Question

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

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

Answer 1

The correct option is C $\frac{2}{3}$

We have,

$x=\frac{y-1}{2}=\frac{z-3}{\lambda }x+2y+3z=4$

$={\cos }^{-1}\left(\frac{\sqrt{5}}{14}\right)\lambda =?$

$=\cos =\frac{\sqrt{5}}{14}$

$\theta ={\sin }^{-1}\sqrt{1-\frac{5}{14}}={\sin }^{-1}\left(\frac{\sqrt{9}}{14}\right)$

$\sin \theta =\sqrt{\frac{9}{14}}$

$\Rightarrow \frac{1+4+3\lambda }{\sqrt{5+{\lambda }^2}\sqrt{14}}=\frac{3}{\sqrt{14}}$..................obtained from the cross product relation of two normal vectors of the line and the plane.

$\Rightarrow \frac{5+3\lambda }{\sqrt{5+{\lambda }^2}\sqrt{14}}=\frac{3}{\sqrt{14}}$

$\Rightarrow 25+9{\lambda }^2+30\lambda =9\left(5+{\lambda }^2\right)$

$\Rightarrow 25+30\lambda =45$

$20=30\lambda$

$\lambda =\frac{2}{3}$

Then,

Option $C$ is correct answer.