Question

Find the distance between the lines l₁ and l₂ given by  \[\overrightarrow{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k}  \right) \text{ and } , \overrightarrow{r} = 3 \hat{i} + 3 \hat{j}  - 5 \hat{k}  + \mu\left( 2 \hat{i} + 3 \hat{j}  + 6 \hat{k}  \right)\]

 

 

Answer 1

\[\overrightarrow{r} = \hat{i} + 2 \hat{j}  - 4 \hat{k} + \lambda\left( 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k}  \right) \]
\[ \overrightarrow{r} = 3 \hat{i}  + 3 \hat{j}  - 5 \hat{k}  + \mu\left( 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k}  \right)\] 

These two lines pass through the points having position vectors

\[\overrightarrow{a_1} = \hat{i} + 2 \hat{j} - 4 \hat{k}  \text{ and }  \overrightarrow{a_2} = 3 \hat{i}+ 3 \hat{j} - 5 \hat{k} \]  and are parallel to the vector 

\[\overrightarrow{b} = 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k} \]

Now, 

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

and

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) \times \left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k}  \right)\]
\[ = \begin{vmatrix}\hat{i}  & \hat{j} & \hat{k}  \\ 2 & 1 & - 1 \\ 2 & 3 & 6\end{vmatrix}\]
\[ = 9 \hat{i} - 14 \hat{j} + 4 \hat{k} \]
\[ \Rightarrow \left| \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} \right| = \sqrt{9^2 + \left( - 14 \right)^2 + 4^2}\]
\[ = \sqrt{81 + 196 + 16}\]
\[ = \sqrt{293}\]
\[\text{ and } \left| \overrightarrow{b} \right| = \sqrt{2^2 + 3^2 + 6^2}\]
\[ = \sqrt{4 + 9 + 36}\]
\[ = 7\]
The shortest distance between the two lines is given by

\[\frac{\left| \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} \right|}{\left| \overrightarrow{b} \right|} = \frac{\sqrt{293}}{7}\]