Let the vectors and be such that and , then is a unit vector, if the angle between and is (A) (B) (C) (D)
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 → | sin θ n ^ | a → × b → | = | | a → | | b → | sin θ n ^ | | a → × b → | = | a → | | b → | sin θ | n ^ |
Since, | n ^ | = 1 as n ^ is a unit vector, therefore,
| a → × b → | = | a → | | b → | sin θ × 1 | a → × b → | = | a → | | b → | sin θ
Substitute the given value of | a → | and | b → | in the above equation.
1 = 3 × 2 3 sin θ sin θ = 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).
The given vectors are
As given
The formula for cross product of
Since,
Substitute the given value of
The angle between the vectors
Therefore, the correct option is (B).
