Question

If $\theta$ the angle 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 the value $\lambda$ is

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

Answer 1

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

Given, equation of line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ for which $a_1=1,b_1=2,c_1=2$ and the plane $2x-y+\sqrt{\lambda }z+4=0$ for which $a_2=2,b_2=-1$ and $c_2=\sqrt{\lambda }$.

Now, given $\sin \theta =\frac{1}{3}$
Therefore, $\sin \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$
$\Rightarrow \frac{1}{3}=\left|\frac{2-2+2\sqrt{\lambda }}{\sqrt{1+4+4}\sqrt{4+1+\lambda }}\right|$
$\Rightarrow \frac{1}{3}=\frac{2\sqrt{\lambda }}{\sqrt{9}\sqrt{5+\lambda }}$
$\Rightarrow \sqrt{5+\lambda }=2\sqrt{\lambda }$
$\Rightarrow 5+\lambda =4\lambda$
$\Rightarrow 3\lambda =5$
$\Rightarrow \lambda =\frac{5}{3}$