Question

If angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2 x-y+\sqrt{\lambda} z+4=0$ 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
$2 x-y+\sqrt{\lambda} z+4=0$
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}$