If $A=(1,-2,-1),B=(4,0,-3),C=(1,2,-1)$ $D=(2,-4,-5)$ find the distance between $AB$and $CD$lines.

Given points are $A=(1,-2,-1),B=(4,0,-3)$

$C=(1,2,-1),D=(2,-4,-5)$

Now $\overline{OA}=\overline{i}-2\overline{j}-\overline{k},\overline{OB}=4\overline{i}-3\overline{k}$

$\overline{OC}=\overline{i}+2\overline{j}-\overline{k},\overline{OD}=2\overline{i}-4\overline{j}-5\overline{k}$

$\therefore \overline{AB}=\overline{OB}-\overline{OA}=4\overline{i}-3\overline{k}-(\overline{i}-2\overline{j}-\overline{k})$

$=3\overline{i}+2\overline{j}-2\overline{k}$

AB line $\overline{r}=(1,-2,-1)+t(3,2,-2)\dots \left(1\right)$

$\overline{CD}=\overline{OD}-\overline{OC}=2\overline{i}-4\overline{j}-5\overline{k}-(\overline{i}+2\overline{j}-\overline{k})$

$CD$line $\overline{r}=(1,2,-1)+s(1,-6,-4)......\left(2\right)$

Let $\overline{OA}=\overline{a}=\overline{i}-2\overline{j}-\overline{k},\overline{AB}=\overline{b}=3\overline{i}+2\overline{j}-2\overline{k}$

$\overline{OC}=\overline{c}=\overline{i}+2\overline{j}-\overline{k},\overline{CD}=\overline{d}=\overline{i}-6\overline{j}-4\overline{k}$

$\overline{a}-\overline{c}=(\overline{i}-2\overline{j}-\overline{k})-(\overline{i}+2\overline{j}-\overline{k})=-4\overline{j}$

$\overline{b}\times \overline{d}=\left|\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ 3& 2& -2\\ 1& -6& -4\end{array}\right|$

$=\overline{i}[-8-(12\left)\right]-\overline{j}[-12-(-2\left)\right]+\overline{k}[-18-(2\left)\right]$

$=-20\overline{i}+10\overline{j}-20\overline{k}$

$|\overline{b}\times \overline{d}|=\sqrt{(-20{)}^2+(10{)}^2+(-20{)}^2}$

$=\sqrt{400+100+400}=\sqrt{900}=30$

$[ \overline{a}-\overline{c} \overline{ b} \overline{d} ]=\left|\begin{array}{ccc}0& -4& 0\\ 3& 2& -2\\ 1& -6& -4\end{array}\right|$

$=0[-8-(12\left)\right]+4[-12-(-2\left)\right]+0[-18-2]$

$=4(-10)=-40$

$\therefore$ The shortest distance between $AB$and $CD$

$=\frac{\left|\left[\overline{a}-\overline{c} \overline{b} \overline{d }\right]\right|}{|\overline{\overline{b}}\times \overline{d}|}=\frac{40}{30}=\frac{4}{3}\text{ units }$