Question

Consider a plane $x+y-z=1$ and point $A(1,2,-3)$. A line $L$ has the equation $x=1+3r,y=2-r$ and $z=3+4r$, then the distance between the points on the line which are at a distance of $4\sqrt{3}$ from the plane is,

• A. $4\sqrt{26}$
• B. $20$
• C. $10\sqrt{13}$
• D. none of these

Answer 1

The correct option is D none of these

Let the point 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, $|2r+1|=12$

$2r+1=12$ and $2r+1=-12$

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

Therefore the points which lie on the line are ,

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

Hence, the distance between the two points are,

$=\sqrt{{36}^2+{12}^2+{48}^2}$

$=12\sqrt{26}$ units.