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

Answer 1

Answer: (B)

$\frac{5}{3}$

Angle between the line and plane is same as the angle between the line and normal to the plane
$\therefore \cos ({90}^0-\theta )=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\therefore \frac{1}{3}=\frac{(1\times 2+2\times (-1)+2\sqrt{\lambda })}{\sqrt{1^2+2^2+2^2}\sqrt{2^2+1^2+\lambda }}$

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