Question

Find the shortest distance between the following pairs of parallel lines whose equations are:
(i) $\overrightarrow{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)\mathrm{and}\overrightarrow{r}=\left(2\hat{i}-\hat{j}-\hat{k}\right)+\mu \left(-\hat{i}+\hat{j}-\hat{k}\right)$

(ii) $\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)+\lambda \left(2\hat{i}-\hat{j}+\hat{k}\right)\mathrm{and}\overrightarrow{r}=\left(2\hat{i}+\hat{j}-\hat{k}\right)+\mu \left(4\hat{i}-2\hat{j}+2\hat{k}\right)$

Answer 1

(i) The vector equations of the given lines are

$$\begin{gathered} \overrightarrow{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)...\left(1\right) \\ \overrightarrow{r}=\left(2\hat{i}-\hat{j}-\hat{k}\right)+\mu \left(-\hat{i}+\hat{j}-\hat{k}\right) \\ =\left(2\hat{i}-\hat{j}-\hat{k}\right)-\mu \left(\hat{i}-\hat{j}+\hat{k}\right)...\left(2\right) \end{gathered}$$

These two lines pass through the points having position vectors $\overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k}\mathrm{and}\overrightarrow{a_2}=2\hat{i}-\hat{j}-\hat{k}$ and are parallel to the vector $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$.

Now,
$\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-3\hat{j}-4\hat{k}$
and
$$\begin{gathered} \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}=\left(\hat{i}-3\hat{j}-4\hat{k}\right)\times \left(\hat{i}-\hat{j}+\hat{k}\right) \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& -4\\ 1& -1& 1\end{array}\right| \\ =-7\hat{i}-5\hat{j}+2\hat{k} \\ \Rightarrow \left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}\right|=\sqrt{{\left(-7\right)}^2+{\left(-5\right)}^2+2^2} \\ =\sqrt{49+25+4} \\ =\sqrt{78} \end{gathered}$$

The shortest distance between the two lines is given by

$\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}\right|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$

(ii) $\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)+\lambda \left(2\hat{i}-\hat{j}+\hat{k}\right)\mathrm{and}\overrightarrow{r}=\left(2\hat{i}+\hat{j}-\hat{k}\right)+\mu \left(4\hat{i}-2\hat{j}+2\hat{k}\right)\mathrm{or}\overrightarrow{r}=\left(2\hat{i}+\hat{j}-\hat{k}\right)+2\mu \left(2\hat{i}-\hat{j}+\hat{k}\right)$

These two lines pass through the points having position vectors $\overrightarrow{a_1}=\hat{i}+\hat{j}\mathrm{and}\overrightarrow{a_2}=2\hat{i}+\hat{j}-\hat{k}$ and are parallel to the vector $\overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}$.

Now,
$\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{k}$
and
$$\begin{gathered} \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}=\left(\hat{i}-\hat{k}\right)\times \left(2\hat{i}-\hat{j}+\hat{k}\right) \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& -1\\ 2& -1& 1\end{array}\right| \\ =-\hat{i}-3\hat{j}-\hat{k} \\ \Rightarrow \left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}\right|=\sqrt{{\left(-1\right)}^2+{\left(-3\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{1+9+1} \\ =\sqrt{11} \end{gathered}$$

The shortest distance between the two lines is given by

$\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}\right|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{11}}{\sqrt{6}}$