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Question
Write the vector equations of the following lines and hence determine the distance between them \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]
Solution
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
\[\frac{\left| \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} \right|}{\left| \overrightarrow{b} \right|} = \frac{\sqrt{293}}{7} \text{ units } \]
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