Question

The angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$ is ${\cos }^{-1}\frac{p}{21}-\frac{\pi }{2},$ then the value of ${}^{\prime }{p}^{\prime }$

Answer 1

Given :
Line: $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6},$
$D.r^{\prime }s$ of line $(a_1,b_1,c_1)=(2,3,6)$
Plane: $10x+$ $2y-11z=3$
$D.r^{\prime }s$ of normal to the plane $(a_2,b_2,c_2)=(10,2,-11)$

If the angle between the line and plane is $\theta$
$\Rightarrow \sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

$\cos (90-\theta )=$

$\frac{2\times 10+3\times 2+6\times (-11)}{\sqrt{(2^2+3^2+6^2)({10}^2+2^2+{11}^2)}}=\frac{-8}{21}$

$\therefore \theta ={90}^{\circ }-{\cos }^{-1}\left(\frac{-8}{21}\right)$

$\text{or}\theta =-\frac{\pi }{2}+{\cos }^{-1}\left(\frac{8}{21}\right)$