If `theta` is the angle between the vectors `a=2hati+2hatj-hatk and b=6hati-3hatj+2hatk`, then

Options

A. `cos^(-1).(3)/(11)`
B. `cos^(-1).(2)/(11)`
C. `cos^(-1).(4)/(21)`
D. `cos^(-1).(3)/(22)`

Answer

C

Explanation

The angle between a and b is `theta`.
So, `cos theta=(a.b)/(|a||b|)`
`=((2hat(i)+2hat(j)-hat(k)).(6hat(i)-3hat(j)+2hat(k)))/(sqrt(4+4+1).sqrt(36+9+4))=4/(3xx7)`
`:. theta="cos"^(-1) 4/21`