Question

If 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. (A) \(\frac{ - 3}{5}  \)
• B. (B) \(\frac{5}{3}  \)
• C. (C) \(\frac{ - 4}{3}  \)
• D. (D ) \(\frac{ 3}{4}  \)

Answer 1

The Correct Option is B

Explanation:
Given:Equation of line, $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\dots (i)$and equation of the plane, $2x-y+\sqrt{\lambda }z+4=0\dots (ii)$We have to find value of $\lambda$ such that the angle $\theta$ between line and plane is given by $\sin \theta =\frac{1}{3}$The angle $\theta$ between the line (i) and plane (ii) is given by $\sin \theta =\left|\frac{1.2+2\cdot (-1)+2\cdot (\sqrt{\lambda })}{\sqrt{1+4+4}\cdot \sqrt{4+1+\lambda }}\right|$[Using formula of Angle Between a Line and a Plane] $\Rightarrow \frac{1}{3}=|\frac{2\sqrt{\lambda }}{3\cdot \sqrt{4+1+\lambda }}|[\sin \theta =\frac{1}{3}($ Given $)]$$\Rightarrow \sqrt{5+\lambda }=2\sqrt{\lambda }$$\Rightarrow 5+\lambda =4\lambda$ [Squaring both sides] $\Rightarrow 3\lambda =5$$\Rightarrow \lambda =\frac{5}{3}$Hence, the correct option is (B).