VECTOR ANALYSIS. 27 



is also an identical equation. (The reader will observe that in each 

 of these equations the second member may be expressed as a deter- 

 minant.) 



From these transformations, with those already given, it follows 

 that a product formed of any number of letters (representing vectors 

 and scalars), combined in any possible way by scalar, direct, and skew 

 multiplications, may be reduced to a sum of products, containing 

 each the sign x once and only once, when the original product 

 contains it an odd number of times, or entirely free from the sign, 

 when the original product contains it an even number of timea 



39. Scalar equations of the first degree with respect to an unknown 

 vector. It is easily shown that any scalar equation of the first 

 degree with respect to an unknown vector p, in which all the other 

 quantities are known, may be reduced to the form 



p.a = a, 



in which a and a are known. (See No. 35.) Three such equations 

 will afford the value of p (by equation (8) of No. 37, or equation (3) 

 of No. 38), which may be used to eliminate p from any other equation 

 either scalar or vector. 



When we have four scalar equations of the first degree with 

 respect to p, the elimination may be performed most symmetrically 

 by substituting the values of p. a, etc., in the equation 



which is obtained from equation (8) of No. 37 by multiplying 

 directly by S. It may also be obtained from equation (5) of No. 37 

 by writing S for p, and then multiplying directly by p. 



40. Solution of a vector equation of the first degree with respect to 

 the unknown vector. It is now 1 easy to solve an equation of the 



form 



p), (1) 



where a, /3, y, 8, X, yu, and v represent known vectors. Multiplying 

 directly by /3xy, by yxa, and by ax/5, we obtain 



or 



where a', {, y are the reciprocals of a, ft, y. Substituting these 

 values in the identical equation 



in which X', /z', y are the reciprocals of X, /z, v (see No. 38), we have 



p = XV-<5)4y(/3'.<$)+''(y'.<$), (2) 



which is the solution required. 



