Question

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

• A. ${\cos }^{-1}\left(\frac{3}{2\sqrt{10}}\right)$
• B. ${\sin }^{-1}\left(\frac{3}{2\sqrt{10}}\right)$
• C. ${\cos }^{-1}\left(\frac{8}{21}\right)$
• D. ${\sin }^{-1}\left(\frac{8}{21}\right)$

Answer 1

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

Let $\theta$ be the angle between the line and the plane.
$\text{Then}\sin \theta =\left|\frac{2\times 10+3\times 2+6\times (-11)}{\sqrt{2^2+3^2+6^2}\sqrt{{10}^2+2^2+(-11{)}^2}}\right|$
$\Rightarrow \sin \theta =\left|\frac{-40}{7\times 15}\right|$
$\Rightarrow \sin \theta =\frac{8}{21}$
$\Rightarrow \theta ={\sin }^{-1}\left(\frac{8}{21}\right)$