Find the angle between the vectors
The given vectors are a → = i ^ − 2 j ^ + 3 k ^ and b → = 3 i ^ − 2 j ^ + k ^ .
Magnitude of vector a → is,
| a → | = 1 2 + ( − 2 ) 2 + 3 2 = 14
Magnitude of vector b → is,
| b → | = 3 2 + ( − 2 ) 2 + 1 2 = 14
Dot product of a → and b → is,
a → ⋅ b → = ( i ∧ − 2 j ∧ + 3 k ∧ ) ⋅ ( 3 i ∧ − 2 j ∧ + k ∧ ) = 1 ⋅ 3 + ( − 2 ) ⋅ ( − 2 ) + 3 ⋅ 1 = 3 + 4 + 3 = 10
Formula used for dot product is,
a → ⋅ b → = | a → | | b → | cos θ
Here, the angle between a → and b → is θ .
Substitute the values in the above formula.
10 = 14 × 14 × cos θ 10 14 = cos θ cos θ = 5 7 θ = cos − 1 ( 5 7 )
Thus, the angle between vectors is cos − 1 ( 5 7 ) .
The given vectors are
Magnitude of vector
Magnitude of vector
Dot product of
Formula used for dot product is,
Here, the angle between
Substitute the values in the above formula.
Thus, the angle between vectors is
