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

Answer 1

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

The given line and plane are $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda }$ and $x+2y+3z=4$.

Angle between the line and plane will be,

$\theta ={\sin }^{-1}\left(\frac{1.1+2.2+3.\lambda }{\sqrt{1+4+{\lambda }^2)}\sqrt{1+4+9}}\right)$

$\theta ={\sin }^{-1}\left(\frac{5+3\lambda }{\sqrt{14}\sqrt{5+{\lambda }^2}}\right)$

$\theta ={\cos }^{-1}\left(\sqrt{1-\frac{(5+3\lambda {)}^2}{14(5+{\lambda }^2)}}\right)={\cos }^{-1}\left(\sqrt{\frac{5}{14}}\right)$ ......(given)
$\therefore \lambda =\frac{2}{3}$