Question

If the angle between the line $x = \frac{y -1}{2} = \frac{z - 3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1} \left( \sqrt{\frac{5}{14}} \right),$ then $\lambda$ equals

• A. $\frac{3}{2}$
• B. $\frac{2}{5}$
• C. $\frac{5}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (D)

If $\theta$ be the angle between the given line and plane, then
$\sin \theta = \frac{1\times1+2\times2+\lambda\times3}{\sqrt{1^{2} + 2^{2}+\lambda^{2}} . \sqrt{1^{2}+2^{2}+3^{2}}}$
$ = \frac{5+3\lambda}{\sqrt{14}.\sqrt{5+\lambda^{2}} }$
$ \Rightarrow \cos\theta = \sqrt{1- \frac{\left(5+2\lambda\right)^{2}}{14\left(5+\lambda^{2}\right)}} $
$\therefore \theta= \cos^{-1} \sqrt{1- \frac{\left(5+3\lambda\right)^{2}}{14\left(5+\lambda^{2}\right)} } $
But it is given that $\theta = \cos^{-1}\sqrt{\frac{5}{14}} $
$\therefore \sqrt{1- \frac{\left(5+3\lambda\right)^{2}}{14\left(5+\lambda^{2}\right)} } = \sqrt{\frac{5}{14}} $
$\Rightarrow \lambda = \frac{2}{3} $