What is the angle between two vectors `vec P =2hati +3hatj +hatk and vec Q =-3hati +6hatk`.

Options

A. `0^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

D

Explanation

`cos theta = (vec P cdot vec Q)/(PQ)`
`vec P cdot vec Q = P_(x) Q_(x) + P_(y) O_(y) + P_(z) Q_(z)`
`=[2xx (-3) +3xx0+1xx6]=-6+6=0`
`therefore cos theta =(0)/(PQ) =0" "therefore theta =90^(@)`