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If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}$ $=\frac{z-2}{2}$ and the plane $2 x-y+\sqrt{\lambda} z+4=0$ is such that $\sin \theta=\frac{1}{3}$ then the value of $\lambda$ is
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Verified Answer
The correct answer is:
$\frac{5}{3}$
If $\theta$ is the angle between line and plane then $\left(\frac{\pi}{2}-\theta\right)$ is the angle between line and normal to plane given by
$$
\begin{aligned}
& \cos \left(\frac{\pi}{2}-\theta\right)=\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\sqrt{\lambda} \hat{k})}{3 \sqrt{4+1+\lambda}} \\
& \cos \left(\frac{\pi}{2}-\theta\right)=\frac{2-2+2 \sqrt{\lambda}}{3 \times \sqrt{5}+\lambda} \\
& \Rightarrow \sin \theta=\frac{2 \sqrt{\lambda}}{3 \sqrt{5}+\lambda}=\frac{1}{3} \Rightarrow 4 \lambda=5+\lambda \Rightarrow \lambda=\frac{5}{3} .
\end{aligned}
$$
$$
\begin{aligned}
& \cos \left(\frac{\pi}{2}-\theta\right)=\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\sqrt{\lambda} \hat{k})}{3 \sqrt{4+1+\lambda}} \\
& \cos \left(\frac{\pi}{2}-\theta\right)=\frac{2-2+2 \sqrt{\lambda}}{3 \times \sqrt{5}+\lambda} \\
& \Rightarrow \sin \theta=\frac{2 \sqrt{\lambda}}{3 \sqrt{5}+\lambda}=\frac{1}{3} \Rightarrow 4 \lambda=5+\lambda \Rightarrow \lambda=\frac{5}{3} .
\end{aligned}
$$
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