Question

Let the plane $P:8x+{\alpha }_{1}y+{\alpha }_{2}z+12=0$ 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: $8x+{\alpha }_{1}y+{\alpha }_{2}z+12=0$
$L:\frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$
$\because P$ 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 $2x-3y+z+3=0$
$\Rightarrow$ Distance of line $L$ from Plane $P$ is
$=\left|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|$
$=\sqrt{14}$