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Question
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(1, 3, 0) and (0, 3, 0)
Solution
The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is
\[\frac{x - 1}{0 - 1} = \frac{y - 3}{3 - 3} = \frac{z - 0}{0 - 0}\]
\[ = \frac{x - 1}{- 1} = \frac{y - 3}{0} = \frac{z}{0}\]
Since the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]
\[\text{ Here }, \]
\[ \overrightarrow{a_1} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} + 0 \hat{j}+ 2 \hat{k} \]
Also, the second line passes through the point (1, 3, 0) and has direction ratios proportional to -1, 0 , 0 .
Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} . . . (2) \]
\[\text{ Here }, \]
\[ \overrightarrow{a_2} = \hat{i} + 3 \hat{j} + 0 \hat{k} \]
\[ \overrightarrow{b_2} = - \hat{i} + 0 \hat{j} + 0 \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} + 3 \hat{j} + 0 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ - 1 & 0 & 0\end{vmatrix} = 0 \hat{i} - 2 \hat{j} + 0 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{0^2 + \left( - 2 \right)^2 + 0^2}\]
\[ = \sqrt{0 + 4 + 0}\]
\[ = 2\]
\[\text{ and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{i} + 3 \hat{j} + 0 \hat{k} \right) . \left( 0 \hat{i} - 2 \hat{j} + 0 \hat{k} \right)\]
\[ = - 6\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{- 6}{2} \right|\]
\[ = 3 \]
\[ \therefore d = \text{ 3 } units\]
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