You are given two vectors `vec P=2hati-3hatj-hatk and vec Q =-6hati+9hatj+3hatk` From their vector product, we find that

Options

A. `vec P bot vec Q`
B. `vec P || vec Q`
C. Angle between `vec P and vec Q" is "30^(@)`
D. Angle between `vec P and vec Q" is "60^(@)`

Answer

B

Explanation

The vector product of P and Q is given by
`vec R = vec P xx vec Q`
`=|{:(hati, hatj, hatk),(P_(x),P_(y),P_(z)),(Q_(x),Q_(y),Q_(z)):}|=|{:(hati, hatj, hatk),(2,-3,-1),(-6,9,3):}|`
`therefore vec R = hati (-9+9) + hatj (6-6) + hatk (18-18)=0`
`therefore vec R =0`
`because` Their vector product is zero, the vector `vec P and vec Q` are parallel to each other.