Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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 $\bar{b}\cdot \bar{n}=(\hat{i}+\hat{j}+\hat{k})\cdot (2\hat{i}-\hat{j}+\hat{k})=2-1+1=2$ $\therefore \sin \theta =\frac{2}{\sqrt{3}\sqrt{6}}=\frac{2}{\sqrt{3}\times \sqrt{3}\times \sqrt{2}}\Rightarrow \sin \theta =\frac{\sqrt{2}}{3}\Rightarrow \theta ={\sin }^{-1}\frac{\sqrt{2}}{3}$