Question

Let the vectors and be such that and , then is a unit vector, if the angle between and is (A) (B) (C) (D)

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Solution

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).

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