The acute angle between the line $\bar{r}=(\hat{\imath}+2…
The acute angle between the line $\bar{r}=(\hat{\imath}+2 \hat{\jmath}+\hat{k})+\lambda(\hat{\imath}+\hat{\jmath}+\hat{k})$ and the plane
$\bar{r} \cdot(2 \hat{\imath}-\hat{\jmath}+\hat{k})=5$ is
$\sin ^{-1}\left(\frac{\sqrt{2}}{3}\right)$
$\sin ^{-1}\left(\frac{2}{3}\right)$
$\sin ^{-1}\left(\sqrt{\frac{2}{3}}\right)$
$\sin ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
Solution
Given $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and the plane $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})=5$
The angle between the line $\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ and the plane $\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=p$ is given by $\sin \theta=\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}}| \overline{\mathrm{n}} \mid}$
Here $\bar{b}=\hat{i}+\hat{j}+\hat{k} \Rightarrow|\bar{b}|=\sqrt{3}$ and $\bar{n}=2 \hat{i}-\hat{j}+\hat{k} \Rightarrow|\bar{n}|=\sqrt{6}$
Here $\bar{b} \cdot \bar{n}=(\hat{i}+\hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})=2-1+1=2$
$\therefore \sin \theta=\frac{2}{\sqrt{3} \sqrt{6}}=\frac{2}{\sqrt{3} \times \sqrt{3} \times \sqrt{2}} \Rightarrow \sin \theta=\frac{\sqrt{2}}{3} \Rightarrow \theta=\sin ^{-1} \frac{\sqrt{2}}{3}$