434 HITCHCOCK. 



This process assumes that Va2az, Vazai, and Vaia2 are diplanar, 

 since <^p and Op were assumed in terms of them. This, in turn, implies 

 that ai, ao, and az are diplanar, in general true, but not necessary, 

 even in what I have called a vector of the first, or general type. For 

 it was shown that if Fp has three zeros, and if, of the four remaining 

 axes, three are coplanar, the last or seventh axis will also be a zero, 

 (cf. Art. 6), and the vector itself will be always in one plane, i. e. ai, 

 a.2, as will be coplanar. 



The possibility of the normal form (197) therefore depends, when 

 sets of coplanar axes exist, upon so selecting the zeros that, of the 

 four other axes, no three shall be coplanar, while the three zeros them- 

 selves cannot be coplanar. Each relation of coplanarity between sets 

 of three vectors diminishes the number of possibilities. Thus if no 

 three axes are coplanar, three can be chosen in thirty-five ways, each 

 leading to a separate form like the right member of (201). If three 

 axes are coplanar, the number of possibilities reduces to thirty-one, 

 since we must now exclude any choice of three zeros from the four 

 non-coplanar axes, and three can be chosen from four in four ways. 



That the choice of three diplanar axes, so that of the four that 

 remain no three shall be coplanar, will always be possible may be 

 proved by exhaustion as follows. Let |3i, ^2, and jSz be chosen as any 

 three diplanar axes. Let ^i, jSs, and jSe be any remaining three which 

 are not coplanar. If jSt is coplanar with two of this latter three, 

 suppose (567) = 0. If of the remaining four vectors jSi, 182, (3 3, 64, no 

 three are coplanar, the problem is solved, since (456) is not zero. 



Of these four, (123) does not vanish by hypothesis. If /Sy is coplanar 

 with two out of the three ,81, ^2, jSs, suppose (237) = 0. Then (127) 

 is not zero, since no four axes are coplanar. 



Choose /3i, 182, and /Sy to be zeros. If of the four others jSs, ^i, jSs, 

 jSe, no three are coplanar, the problem is solved. Of the four, (456) 

 is not zero by hypothesis, and (356) is not zero for (567) = and no 

 four axes are coplanar. If jSs and (84 are coplanar with either /Ss or /Se 

 (so far treated alike), suppose (345) = 0. 



Choose /Ss, ^5, and |3g to be zeros. If of the four others /3i, 182, ^i, ^t, 

 no three are coplanar, the problem is solved. Of these four, (127) 

 and (247) can neither be zero, since (237) = 0. Now (124) and (147) 

 cannot both be zero. Two cases are to be distinguished, — 



Case 1. (124) = 0. Choose /3], ^i, and ^^ to be zeros. Of the 

 four others, jSo, ^z, jSs, /Se, (235) and (236) are not zero since (237) = 0. 

 And (256) and (356) are not zero, since (567) = 0. The problem is 

 therefore solved. We may still have (163) = 0, when the seven axes 



