Question

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

Line : $\frac{x}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda }$

and the plane $x+2y+3z=4$

$\therefore \sin \theta =\frac{1+4+3\lambda }{\sqrt{1+4+{\lambda }^2}\sqrt{1+4+9}}$

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

$co{s}^{-1}\sqrt{\frac{5}{14}}=si{n}^{-1}\frac{3}{\sqrt{14}}$

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

\sqrt{14}}$

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

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

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