THE AXES OF A QUADRATIC VECTOR. 343 



5. If VpFp be divided by the product SyipSy2pS'YzpSy4p the quo- 

 tient is an irrotational vector; it is normal to the family of cones (<S7ip)^* 

 {Sy2py^(Sy3py^{Syipy^ = const. 



If, on the other hand, Fp is to have a central axis, and four coplanar 

 sets, we may begin with case 2° of Art. 5 and let An be zero. The 

 only new property introduced, so far as I am aware, is that the 



system -^ = — = — can now be fully solved by quadratures. The 



proof may be carried out by the aid of the theorems already 

 established. 



8. Other Special Types. 



There are but two remaining types depending on coplanarity of axes. 

 These are first, a vector sharing the properties of the two vectors of 

 the last article, possessing, therefore, five coplanar sets, and having 

 two central axes; second, a vector having six coplanar sets and four 

 central axes. They are easily obtained from the types which precede. 

 The vector having four central axes bears an interesting resemblance 

 to the self -con jugate linear vector function. The investigation may 

 be left to the reader. 



9. The Converse Problem. 



In what precedes it has been shown vectors of assigned type may 

 be written down by giving proper values to the ^'s. I now propose 

 to examine the converse problem: given any values of the nine A's, 

 to test ^vhether the vector 2)ossesses three coplanar axes. 



Evidently, if the vector should happen to fall under one of the 

 forms which have been explicitly written out, we would at once recog- 

 nize its type; but these vectors have been merely "normal forms'^ 

 of their respective types, hence not sufficient to serve as tests for com- 

 parison. 



The necessary and sufficient condition for coplanarity of axes of 

 the vector (1) may be obtained as follows. A set of three coplanar 

 axes exists if, and only if, a value of 5 can be found which makes the 

 right side of (1) a binomial, (proved C. Q. V. p. 384). Let y be nor- 

 mal to the plane of this binomial. Then 



SyFp = (40) 



identically. Let 7 be expanded as 



7 = PiV^2^z + 2>2F)S3^i + 3>3F/3u82 (41) 



