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

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 $|\bar{n}|=\sqrt{4+1+1}=\sqrt{6}=\sqrt{3}\times \sqrt{2}$ $\therefore \sin \theta =\frac{4}{\sqrt{3}\times \sqrt{3}\times \sqrt{2}}=\frac{2\sqrt{2}}{3}\Rightarrow \theta ={\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$