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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)
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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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