Question

Find the shortest distance between line $l_1$ and $l_2$ whose vector equations are
$\overrightarrow{r}=\hat{i}+\hat{j}+\mu (2\hat{i}-\hat{j}+\hat{k})$ ................ (i)
and $\overrightarrow{r}=2\hat{i}+\hat{i}-\hat{k}+\mu (3\hat{i}-5\hat{j}+2\hat{k})$ ............. (ii)

• A. $\frac{20}{\sqrt{59}}$
• B. $\frac{10}{\sqrt{59}}$
  
• C. $\frac{59}{\sqrt{10}}$
• D. $\frac{59}{\sqrt{20}}$

Answer 1

The correct option is B $\frac{10}{\sqrt{59}}$

Comparing (i) and (ii) with $\overrightarrow{r}=\overrightarrow{a}+t\overrightarrow{b}$ we get,
$\overrightarrow{a_1}=\hat{i}+\hat{j},\overrightarrow{b_1}=2\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{a_2}=2\hat{i}+\hat{j}-\hat{k},\overrightarrow{b_2}=3\hat{i}-5\hat{j}+2\hat{k}$
Therefore $\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{k}$
and $\overrightarrow{b_1}\times \overrightarrow{b_2}=(2\hat{i}-\hat{j}+\hat{k})\times (3\hat{i}-5\hat{j}+2\hat{k})$
$\hat{i}\hat{j}\hat{k}$
$=|2-11|=3\hat{i}-\hat{j}-7\hat{k}$
$3-52$
So $|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{9+1+49}=\sqrt{59}$
Hence, the shortest distance between the given lines is given by
$d=|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2})\cdot (\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}|=\frac{|3-0+7|}{\sqrt{59}}=\frac{10}{\sqrt{59}}$