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

The acute angle $\theta$ between line $\bar{a}+\lambda \bar{b}$ and the plane $\overline{\mathrm{r}}\cdot \overline{\mathrm{n}}=\mathrm{p}$ is given by $\sin \theta =\left|\frac{\overline{\mathrm{b}}\cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}}\cdot \cdot \cdot \overrightarrow{\mathrm{n}}|}\right|$ Here, $\overline{\mathrm{b}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}$ and $\overline{\mathrm{n}}=\hat{\mathrm{i}}-2\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}$ Also, $\theta ={\cos }^{-1}\left(\frac{2\sqrt{2}}{3}\right)={\sin }^{-1}\left(\frac{1}{3}\right)$ $\begin{array}{rl}& \therefore \text{ (i) }\Rightarrow \sin \left({\sin }^{-1}\left(\frac{1}{3}\right)\right)=\left|\frac{2-2+2\lambda }{\sqrt{4+1+4}\sqrt{1+4+{\lambda }^2}}\right|\\ & \Rightarrow \frac{1}{3}=\left|\frac{2\lambda }{3\sqrt{5+{\lambda }^2}}\right|\\ & \Rightarrow 5+{\lambda }^2=4{\lambda }^2\\ & \Rightarrow {\lambda }^2=\frac{5}{3}\\ & \Rightarrow \lambda =\sqrt{\frac{5}{3}}\end{array}$