The angle betwene the line `vecr=(1+2mu)hati+(2+mu)hatj+(2m-1)hatk` and the plane `3x-2y=6z=0` where `mu` is a scalar is (A) `sin^-1(15/21)` (B) `cos^-1(16/21)` (C) `sin^-1(16/21)` (D) `pi/2`

Options

A. `sin^(-1)((15)/(21))`
B. `cos^(-1)((16)/(21))`
C. `sin^(-1)((16)/(21))`
D. `(pi)/(2)`

Answer

C

Explanation

The given line is
`r = (1+2mu)hati+(2+mu) hatj+(2mu01)hatk`
` = (hati + 2hatj - hatk) + mu(2hati + hatj + 2hatk)`
Vector equation of line written in cartesian form is
`(x-1)/(2) = (y-2)/(1) = (z+1)/(2)`
`therefore` Angle between line and a plane is given by
`sin theta = (a_(1)a_(2) +b_(1)b_(2) +c_(1)c_(2))/(sqrt(a_(1)^(2) +b_(1)^(2)+c_(1)^(2))sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))`
` = (2xx 3 +1x(-2)+2xx6)/(sqrt(4+1+4)sqrt(9+4+36))=(16)/(21)`
` = rArr theta = sin^(-1)((16)/(21))`