Find the angle between the line \overrightarrow r  = (\widehat i + 2\widehat k - \widehat k) + \lambda (\widehat i  - \widehat j + \widehat k) and the plane \overrightarrow r .(2\widehat i - \widehat j + \widehat k) = 4
Find the angle between the line \overrightarrow r  = (\widehat i + 2\widehat k - \widehat k) + \lambda (\widehat i  - \widehat j + \widehat k) and the plane \overrightarrow r .(2\widehat i - \widehat j + \widehat k) = 4

Answer:

The equation of line is:

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

Here, \vec{b}=\hat{i}-j+\vec{k}

The equation of plane is:

r·(2i^j^+k^)=4

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

Here, \overrightarrow{\mathrm{n}}_{1}=2 i-j+\hat{k}

\vec{b} is parallel to line and

\vec{n} is perpendicular to plane.

Let’ \theta ‘ is the angle between line and plane. It is also the angle between \vec{b} and \vec{n}.

sinθ=(ij+k)·(2ij+k)1+1+14+1+1sinθ=2+1+136=418=432=223sinθ=223θ=sin1223

\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}