Find the angle between the vertors `vec(A) = hati + 2hatj - hatk` and `vec(B) = - hati +hatj - 2hatk`.

Options

A. `pi`
B. `(pi)/(3)`
C. `(pi)/(2)`
D. 0

Answer

B

Explanation

`vec A = hati+2hatj-hatk, vec B =hati+hatj -2hatk`
Angle between `vec A and vec B` is given by `cos theta =(vec A cdot vec B)/(AB)`
Thus, `A = |A|=sqrt (1^(2)+2^(2)+1^(2))=sqrt(6)`
`B = |B| = sqrt (1^(2) +1^(2)+2^(2))=sqrt(6) and vec A cdot vec B =(hati +2hatj - hatk) cdot (-hati + hatj -2hatk)`
`=-1+2+2=3`
`therefore cos theta =(3)/(sqrt(6)xxsqrt(6))=(3)/(6) =(1)/(2)`
`therefore theta =60^(@) =(pi)/(3)`