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. $2/5$
• B. $5/3$
• C. $2/3$
• D. $3/2$

Answer 1

The correct option is C $2/3$

Angle between line and normal to plane would be $\frac{\pi }{2}-{\cos }^{-1}\sqrt{\frac{5}{14}}={\sin }^{-1}\sqrt{\frac{5}{14}}={\cos }^{-1}\frac{3}{\sqrt{14}}$

$\Rightarrow \frac{(\hat{i}+2\hat{j}+\lambda \hat{k}).(\hat{i}+2\hat{j}+3\hat{k})}{|\hat{i}+2\hat{j}+\lambda \hat{k}||\hat{i}+2\hat{j}+3\hat{k}|}=\frac{3}{\sqrt{14}}$

$\Rightarrow \frac{1+4+3\lambda }{\sqrt{5+{\lambda }^2}\sqrt{1+4+9}}=\frac{3}{\sqrt{14}}$

$\Rightarrow 3\lambda +5=3\sqrt{5+{\lambda }^2}\Rightarrow {(3\lambda +5)}^2=9({\lambda }^2+5)$

$\Rightarrow 9{\lambda }^2+25+30\lambda =9{\lambda }^2+45\Rightarrow 30\lambda =20\Rightarrow \lambda =\frac{2}{3}$