The given vectors are a → and b → such that | a → |=3 and | b → |= 2 3 .
As given a → × b → is a unit vector. Then the magnitude of ( a → × b → )=| a → × b → |=1.
The formula for cross product of a → and b → is,
a → × b → =| a → || b → |sinθ n ^ | a → × b → |=| | a → || b → |sinθ n ^ | | a → × b → |=| a → || b → |sinθ| n ^ |
Since, | n ^ |=1as n ^ is a unit vector, therefore,
| a → × b → |=| a → || b → |sinθ×1 | a → × b → |=| a → || b → |sinθ
Substitute the given value of | a → | and | b → |in the above equation.
1=3× 2 3 sinθ sinθ= 1 2 θ= sin −1 ( 1 2 ) θ= π 4
The angle between the vectors a → and b → is π 4 .
Therefore, the correct option is (B).