Question

Find the perpendicular distance between the lines $x=3-2k,y=k,z=3-k,\in R$ and $x=2k-3,y=2-k,z=7+k,k\in R$.

Answer 1

$x=3-2k,y=k,z=3-k\Rightarrow \frac{x-3}{-2}=k,\frac{y-0}{1}=k,\frac{z-3}{-1}=k$

$\therefore \frac{x-3}{-2}=\frac{y}{1}=\frac{z-3}{-1}$ is equation of line 1.

Similarly,

$\frac{x+3}{2}=\frac{y-2}{-1}=\frac{z-7}{1}$ -Line 2 (Both lines are parallel)

Perpendicular shortest distane of $(3,0,3)$ to line 2$=$ Perpendicular distance between lines.

Foot of $\perp$ be $F(2r-3,-r+2,r+7)$

$\therefore (2r-3-3),(-r+2-0),(r+7-3)$ is $\perp$ to $(2,-1,1)$

$\therefore (2r-6)2+(-r+2)(-1)+(r+7-3)1=0\Rightarrow r=\frac{5}{3}\therefore F(\frac{1}{3},\frac{1}{3},\frac{26}{3})$

$\therefore$ distance $=\sqrt{{(\frac{1}{3}-3)}^2+{\left(\frac{1}{3}\right)}^2+{(\frac{26}{3}-3)}^2}=\frac{\sqrt{354}}{3}$