Question

The distance between the lines $\frac{x+8}{4}=\frac{4y+24}{8}=\frac{z-2}{6}$ and $\frac{x+6}{2}=\frac{y+10}{1}=\frac{z-6}{3}$ is

Answer 1

We can see that dr's are proportional, so the lines are parallel.
Any point on the line ,
$\frac{x+8}{4}=\frac{4y+24}{8}=\frac{z-2}{6}$ is of the form
$(4\lambda -8,2\lambda -6,6\lambda +2)$
Draw a perpendicular line from $(-6,-10,6)$ to the line $\frac{x+8}{4}=\frac{4y+24}{8}=\frac{z-2}{6}$
Then $4(4\lambda -8+6)+8(2\lambda -6+10)+6(6\lambda +2-6)=0$
$\Rightarrow \lambda =0$
$\therefore$ foot of the perpendicular from
$(-6,-10,6)$ is $(-8,-6,2)$
So the distance is,
$=\sqrt{2^2+4^2+4^2}=6$