Let a and b be two vectors such that a = 2 i - j + k , a b = 1 , and a b = - i + j + 3 k . If P is the square of the projection of ( b - a ) on ( b + a ) , then the value of 10P is equal to

Step-by-step solution

Given a = 2 i - j + k , we have | a |^2 = 2^2 + (-1)^2 + 1^2 = 6 . Also, a b = - i + j + 3 k , so | a b |^2 = (-1)^2 + 1^2 + 3^2 = 11 . Using Lagrange's identity: | a |^2 | b |^2 = ( a b )^2 + | a b |^2 Substitute the known values ( a b = 1 ): 6 | b |^2 = (1)^2 + 11 = 12 | b |^2 = 2 The projection of ( b - a ) on ( b + a ) is given by: ( b - a ) ( b + a ) | b + a | = | b |^2 - | a |^2 | b + a | We have | b |^2 - | a |^2 = 2 - 6 = -4 . Now, find | b + a |^2 : | b + a |^2 = | b |^2 + | a |^2 + 2( a b ) = 2 + 6 + 2(1)