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 the $\sin \theta =\frac{1}{3}$ then the value of $\lambda$ is

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

Answer 1

The correct option is B $\frac{5}{3}$

$(2,-1,\sqrt{\lambda })$ are the direction ratios perpendicular to the plane.

and,

$(1,2,2)$ is the direction ratio of the line.

Then,

$\overrightarrow{l}=\hat{i}+\widehat{2j}+\widehat{2k}$

$\overrightarrow{p}=\widehat{2i}-\hat{j}+\sqrt{\lambda }\hat{k}$

Since,

$\because cos{\theta }^{\prime }=\frac{\overline{l}\cdot \overline{p}}{\overline{\left|l\right|}.\overline{\left|p\right|}}=\frac{2-2+2\sqrt{\lambda }}{3\times \sqrt{5+\lambda }}$

$\Rightarrow cos{\theta }^{\prime }=\frac{2\sqrt{\lambda }}{3\sqrt{5+\lambda }}=\frac{2}{3}\sqrt{\frac{\lambda }{5+\lambda }}$

$\Rightarrow {\theta }^{\prime }=co{s}^{-1}\frac{2}{3}\sqrt{\frac{\lambda }{5+\lambda }}$

$\Rightarrow \theta =\frac{\Pi }{2}-{\theta }^{\prime }=si{n}^{-1}\frac{2}{3}\sqrt{\frac{\lambda }{5+\lambda }}$

$sin\theta =\frac{2}{3}\sqrt{\frac{\lambda }{5+\lambda }}$ $\to \left(1\right)$

But from the question,

$sin\theta =\frac{1}{3}$ $\to \left(2\right)$

Equation (1) & (2)

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

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

$\Rightarrow \frac{1}{4}=\frac{\lambda }{5+\lambda }$

$5+\lambda =4\lambda$

$5=3\lambda$

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

So, (B) is the correct answer.