450 HITCHCOCK. 



Case 3. Let r\ = 0. We cannot also have p'l = 0; for if so, we 

 have 2/ = 0, (by 2), and r = 0, (by 7, since q\ is not zero by 9); then 

 r"i = 0, (by 14), and z = 0, (by 13), but z and y are not both zero. 



We then must have, by 8, r'l = r = 0. By7,y = 0. By 2, r' = 0. 

 By 1, bi = 0, which is not true. Hence r\ cannot be zero. 



Thus the vector (241) cannot be written in the form 



Vippdp -\- pS8p. 



This completes the examination of (185) when u — 0. The factoriza- 

 tion (238) also 'fails when St/jl^i = 0. If there exists an axis not in 

 the plane .T2 = 0, we may, as before, subtract the term apx2, with 

 p. = a02- If a does not vanish we may factor as in (203). We can- 

 not now conclude that a is not zero, since we have not w = 0. If 

 a = 0, and Sir8i^2 = 0, the vector (185) becomes 



(ci/3i + 02182) {xixo + gx^ + g'x^xz) + u^iXiXz (242) 



where ci and C2 are constants, and c^, at least, is not zero since tt is 

 not parallel to /3i. If Ci is not zero, we may add the term 



when the quadratic vector may be written 



181 {01X1X2 + cigx^ + (ci^i + u)x2Xz — zxiXz] 



+ fii{c2XiX2 + C2gXi^ + C2sriX2.r3 — 2x2X3} — /Sszxa^, (243) 



where z has been written for — . If we then take 



</)p = C2X3F/32|(33 — ClXsF/SsiSi — C2XiFj8ii32, 



ep = F/32^3 + F/33/31 1 ^^ - ^ I + F^,82 j ^X3 + L + "jx2 1 , (244) 



we find (243) identically equal to Vcppdp, aside from the factor S^i^2^3- 

 This method fails if Ci = 0. If so, the vector (242) becomes, aside 

 from a scalar factor, 



/32(axiX2 + gxz^ + 5^1X2X3) + u^ 1X2X3, (245) 



where a is a constant which is not zero. This quadratic vector has /3i 



