The angle between the line `vecr=(hati+hatj-3hatk)+lambda(2hati+2hatj+hatk)` and the plane `vecr.(6hati-3hatj+2hatk)=5`, is

Options

A. `cos^(-1)(5/14)`
B. `cos^(-1)(5/21)`
C. `sin^(-1)(5/21)`
D. `sin^(-1)(8/21)`

Answer

d

Explanation

`sintheta=|vecb.vecn|/(|vecb||vecn|)`
Here `vecb=(2hati+2hatj+hatk)` and `vecrn=(6hati-3hatj+2hatk)`.
`therefore sintheta=|(2hati xx 2hatj+hatk).(6hati-3hatj+2hatk)|/(sqrt(2^(2)+2^(2)+1^(2)).sqrt(6^(2)+(-3)^(2)+2^(2))`
`|(12-6+2)|/(sqrt(9)sqrt(49))= 8/(3 xx 7) = 8/21 rArr theta= sin^(-1)(8/21)`.