Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 8)

503] 
THE VERTICES OF CONES WHICH SATISFY SIX CONDITIONS. 
115 
{Surface abedaß.} 
15—2 
(1) The line ab is a 4-tuple line. To show this, observe in the first instance, 
that we may obtain the 8 factors of the norm by giving to the radical 
Vpaa.paß the sign +, and to the other three radicals the signs + , —, 
at pleasure. For a point on the line in question, we have pdab = 0, 
pabc = 0; hence the norm is the product of the four equal factors 
Vpaa .paß.pbcd — \ipba. pbß. pcda, 
and the other four equal factors obtained by writing herein + instead of —. 
Now for a point on the line ab, we may write for x, y, z, w the values 
ux a + vx b , uy a + vy b , uz a + vz b , uw a + vw b , where u, v are arbitrary coefficients. We have 
pa,a — u . aaa + v . baa = v . baa = — v. aba, 
paß — v. baß = — v. abß, 
pba = u. aba + v. bba — u. aba, 
pbß = u. abß, 
pbcd = u. abed + v . bbed = u. abed, 
pcda= u. aeda + v. beda = v . beda = — v. abed, 
where aba = 0 is the condition that the points a, b and the line a may be in the 
same plane (or, what is the same thing, that the lines ab and a may intersect), viz. 
baa is = P a x b + Q a y b + R a z b + S a tu b . And similarly abcd = 0 is the condition that the 
four points a, b, c, d may be in a plane; viz. we have 
(Z a , 
Va, 
z a> 
W a 
x b , 
Vb, 
z b > 
W b 
¿'c > 
Ve, 
z c, 
w c 
æ d> 
Vd, 
z d, 
Wd 
Substituting, we have Vpaa .paß. pbcd and \fpba.pbß .pcda, each equal (save as to 
sign) to uv Vaba. abß . abed; that is, the four equal factors of one set will vanish. The 
vanishing factors are of the form 0 1 , and the norm is 0 4 , that is, the line in question, 
ab, is a 4-tuple line. 
(2) The line a is a 4-tuple line; in fact, for any point of the line we have 
paa = 0, pba = 0, pea = 0, pda = 0; each factor of the norm is therefore evanescent, of 
the form 0^, and the norm itself is thus =0 4 . 
29. (5) The line (ab, cd, a, ß) is a double line. To show this, take z = 0, w=0 
as the equations of the line in question; then we have h a = 0, hß = 0, z a w b — z b w a = 0, 
or say w a = \z a , w b = \z b : and z c w d — z d w c = 0; or say w c = pz c , w d = pz d (A and p 
arbitrary coefficients). Putting for shortness 
I=(g-\a)x-(f+\b)y, J=(g — pd)x-(f + fib)y;
	        
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