Question

Find the angle $\theta$, between the line $\frac{x-2}{3}=\frac{y-3}{5}=\frac{z-4}{4}$ and the plane $2x-2y+z-5=0$

Answer 1

Angle between line and plane is =($90$-Angle between line and normal to plane)

$=90-{\cos }^{-1}\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{c}|}\right)$

Where

$\overrightarrow{a}=3\hat{i}+5\hat{j}+4\hat{k}$

$\overrightarrow{b}=2\hat{i}-2\hat{j}+\hat{k}$

Angle =$90-{\cos }^{-1}\left(\frac{6-10+4}{\sqrt{3^2+5^2+4^2}\sqrt{2^2+(-2{)}^2+1^2}}\right)$

$=90-{\cos }^{-1}\left(0\right)$

$={90}^0-{90}^0$

$=0^0$