Find the angle between the line and the plane
Find the angle between the line and the plane

The equation of line is:

Here,

The equation of plane is:

$\stackrel{\to }{r}·\left(2\stackrel{^}{i}–\stackrel{^}{j}+\stackrel{^}{k}\right)=4$

\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=4

Here,

is parallel to line and

is perpendicular to plane.

Let’ ‘ is the angle between line and plane. It is also the angle between and .

$\begin{array}{l}\mathrm{sin}\theta =\left|\frac{\left(i–j+k\right)·\left(2i–j+k\right)}{\sqrt{1+1+1}\sqrt{4+1+1}}\right|\\ \therefore \mathrm{sin}\theta =\frac{2+1+1}{\sqrt{3}\sqrt{6}}=\frac{4}{\sqrt{18}}=\frac{4}{3\sqrt{2}}=\frac{2\sqrt{2}}{3}\\ \therefore \mathrm{sin}\theta =\frac{2\sqrt{2}}{3}\\ \therefore \theta ={\mathrm{sin}}^{–1}\left(\frac{2\sqrt{2}}{3}\right)\end{array}$

\begin{array}{l}

\sin \theta=\left|\frac{(i-j+k) \cdot(2 i-j+k)}{\sqrt{1+1+1} \sqrt{4+1+1}}\right| \\

\therefore \sin \theta=\frac{2+1+1}{\sqrt{3} \sqrt{6}}=\frac{4}{\sqrt{18}}=\frac{4}{3 \sqrt{2}}=\frac{2 \sqrt{2}}{3} \\

\therefore \sin \theta=\frac{2 \sqrt{2}}{3} \\

\therefore \theta=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)

\end{array}