Question

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

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

Answer 1

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

We know that,

angle between line $\frac{x-x_1}{A_1}=\frac{y-y_1}{B_1}=\frac{z-z_1}{C_1}$

and plane $A_{2}x+B_{2}y+C_{2}z+D_2=0$ is given by $\theta$, where, $\sin \theta =\frac{A_1{A}_2+B_1{B}_2{C}_1{C}_2}{\sqrt{A_1^1+B_1^2+C_1^2}.\sqrt{A_2^2+B_2^2+C_2^2}}\Rightarrow \sin \theta =\frac{1\left(1\right)+2\left(2\right)+\lambda \left(3\right)}{\sqrt{1^2+2^2+{\lambda }^2}.\sqrt{1^2+2^2+{\lambda }^2}}\Rightarrow \sin \theta =\frac{3\lambda +5}{\sqrt{{\lambda }^2+5}.\sqrt{14}}....eq.1$

As, $\theta ={\cos }^{-1}\sqrt{\frac{5}{14}}\Rightarrow \cos \theta =\sqrt{\frac{5}{14}}\Rightarrow co{s}^2\theta =\frac{5}{14}\Rightarrow 1-si{n}^2\theta =\frac{5}{14}(\because {\sin }^2\theta +{\cos }^2\theta =1)\Rightarrow {\sin }^2\theta =\frac{9}{14}\Rightarrow \sin \theta =\sqrt{\frac{9}{14}}....eq.2$

Equating eq.1 and eq.2 ,

$\frac{3\lambda +5}{\sqrt{{\lambda }^2+5}.\sqrt{14}}=\sqrt{\frac{9}{14}}$

$\frac{3\lambda +5}{\sqrt{{\lambda }^2+5}}=\sqrt{9}$

squaring on both sides,

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