384 HITCHCOCK. 



(P4P5PP) = (123)2 I (12p) (34p) (235) (415) - (125) (345) (23p) 



(41p) } (32) 



The expression in braces may be regarded as a function of the four 

 vectors /3i, ^o, /S3, /S^, taken in cyclic order, and of the two vectors 

 p and jSs. It may be abbreviated C'i234(5, p). It is homogeneous and 

 quadratic in each of the six vectors, and vanishes if any two coincide. 

 Therefore, if equated to zero, it denotes a quadric cone through the 

 five vectors |8. If we write ^7 instead of p, the result cannot vanish, 

 since, by hypothesis, no six axes lie on a quadric cone. That is, 

 the third denominator in (30) does not vanish. Similarly, neither 

 (PsPePr) nor (P6P4P7) can vanish. Theorem I is therefore proved. 



6. We may regard (31) as a normal or model form for a vector of 

 type I. A variety of results follow immediately, either by inspection, 

 or by using identities like (32) . 



For example, if a quadratic vector F(p) has three axes which are 

 coplanar, it may, by the addition of a properly chosen term p{px -\-qy 

 + rz) be reduced to a binomial. For let the coplanar axes be num- 

 bered 4, 5, 7, and proceed as in (31). The last term vanishes, and Pop 

 is a sum of scalar multiples of two axes. The resulting binomial 

 vector has /3i, /So, jSs, and /Se for zeros. 



Conversely, if F{p), being of type I, has four zeros, it has the other 

 three axes coplanar. For choose three of the zeros to be the /3i, ^o, ^3 

 of the foregoing discussion. Determine Po(p) as in (31). /Sy cannot 

 be a zero, for, as has been shown, h cannot be zero. If either jS^, /Ss, 

 or /Se is a zero, we have ^4, J^b, or ke, respectively, zero, entailing the 

 vanishing of either (457), (567), or (647). That is, a set of three axes 

 lie in a plane, not including one of the four zeros. 



Again, let there be three coplanar axes, and number them 1, 2, and 

 7. By expanding as in (32) we see that each of the denominators in 

 the model form (31) becomes a product of three-row determinants. 



If a A'ector of type I has two sets of coplanar axes, the two sets must 

 have one, and only one, axis in common, since no six lie on any quadric 

 cone. Suppose (127) = (457) = 0. (31) becomes, aside from a scalar 

 factor, 



. (567)»Ci23o(6,p) ^ (647) ■ (7x236 (4, p) ,3 . 



(126) (356) (517) (124) (364) (617) ^' ^ 



where C'i235(6, p) vanishes on the cone through /3i, ^2, ^3, 185 and fie, 

 and is obtained as in (32). There is nothing to prevent us from inter- 



