Find the angle between the lines
`vecr=3hati-2hatj+6hatk+lamda(2hati+hatj+2hatk)` and `vecr=(2hatj-5hatk)+mu(6hati+3hatj+2hatk)`.

Options

A. `theta=cos^(-1)(21/19)`
B. `theta=cos^(-1)(19/21)`
C. `theta=cos^(-1)(6/21)`
D. `theta=cos^(-1)(1/19)`

Answer

B

Explanation

We have `vecr=3hati-2hatj|6hatk+lamda(2hati+hatj+2hatk)`
and `vecr=(2hatj-5hatk)+mu(6hati+3hatj+2hatk)`
where `veca_(1)=3hati-2hatj+6hatk,vecb_(1)=2hati+hatj+2hatk`
and `veca_(2)=2hatj-5hatk, vecb_(2)=6hati+3hatj+2hatk`
If `theta` is angle between the lines, then
`costheta=(|vec(b_(1)).vecb_(2)|)/(|vec(b_(1))|.|vec(b_(2))|)`
`=(|(2hati+hatj+2hatk).(6hati+3hatj+2hatk)|)/(|2hati+hatj+2hatk||6hati+3hatj+2hatk|)`
`=(|12+3+4|)/(sqrt(9)sqrt(49))=19/21`
`:. theta=cos^(-1)(19/21)`