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 $z=3+4r.$

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 A $4\sqrt{26}$

The distance of point $(1+3r,2-r,3+4r)$ from the plane is $\left|\frac{1+3r+2-r-3-4r-1}{\sqrt{1+1+1}}\right|=\frac{4}{\sqrt{3}}$
$\Rightarrow \frac{|2r+1|}{\sqrt{3}}=\frac{4}{\sqrt{3}}$
$\Rightarrow r=\frac{3}{2},-\frac{5}{2}$
Hence, the points are $A(\frac{11}{2},\frac{1}{2},\frac{18}{2})$ and $B(\frac{-13}{2},\frac{9}{2},\frac{-14}{2})$
$\Rightarrow AB=\sqrt{416}=4\sqrt{26}$