Let a = i - 2 j + 2 k and b = 2 i + j - k . Let v be a vector such that v a = b a and the angle between v and a is acute. If the length of the projection of v on b is 10 6 , then | v |^2 is equal to

Step-by-step solution

Given v a = b a , we can write ( v - b ) a = 0 . This implies that v - b is collinear with a , so v = b + a for some scalar . The length of the projection of v on b is | v b | | b | . We have v b = ( b + a ) b = | b |^2 + ( a b ) . Here, | b |^2 = 2^2 + 1^2 + (-1)^2 = 6 and a b = (1)(2) + (-2)(1) + (2)(-1) = -2 . Thus, v b = 6 - 2 . Given the projection length is 10 6 , we have |6 - 2 | 6 = 10 6 |6 - 2 | = 10 . This gives 6 - 2 = 10 = -2 or 6 - 2 = -10 = 8 . We are given that the angle between v and a is acute, whi