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 $ $\lambda + 5 = 4 \lambda \, \Rightarrow \, 3 \lambda = 5 \, \Rightarrow \, \lambda = \frac{5}{3}$