The angle between the line $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and plane $x-2 y-\lambda z=3$ is…
The angle between the line $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and plane $x-2 y-\lambda z=3$ is $\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$, then value of $\lambda$ is
$\sqrt{\frac{3}{5}}$
$\frac{5}{\sqrt{3}}$
$\sqrt{\frac{5}{3}}$
$\frac{1}{\sqrt{3}}$
Solution
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, $\theta=\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)=\sin ^{-1}\left(\frac{1}{3}\right)$
$\begin{aligned}
& \therefore \quad \text { (i) } \Rightarrow \sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)=\left|\frac{2-2+2 \lambda}{\sqrt{4+1+4} \sqrt{1+4+\lambda^2}}\right| \\
& \Rightarrow \frac{1}{3}=\left|\frac{2 \lambda}{3 \sqrt{5+\lambda^2}}\right| \\
& \Rightarrow 5+\lambda^2=4 \lambda^2 \\
& \Rightarrow \lambda^2=\frac{5}{3} \\
& \Rightarrow \lambda=\sqrt{\frac{5}{3}}
\end{aligned}$