Question

The angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. ${\sin }^{-1}\left(\frac{8}{21}\right)$

Answer 1

The correct option is D ${\sin }^{-1}\left(\frac{8}{21}\right)$

Lets $\theta$ be the angle between the line and the plane, we have
$a=2,b=3,c=6$
and $a_1=10,b_1=2,c_1=-11$
$\therefore \cos (90-\theta )=\sin \theta$
$=\left|\frac{a{a}_1+b{b}_1+c{c}_1}{\sqrt{a^2+b^2+c^2}\sqrt{a_1^2+b_1^2+c_1^2}}\right|$
$=\left|\frac{\left(2\right)\left(10\right)+\left(3\right)\left(2\right)+\left(6\right)(-11)}{\sqrt{2^2+3^2+6^2}\sqrt{{10}^2+2^2+(-11{)}^2}}\right|$
$=\left|\frac{20+6-66}{\sqrt{49}\cdot \sqrt{225}}\right|=\left|\frac{-40}{7\times 15}\right|=\left|\frac{8}{21}\right|$

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