Question

The shortest distance between the parallel lines:

$\frac{x-3}{4}=\frac{y-1}{2}=\frac{z-5}{-3}$ and $\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-3}{-3}$ is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $3$

A point on the first line can be written in the parametric form as $(4t+3,2t+1,-3t+5)$, while $(1,2,3)$ lies on the second line.

Connecting a line between these two points, its direction would be $(4t+2,2t-1,-3t+2)$

This has to be perpendicular to the line direction, which is $(4,2,-3)$

$\therefore (4t+2)\times 4+(2t-1)\times 2+(-3t+2)\times -3=0$

$\therefore 16t+8+4t-2+9t-6=0$

$\therefore 29t=0$ or $t=0$

The point on the first line thus becomes $(3,1,5)$

Thus, distance between the points $(3,1,5)$ and $(1,2,3)$ is the shortest distance between the lines, which is

$\sqrt{(3-1{)}^2+(1-2{)}^2+(5-3{)}^2}=\sqrt{4+1+4}=3$