Question

Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \right)\]

Answer 1

\[\overrightarrow{r} = \left( \hat{i} + \hat{j}  \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{k}  \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \right) o \overrightarrow{r} = \left( 2 \hat{i}  + \hat{j} - \hat{k}  \right) + 2\mu\left( 2 \hat{i} - \hat{j}+ \hat{k} \right)\]

These two lines pass through the points having position vectors

\[\overrightarrow{a_1} = \hat{i} + \hat{j}  \text{ and }  \overrightarrow{a_2} = 2 \hat{i} + \hat{j} - \hat{k} \]  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

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} = \left( \hat{i} - \hat{k} \right) \times \left( 2 \hat{i} - \hat{j} + \hat{k} \right)\]

\[ = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k} \\ 1 & 0 & - 1 \\ 2 & - 1 & 1\end{vmatrix}\]

\[ = - \hat{i} - 3 \hat{j} - \hat{k}  \]

\[ \Rightarrow \left| \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} \right| = \sqrt{\left( - 1 \right)^2 + \left( - 3 \right)^2 + \left( - 1 \right)^2}\]

\[ = \sqrt{1 + 9 + 1}\]

\[ = \sqrt{11}\]

The shortest distance between the two lines is given by

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