QUADRATIC VECTORS. 397 



yields not more than one distinct scalar equation; for this is the same 

 as saying that the cubic cones (3) are either tangent at 1S4 or have a 

 double line there. The resulting vector equation is 



F/34|85-A:5rio36(4, 4P)+ Vl340,'hTnzA(5,ip) + Fp^4-/^4Cm5(6,4) = 0. 



(81) 



If this is equivalent to one scalar equation, and if we put for p, in suc- 

 cession, any two distinct vectors, the coefficients of V^i^^ and of VjSid^ 

 in the two results must be in proportion. Take as the two vectors 

 /Ss and jSe. The proportionality of the coefficients is given by the van- 

 ishing of the determinant whose elements are these coefficients, viz. 



^'5 7'l236(4, 45) — kiCu35(Q, 4), k6Ti234{5, 45) 



^5 7'l236(4, 46) , A'6 7'l234(5, 46) — ^*4C'i235(6, 4) 



(82) 



whose vanishing determines that ^i shall be a double axis. As a 

 verification, we may note that if we write ki = —SKJSi, etc., this 

 result differs from (79) only in the numbering of the axes. 



13. The two methods above given for obtaining a multiple axis 

 depended upon applying the theory of polars to the fimction C of six 

 vectors. The determinants (79) and (82) are, in fact, symmetrical 

 fimctions of the five single axes, and give a general relation between k, 

 the double axis, and the other five. They may be transformed in 

 many ways by identities similar to those already used. It is desirable, 

 however, to have a normal form for a quadratic vector with a multiple 

 axis in which the tangent plane to the cones (3) at the double axis 

 shall appear explicitly. This may easily be foimd as follows, — 

 Start with the general normal form (31). Write ^i = ma + WjSs. 

 By this substitution {PiP^Pp) becomes, with the aid of (32), 



m.,{PaP,Pp) + 7;m(123)2 { (12p) (35p) (235) (al5) 



- (125) (3a5) (23p) (51p)} 



The expression in braces is quadratic in /3i, 182, ^3 and p. It expresses, 

 by its vanishing, a quadric cone through /3i, ^2, /Ss, and /Ss, with the 

 tangent plane at /Ss given by (5ap) = 0; to prove this, take the polar 

 with respect to ^5, 



(125) (35p) (235) (al5) - (125) (3a5) (235) (51p) 



