Question

Find the distance between the lines $l_1$ and $l_2$ given by

$\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$ and $\overrightarrow{r}=3\hat{i}+3\hat{j}-5\hat{k}+\mu (2\hat{i}+3\hat{j}+6\hat{k})$

Answer 1

$\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$

$\overrightarrow{r}=3\hat{i}+3\hat{j}-5\hat{k}+\mu (2\hat{i}+3\hat{j}+6\hat{k})$

These two lines pass through the points having position vectors

$\overrightarrow{a_1}=\hat{i}+2\hat{j}-4\hat{k}$ and $\overrightarrow{a_2}=3\hat{i}+3\hat{j}-5\hat{k}$ are parallel to the vector $\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}$

Now,$\overrightarrow{a_2}-\overrightarrow{a_1}=3\hat{i}+3\hat{j}-5\hat{k}-\hat{i}-2\hat{j}+4\hat{k}=2\hat{i}+\hat{j}-\hat{k}$

and $(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}=(2\hat{i}+\hat{j}-\hat{k})\times (2\hat{i}+3\hat{j}+6\hat{k})$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 2& 3& 6\end{array}\right|$

$=\hat{i}(6+3)-\hat{j}(12+2)+\hat{k}(6-2)$

$=9\hat{i}-14\hat{j}+4\hat{k}$

$|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|=\sqrt{9^2+{(-14)}^2+4^2}=\sqrt{81+196+16}=\sqrt{293}$

and $\left|\overrightarrow{b}\right|=\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=7$

The shortest distance between the two lines is given by $\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{293}}{7}$