450 ■ HITCHCOCK. 



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

 have y = 0, (by 2), and r = 0, (by 7, since q'l is not zero by 9); then 

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



We then must have, by 8, r'l = r = 0. By 7, y = 0. By 2, r' = 0. 

 By 1, ^1 = 0, which is not true. Hence r'l 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/x/Si = 0. If there exists an axis not in 

 the plane Xi = 0, we may, as before, subtract the term apxo, with 

 H = 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 u = 0. If 

 a = 0, and St3i^2 = 0, the vector (185) becomes 



(ci|3i + C2J82) (xixo + gx-f + (j'xoxs) + 11^1X2X3 (242) 



where Ci and C2 are constants, and C2, at least, is not zero since x is 

 not parallel to j3i. If Ci is not zero, we may add the term 



when the quadratic vector may be written 



^l{ciXiX2 + CiQXi^ + (Ci^fi + U)X2X3 .— ZX1X3} 



+ ^'>{C2XIX2 + C2gX3^ + C2giX2Xz — ZX2X3} — I33ZX3-, (243) 



where z has been written for . If we then take 



<I>P = C2X3V^2^3 — C1.T3F/33JS1 — C2.TlF/3i/32, 



dp = F^2/33 + F/333i I ^' - ^ ( + ^^^^- ) 3'-"^ + U + '^y^ \ ' (244) 



we find (243) identically equal to V(t)pdp, 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, 



^2(.axiX2 + gx3^ + giX2X3) + 11^1X2X3, (245) 



where a is a constant which is not zero. This quadratic vector has j8i 



