Question

Find the shortest distance between the lines whose vector equations are :
$\overrightarrow{r}=(1-p)\hat{i}+(p-2)\hat{j}+(3-2p)\hat{k}$
$\overrightarrow{r}=(q+1)\hat{i}+(2q-1)\hat{j}-(2q-1)\hat{k}$

Answer 1

The given lines are
$\overrightarrow{r}=(1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}$
$\Rightarrow$ $\overrightarrow{r}=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-2\hat{k})......\left(1\right)$
$\overrightarrow{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}$
$\Rightarrow$ $\overrightarrow{r}=(\hat{i}-\hat{j}+3\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k}).......\left(2\right)$
It is known that the shortest distance between the lines,
$\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$, is given by
$d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2})(\overrightarrow{a_1}-\overrightarrow{a_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|.....\left(1\right)$
For the given equations,
$\overrightarrow{a_1}=\hat{i}-2\hat{j}+3\hat{k}$
$\overrightarrow{b_2}=-\hat{i}+\hat{j}-2\hat{k}$
$\overrightarrow{a_2}=\hat{i}-\hat{j}-\hat{k}$
$\overrightarrow{b_2}=\hat{i}+2\hat{j}-2\hat{k}$
$\overrightarrow{a_2}-\overrightarrow{a_1}=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})=\hat{j}-4\hat{k}$
$(\overrightarrow{b_1}\times \overrightarrow{b_2})(\overrightarrow{a_1}-\overrightarrow{a_2})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& -2\\ 1& 2& -2\end{array}\right|$
$=(-2+4)\hat{i}-(2+2)\hat{j}+(2-1)\hat{k}$
$\Rightarrow$ $|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{{\left(2\right)}^2+{(-4)}^2+{(-3)}^2}=\sqrt{4+16+9}=\sqrt{29}$
$\therefore$ $(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_1}-\overrightarrow{a_2})=(2\hat{i}-4\hat{j}-3\hat{k}).(\hat{j}-4\hat{k})=-4+12=8$
Substituting all the values in equation (3), we obtain,
$d=\left|\frac{8}{\sqrt{29}}\right|=\frac{8}{\sqrt{29}}$
Therefore, the shortest distance between the lines is $\frac{8}{\sqrt{29}}$ units.