. We have further: 
(a). 04. pors=A.Sp.Viqrs—A.~Vps8.qrs, ce. 
(i) -S:'p. A; Kqir Ks—24A': Sp: 8. gAvrs 
—=—4A.Vp. VigArs=4A.Sp.VpVq. V3. etc., 
=A Sup. heir Re Alp? Megara k 8; de. 
Note.—The first equation of (b) follows from 36 e, thus: A. Kg.r.Ks= 
S.qgArs—(Sq. Ars—Sr.A.qs+Ss.A.qr), and operating by S. p we find (b). 
Note.—From (b), S.s 4. Kp.qKr=—S.sKAp. Ky.7 isanalternate of 
fourth order; ittherefore vanishes when s —p. qg, or r, — in other words Ap. Kq.r 
is perpendicular to the lines p,g, 7. To find the order in space make p, q, r= 
Sein whence A. p. Kq.r=1t. Ki. k=1. 
ep) 8. Kp Ae. Kr ve——S8:. p AWK gir Kea=—D— A. Kpi¢ 
Kyr.s, ete. 
40. Resolve p into p’ —p” respectively in and perpendicular to the space 
4, 7, 8, thus: 
A.Kp.q.Kr.s=—Kp”’.AqKrs, 
whose tensor is altitude & base of the four parallelogram on p, q, 7, sas sides. This 
is the scalar content of the four-parallelogram, positive when in the order 1, 1, 7, & 
since, substituting these values in the alternates, the result is A1.i. A j.k—1. 
41. We have identically A. Kp.q.Kr.s.t=0. 
lei = A.qg. Kr. 2, of =— A. p.Kr.s, 
AC ip . Ke gi's,'s. = Ae ps Ki qr 
Then A. Kp.qg.Kr.s=Sp Kp—Sf Ky, etc., 
and our expanded identity is, 
(a.) tSp’ Kp=pS.tKp’+q8.tKq¢ +rS.tKr+s.St Ks. 
42. Ifos bea linear function, we have 
(a.) 4A.p.Kq.r.¢s=c.Sp’ Kp, wherec—4A.1.Ki.j.ok 
=—[e1il+ieitjoj+kok]. 
