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. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{5}$

Answer 1

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

Angle between straight line $x=\frac{y-1}{2}=\frac{z-3}{\lambda }$ and the plane $x+2y+3z=4$ is $\theta$ is given by

$\sin \theta =\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Here $\overrightarrow{b}=(1,2,\lambda )$ and $\overrightarrow{n}=(1,2,3)$
$\therefore \sin \theta =\frac{(\hat{i}+2\hat{j}+\lambda \hat{k})\cdot (\hat{i}+2\hat{j}+3\hat{k})}{\sqrt{1+4+{\lambda }^2}\sqrt{1+4+9}}\Rightarrow \sin \theta =\frac{5+3\lambda }{\sqrt{{\lambda }^2+5}\cdot \sqrt{14}}$

Given, $\cos \theta =\sqrt{\frac{5}{14}}$
$\therefore \sin \theta =\frac{3}{\sqrt{14}}$
$\Rightarrow \frac{3}{\sqrt{14}}=\frac{5+3\lambda }{\sqrt{{\lambda }^2+5}\cdot \sqrt{14}}$ $\Rightarrow 9({\lambda }^2+5)=9{\lambda }^2+30\lambda +25$ $\Rightarrow 9{\lambda }^2+45=9{\lambda }^2+30\lambda +25$ $\Rightarrow 30\lambda =20$
$\therefore \lambda =\frac{2}{3}$