Question

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

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

Answer 1

Answer: (A)

$\frac{5}{3}$

Angle between line and normal to plane is

$\cos (\frac{\pi }{2}-\theta )=\frac{2-2+2\sqrt{\lambda }}{3\times \sqrt{5+\lambda }}$, where $\theta$ is angle between line a plane

$\Rightarrow \sin \theta =\frac{2\sqrt{\lambda }}{3\sqrt{5+\lambda }}=\frac{1}{3}$

$\Rightarrow 4\lambda =5+\lambda$

$\Rightarrow \lambda =\frac{5}{3}$