Let u and v be unit vectors, and w be a vector such that | w | = 1 2 . If u = 3 w + 2( w v ) , then the value of ^2 , where is the angle between w and v , is equal to:

Step-by-step solution

Given the vector relation: u = 3 w + 2( w v ) Taking the dot product of the vector with itself (squaring both sides): | u |^2 = | 3 w + 2( w v )|^2 Expanding the right side, the cross term is 4 3 w ( w v ) . Since the scalar triple product with repeated vectors is zero, this term vanishes: | u |^2 = 3| w |^2 + 4| w v |^2 We are given that u and v are unit vectors, so | u | = 1 and | v | = 1 . Also, | w | = 1 2 . Substituting these values: 1 = 3 ( 1 2 )^2 + 4| w |^2| v |^2 ^2 1 = 3 ( 1 4 ) + 4 ( 1 4 )(1) ^2 1 = 3 4