# The shortest distance between the lines

Question:

The shortest distance between the lines

$\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z+1=0$

$2 x-y+z+3=0$ is :

1. $\frac{1}{2}$

2. 1

3. $\frac{1}{\sqrt{2}}$

4. $\frac{1}{\sqrt{3}}$

Correct Option: , 4

Solution:

Line of intersection of planes

$x+y+z+1=0$ ........(1)

$2 x-y+z+3=0$ ............(2)

eliminate $y$

$3 x+2 z+4=0$

$x=\frac{-2 z-4}{3}$    .........(3)

put in equaiton (1)

$z=-3 y+1$    ..............(4)

from $(3)$ and $(4)$

$\frac{3 x+4}{-2}=-3 y+1=z$

$\frac{x-\left(-\frac{4}{3}\right)}{-\frac{2}{3}}=\frac{y-\frac{1}{3}}{-\frac{1}{3}}=\frac{z-0}{1}$

now shortest distance between skew lines

$\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$

$\frac{x-\left(-\frac{4}{3}\right)}{-\frac{2}{3}}=\frac{y-\left(\frac{1}{3}\right)}{-\frac{1}{3}}=\frac{z-0}{1}$

S.D. $=\left|\frac{(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})}{|\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}|}\right|$

where $\overrightarrow{\mathrm{a}}=(1,-1,0)$

$\overrightarrow{\mathrm{b}}=\left(-\frac{4}{3}, \frac{1}{3}, 0\right)$

$\overrightarrow{\mathrm{c}}=(0,-1,1)$

$\overrightarrow{\mathrm{d}}=\left(-\frac{2}{3},-\frac{1}{3}, 1\right)$

$\Rightarrow$ S.D $=\frac{1}{\sqrt{3}}$