The angle between the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}$ and the plane $\bar{r} \cdot(6…
The angle between the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}$ and the plane $\bar{r} \cdot(6 \hat{\imath}-2 \hat{\jmath}-3 \hat{k})=5$
is
$\sin ^{-1}\left(\frac{4}{21}\right)$
$\cos ^{-1}\left(\frac{4}{21}\right)$
$\sin ^{-1}\left(\frac{5}{7}\right)$
$\cos ^{-1}\left(\frac{5}{7}\right)$
Solution
Angle between line and plane is given by
$\sin \theta=\frac{a a_{1}+b b_{1}+c c_{1}}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}$
Here $a=2, b=1, c=2$ and $a_{1}=6, b_{1}=-2, c_{1}=-3$
$\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)$