656 MOORE. 



2-vedors, one of which is arbitrary. For, let A be any plane vector, 

 then M—}^A will be a simple plane if 



{M-\AMM-\A) = M^M-2\MxA = 



MxM 



or 



X = 



2i¥xyl 



and we can write 



.j^^t^A-^iM-'^A) 

 2J/x^ ^ ^ 2MxA ' 



We shall now show that the complex 2-vector can always be re- 

 solved in at least one way into the sum of two perpendicular planes. 

 Let 



(8) M = miMi + '"2^/2 



where Mi and il/2 are completely perpendicular unit planes and iiii 

 and Mi numbers. Then indicating ( J/x J/) • M by A we can wTite 



(9) A = {Mx3I)-M = 2mim2(Mi>^M2)-M = 2wi"?2(w2J/i+»hJ/2). 



Solving (8) and (9) for Mi and i/2 we have 



j^^- 2m^M-A 

 2vi2{m2^ - nil-) 

 ^^ ,, 2miKM-A 



^^2 = ^r~r-^ ^ 



2?ni(??h" — '»2 j 

 The values of nii and VI2 can be computed from the relations 



M-M = mi'-hmi' 

 A-M = 4mM2' 

 From which we get 



m = h Vm-M + VM-A + ^ Vm-M - VM-A 

 2 Vm-M -i-VJT^ - i Vm-M -V^T^ 



VH 



_ 1 



If TTJi 9^ ±W2the solution (10) is unique.^ If mi = ±7??2 the solutions 

 become infinite. From (9) we see at once that this relation means 



8 For the non-euclidean geometry considered by Wilson and Lewis this 

 resolution was unique since in that case the denominators contained the 

 factor nir + m-,- instead of m,- — m-r. 



