448 HITCHCOCK. 



On the one hand we have the vector product of these two expressions. 

 On the other, we have to add to (241) a term which we may write 



— (i3i.Ti + ^2X2 + iSs-Ts) (xxi + yX2 + ZXz), 



where x, y, and z are undetermined, and are the components of the 

 vector 5. Muhiplying out, and equating coefficients of Hke terms in 

 the variables xi, X2, xz, we have a system of eighteen equations. It is 

 clear at the start that we may take a; = 0, for this is the same as 

 saying that /3i must be a zero; which follows from the facts: that /3i 

 is a sextuple axis; and Vippdp has 3 zeros. Also, we may take pi = 

 qi = ri = 0. For this is the same as making the coefficient of Xi^ 

 zero, a necessary condition that |3i be a zero. We then have the 

 system of equations, fifteen in number, 



Coef . of X2^. Coef . of xa^. Coef . of xix^ 



1. q'lr'- q'r\= bi 4. p'\q"- p"q"i= 9 - z 7. q\r - qr\= - y 



2. p'r'i- p\r'= -y 5. p"r'\- p'\r"= 8. pr\- p\r = 



3. p\(^- p'q\= 63 6. q"xr"- q"r'\= 9. p\q - pq\ = a 



Coef. of 0:2X3 Coef. oixsXi 



10. q\r" - q'r", + q'\r' - q"r\ = 13. q'\r - q r'\ = - z 



11. p'r'\ - p\r" + p"r\ - p'\r' = -z 14. pr'\ - p'\r = 



12. p\q"- p'q'\+ p'\q'- p"q\= g,- y 15. p'\q - p q'\ = 0. 



Either 4)p or dp may be divided by a scalar which is also multiplied 

 into the other, leaving V(j)pdp unchanged. We may therefore without 

 loss of generality assume that some one letter, as r'l, is either zero or 

 unity. 



Case 1. Let r'l =1. I shall show first that p must be zero. 

 For, if not, we have, (by 8), p = p'lr. Substituting in 14 we have, 

 (since r cannot be zero if p is not zero), p"i = p'ir"i. Then writing 

 for p and for p"i their values in 15 we have, since p\ cannot vanish 

 if p does not, 



T'\q - q'\r = 0. (A) 



Comparing with 13, this gives z = 0. Hence r"i is not zero, for if so, 

 by 13 and 14, (since r is not zero), p"i = and q"i = 0, making dp 

 a monomial and V(f)pdp reducible. We may then write r"i = cr, 

 where c is not zero. By (A), q"i = cq and by 14 p"i = cp. From 

 5 and 6 we have 



p"r - pr" = 0, q"r - qr" = 0, 



