Question

Let the plane $P : 8 x+\alpha_1 y+\alpha_2 z+12=0$ be parallel to the line $L : \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $P$ on the $y$-axis is 1 , then the distance between $P$ and $L$ is :

• A. $\sqrt{\frac{2}{7}}$
• B. $\frac{6}{\sqrt{14}}$
• C. $\sqrt{\frac{7}{2}}$
• D. $\sqrt{14}$

Answer 1

Answer: (D)

P: $8 x+\alpha_1 y+\alpha_2 z+12=0$
$L : \frac{ x +2}{2}=\frac{ y -3}{3}=\frac{ z +4}{5}$
$\because P$ is parallel to $L$
$ \Rightarrow 8(2)+\alpha_1(3)+5\left(\alpha_2\right)=0 $
$ \Rightarrow 3 \alpha_1+5\left(\alpha_2\right)=-16$
Also $y$-intercept of plane $P$ is 1
$\Rightarrow \alpha_1=-12$
And $\alpha_2=4$
$\Rightarrow$ Equation of plane $P$ is $2 x -3 y + z +3=0$
$\Rightarrow$ Distance of line $L$ from Plane $P$ is
$ =\left|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|$
$ =\sqrt{14}$