Question

Write the vector equations of the following lines and hence determine the distance between them  \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]

Answer 1

We have 

\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}\]

\[\frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]

Since the first line passes through the point (1, 2, -4) and has direction ratios proportional to 2, 3, 6, its vector equation is

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]

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

Also, the second line passes through the point (3, 3, -5) and has direction ratios proportional to 4, 6, 12.
Its vector equation is

\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} . . . (2) \]

\[ \Rightarrow \overrightarrow{r} = 3 \hat{i}  + 3 \hat{j} - 5 \hat{k} + \mu\left( 4 \hat{i} + 6 \hat{j} + 12 \hat{k} \right)\]

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

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 

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