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

The correct answer is $sin^{-1}\dfrac{8}{21}$

Let $\theta$ bet the angle between the line and the plane.

∴ $sin \theta$ = $\left| \dfrac{\overrightarrow{b} . \overrightarrow{n}}{|\overrightarrow{b}| . |\overrightarrow{n}|} \right|$


Given $\overrightarrow{b}= 2\hat{i} + 3\hat{j} + 6\hat{k}$

$|\overrightarrow{b}|$ = $\sqrt{4+9+36}$ = $\sqrt{49} = 7$


Given $\overrightarrow{n}= 10\hat{i} + 2\hat{j} - 11\hat{k}$

$|\overrightarrow{b}|$ = $\sqrt{100+4+121}$ = $\sqrt{225} = 15$


$\overrightarrow{b} . \overrightarrow{n}$ = $(2\hat{i} + 3\hat{j} + 6\hat{k})$ . $(10\hat{i} + 2\hat{j} - 11\hat{k})$ = $20+6-66$ = $-40$

∴ $sin \theta$ = $\left| \dfrac{-40}{7 \times 15} \right|$ = $\dfrac{8}{21}$
$\theta$ = $sin^{-1}\dfrac{8}{21}$