The angle between the straight line `r=(2-3t)hati+(1+2t)hatj+(2+6t)hatkand r=1(1+4s)hati+(2-s)hatj+(8s-1)hatk`

Options

A. `cos^(-1)((sqrt41)/(34))`
B. `cos^(-1)((21)/(34))`
C. `cos^(-1)((34)/(63))`
D. None of these

Answer

C

Explanation

Given equation of line are `r=2hati+hatj+2hatk+t(-3hati+2hatj+6hatk) and and r=hati+2hatj-hatk+s (4hati-hatj+8hatk)`
On compairing with `=1+t 1 and r=2+s (2), ` we get `b_(1)=-3hati+2hatj+6hatk and b_(2)=4hati-hatj+8hatk`
`therefore cos theta=(b_(1).b_(2))/(|b_(1)||b_(2)|)`
`=((-3hati+2hatj+6hatk).(4hati-hatj+8hatk))/(sqrt((-3)^(2)+2^(2)+6^(2))sqrt(4^(2)+(-1)^(2)+8^(2)))`
`=(-12-2+48)/(sqrt(9+4+36)sqrt(16+1+64))`
`=(34)/(7xx9)=(34)/(63)`
`theta=cos^(-1)((34)/(63))`