Question

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

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

Answer 1

Answer: (B)

The line is $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane is
$2x-y+\sqrt{\lambda }z+4=0$
If $\theta$ is the angle between the line and the plane, then $90^{\circ }-\theta$ is the angle between the line and normal to the plane. Thus
$\cos \left(90^{\circ }-\theta \right)=\frac{(1)(2)+(2)(-1)+(2)(\sqrt{\lambda })}{\sqrt{1+4+4}\sqrt{4+1+\lambda }}$
or $\sin \theta =\frac{2-2+2\sqrt{\lambda }}{3\sqrt{5+\lambda }}$ or $\frac{1}{3}=\frac{2\sqrt{\lambda }}{3\sqrt{5+\lambda }}$
or $\sqrt{5+\lambda }=2\sqrt{\lambda }$
or $5+\lambda =4\lambda$
or $3\lambda =5$
or $\lambda =\frac{5}{3}$