Let a unit vector u ^ = x i ^ + y j ^ + z k ^ make angles π 2 , π 3 and 2 π 3 with the vectors 1 2 i ^ + 1 2 k ^ , 1 2 j ^ + 1 2 k ^ and 1 2 i ^ + 1 2 j ^ respectively. If v → = 1 2 i ^ + 1 2 j ^ + 1 2 k ^ , then (| u - v |^2 ) is equal to

Step-by-step solution

Unit vector u ^ = x i ^ + y j ^ + z k ^ p → 1 = 1 2 i ^ + 1 2 k ^ , p → 2 = 1 2 j ^ + 1 2 k ^ p → 3 = 1 2 i ^ + 1 2 j ^ Now angle between u ^ and p → 1 = π 2 u ^ · p → 1 = 0 ⇒ x 2 + z 2 = 0 ⇒ x + z = 0 … i Angle between u ^ and p → 2 = π 3 u ^ · p → 2 = | u ^ | · p → 2 cos π 3 u ^ · p → 2 = y 2 + z 2 = 1 2 … ii Angle between u ^ and p → 3 = 2 π 3 u ^ · p → 3 = | u ^ | · p → 3 cos 2 π 3 ⇒ x 2 + y 2 = - 1 2 ⇒ x + y = - 1 2 … iii from equation (i), (ii) and (iii) we get x = - 1 2 , y = 0 , z = 1 2 Thus u ^ - v → = - 1