Question

Let $P,Q$ are the points on the line $\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}$ which are at a distance of $4\sqrt{3}$ from the plane $x+y-z=1.$ Then the value of $|PQ|$ is

• A. $4\sqrt{26}$
• B. $20$
• C. $10\sqrt{13}$
• D. $12\sqrt{26}$

Answer 1

The correct option is D $12\sqrt{26}$

Given line equation is $\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}=r$(say)
Let the variable point on the line be, $(1+3r,2-r,3+4r)$

Then distance of the point from the plane

$=\frac{|1+3r+2-r-3-4r-1|}{\sqrt{3}}=\frac{|-2r-1|}{\sqrt{3}}=4\sqrt{3}$

Hence, $|2r+1|=12$

$2r+1=12\text{or}2r+1=-12$

Hence, $r=\frac{11}{2}$ and $r=\frac{-13}{2}$

Therefore the points which lie on the line are,

$P=(\frac{35}{2},\frac{-7}{2},25)$ and $Q=(\frac{-37}{2},\frac{17}{2},-23)$

Hence, the distance between the two points is,

$PQ=\sqrt{{36}^2+{12}^2+{48}^2}=12\sqrt{26}\text{units}$