422 



HITCHCOCK. 



plane to be .ti = 0. By (96) this gives, (on putting p = ^2, xs = 

 xi = 0), 



An = 0. 

 If we next subtract from the vector the term 



A2lXz(filXi + /32.T2 + ISsXs) 



which is of the form pS8p and does not alter the axes, we shall remove 

 the remaining term in xoxo. If the vector coefficient of Cs^ be called f , 

 the vector may be written 



^l[AuXiX2 + (^12 — ^21)2:3^1] + r^3^ 



The polar vector of VpFp now becomes 



(168) 



Vp{^i[Ai3XiX2 + {A12 - ^21)3:3X1] + txi"^} + Vp^iAi3{xi'x2 + a:ia;2') 

 + Vp^,{Ai2 - A21) {xs'xi + xzxi') + 2Vptx3%. (169) 



If the rule for a triple axis be applied to /So, we shall now put 62 for p' 

 and /Ss for p, that is Xi = X3' = and xi = a;2 = 0. This gives f 

 parallel to the r of the rule of Art. 18. Again, |3i is parallel to the t of 

 the same rule. Hence if ^2 is a triple axis, we must have S/So/Sif = 0. 

 We may therefore take 



r = «ii3i + a2|82 

 and the vector becomes 



fii[AuXiX2 + {A12 — ^21)3:3X1 + aiX3^] + ^2a2Xz^, 



(170) 



which has only two distinct axes, /3i and ^2, quadruple and triple. 

 If ai is zero, ^2 is an inflectional element of the cones (3). We cannot 

 have a2 = or ^13 = if the vector is to be irreducible. 



It appears therefore that all possible quadratic vectors which are 

 irreducible, and have /3i for a quadruple axis of the sort where all cones 

 (3) have double elements at /3i, are included under (165) and (170). 

 It is so far assumed that /Si is not of higher order than four. 



28. For Case 2° we may follow a similar method. Take /3i an 

 axis, supposed at first to be at least triple. Take 1S3 in the tangent 

 plane to (3) at jSi. If ^i is rendered a zero the vector may be written, 

 J82 being at present any vector such that (123) does not vanish. 



