Question

If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-1}{2}$ and the plane $2x-y+\sqrt{\lambda }z+4=0$ is such that sin $\theta =\frac{1}{3}$, then the value of $\lambda$ is :

• A. $-\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $-\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

Answer: (D)

If $\theta$ is the angle between the line and the plane then 90-$\theta$ will be the angle between the line and the normal to the plane.
$\therefore$ $\cos (90^{\circ }-\theta )=\frac{(\hat{i}+2\hat{j}+2\hat{k}),(2\hat{i}-\hat{j}+\sqrt{\lambda }\hat{k})}{\sqrt{1+4+4}\sqrt{4+1+\lambda }}$
$\therefore$ sin $\theta =\frac{2-2+2\sqrt{\lambda }}{2\sqrt{\lambda +5}}=\frac{2\sqrt{\lambda }}{3\sqrt{\lambda +5}}$
$\Rightarrow$ $\frac{1}{3}=\frac{2\sqrt{\lambda }}{3\sqrt{\lambda +5}}$
$\therefore$ $\lambda +5=4\lambda \Rightarrow 3\lambda =5\Rightarrow \lambda =\frac{5}{3}$