Question

The distance of the point $(1,2,1)$ from the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}$ is

• A. $\frac{\sqrt{5}}{3}$
• B. $\frac{2\sqrt{3}}{5}$
• C. $\frac{20}{3}$
• D. $\frac{2\sqrt{5}}{3}$
  

Answer 1

The correct option is D $\frac{2\sqrt{5}}{3}$


$\text{Let the given point be}A(1,2,1).\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}=k$
$(x,y,z)=2k+1,k+2,2k+3$
dr's of $AB:2k,k,2k+2$
dr's of line : $2,1,2$
As both lines are perpendicular to each other
So, $4k+k+4k+4=0$
$k=\frac{-4}{9}$
point on the line is($\frac{1}{9},\frac{14}{9},\frac{19}{9}$)
Distance=$\frac{2\sqrt{5}}{3}$units