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. $\frac{15}{2}$
• B. $\frac{3}{2}$
• C. $\frac{2}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{15}{2}$

equation of line $\frac{x-0}{1}=\frac{y-1}{2}=\frac{2-3}{\lambda }$ where $a_1=1,a_2=2,a_3=\lambda$

equation of plane $n+2y+32=4$

where $b_1=1,b_2=2,b_3=3$

angle between Line and plane

$\cos \theta =\frac{|a_1{b}_1+a_2{b}_2+a_3{b}_3|}{\sqrt{a_1^2+a_2^2+a_3^2}.\sqrt{b_1^2+b_2^2+b_3^2}}$

$\cos \theta =\frac{|1.1+2.2+3.\lambda |}{\sqrt{1+4+{\lambda }^2}.\sqrt{1+4+9}}$

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

given $\theta ={\cos }^{-1}\left(\sqrt{\frac{5}{14}}\right)$

$\cos \theta =\left(\sqrt{\frac{5}{14}}\right)----\left(2\right)$

By eqn $\left(1\right)$ & $\left(2\right)$

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

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

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

$25+9{\lambda }^2+30\lambda =25+5{\lambda }^2$

$4{\lambda }^2+30\lambda =0$

$\lambda (2\lambda +15)=0\Rightarrow \lambda =0,\frac{15}{2}$

but $\lambda \ne 0$

So $\lambda =\frac{15}{2}$ Ans