The angle between the line $\overline{\mathrm{r}}=(\hat{\imath}+2…
The angle between the line
$\overline{\mathrm{r}}=(\hat{\imath}+2 \hat{\mathrm{\jmath}}-\widehat{\mathrm{k}})+\lambda(\hat{\imath}-\hat{\mathrm{\jmath}}+\widehat{\mathrm{k}})$ and the plane
$\overline{\mathrm{r}} \cdot(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})=4 \mathrm{is}$
$\sin ^{-1}\left(\frac{2}{3}\right)$
$\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$
$\sin ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
Solution
The angle $\theta$ between the line $\bar{r}=\bar{a}+\lambda b$ and the plane $\bar{r} \cdot \bar{n}=p$ is given by
$\sin \theta=\frac{\bar{b} \cdot \bar{n}}{|\bar{b}| \cdot|\bar{n}|}$
Here $\overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{n}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\therefore \overline{\mathrm{b}} \cdot \overline{\mathrm{n}}=1(2)+(-1)(-1)+1(1)=4$
$|\bar{b}|=\sqrt{1+1+1}=\sqrt{3}$ and $|\bar{n}|=\sqrt{4+1+1}=\sqrt{6}=\sqrt{3} \times \sqrt{2}$
$\therefore \sin \theta=\frac{4}{\sqrt{3} \times \sqrt{3} \times \sqrt{2}}=\frac{2 \sqrt{2}}{3} \Rightarrow \theta=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$