Three Dimensional Geometry Question 29
Question 29 - 31 January - Shift 2
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 :
(1) $\sqrt{14}$
(2) $\frac{6}{\sqrt{14}}$
(3) $\sqrt{\frac{2}{7}}$
(4) $\sqrt{\frac{7}{2}}$
Show Answer
Answer: (1)
Solution:
Formula: Equation Of A Plane
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 $L$
$\Rightarrow 8(2)+\alpha_1(3)+5(\alpha_2)=0$
$\Rightarrow 3 \alpha_1+5(\alpha_2)=-16$
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
$=|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}|$
$=\sqrt{14}$
