\ 



418 HITCHCOCK. 



{Qa,QM = (123)2 [(21p) (25p) (235) (154) - (23p) (15p) (125) (245)], 



(157) 



whence, writing for /3i its value and taking the limit, 



Lim '^Q-'^'QP^ = (23p) (45p) (245)^ + 6(25p)2(235) (145) 



(Q45Q5Q7) (237) (457) (245)2 + 6(257)^(235) (145)' ^ '^ 



This result is, in form, like (156), but the constants a and b have 

 different meaning; being, respectively, parameters whose vanishing 

 causes the cones (3) to have inflectional elements at ^2 and 185. 

 Collecting results, we have as the limiting form of (155), 



{/34(527) + 185(247)} l(235)2(23p) 45p) + a(236) (425) (52p)2} 



+ 



(235)2(237) (457) + a(236) (425) (527)^ 

 i82l(23p) (45p) (245)2 + 6(25p)2(235) (145) {(457) 



-/35 



(237) (457) (245)2 + 6(257)^(235) (145) 

 (527) (6(523) (235)2(154) + a(236) (425)2(524)} 

 {(523) (237) (23p) (57p) + a(236) (527) (52p) (72p)} 



{(235)2(237) (457) + a(236) (425) (527)2}2 



(159) 



which is the most general form of quadratic vector with two triple 

 axes. It is always possible; for the only limitation on the seven vec- 

 tors is that the determinant (452) shall not vanish, it being always 

 assumed that the vector is irreducible. Now we may choose either 

 triple axis as jSb, with the tangent plane to (3) given by (45p) = 0. If 

 this should pass through 182 we have merely to choose 182 instead of 185 ; 

 for the tangent at /Sa cannot also be tangent at (82, the vector being 

 irreducible. 



If the three axes are coplanar, the determinant (257) vanishes, 

 and the vector as above written becomes a binomial. 



As a simple special case, let i = 182, j = /Ss, and i -\- j = (87. 



Take A; = 183 = 8i, making the tangent planes to (3) at i and j respec- 

 tively y = and x = 0. We cannot have both a and 6 zero, as this 

 would give inflectional elements at both i and j and the vector would 

 become reducible. Take a = 0, giving the cones (3) three common 

 points in the plane y = at i. Then /Se disappears from the formula. 

 Leaving 6 arbitrary, with 0i = i, and noting that all the three-row 

 determinants are either +1 or —1, we easily find 



