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प्रश्न
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \lambda - 1 \right) \hat{i} + \left( \lambda + 1 \right) \hat{j} - \left( 1 + \lambda \right) \hat{k} \text{ and } \overrightarrow{r} = \left( 1 - \mu \right) \hat{i} + \left( 2\mu - 1 \right) \hat{j} + \left( \mu + 2 \right) \hat{k} \]
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उत्तर
\[\overrightarrow{r} = \left( \lambda - 1 \right) \hat{i} + \left( \lambda + 1 \right) \hat{j} - \left( 1 + \lambda \right) \hat{k} \text{ and } \overrightarrow{r} = \left( 1 - \mu \right) \hat{i} + \left( 2\mu - 1 \right) \hat{j} + \left( \mu + 2 \right) \hat{k} \]
The vector equations of the given lines can be re-written as
\[\overrightarrow{r} = - \hat{i} + \hat{j} - \hat{k} + \lambda\left( \hat{i} + \hat{j} - \hat{k} \right) \text{ and } \overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} + \mu\left( - \hat{i} + 2 \hat{j} + \hat{k} \right)\]
Comparing the given equations with the equations
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]
we get ,
\[\overrightarrow{a_1} = - \hat{i} + \hat{j} - \hat{k} \]
\[ \overrightarrow{a_2} = \hat{i} - \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} + \hat{j} - \hat{k} \]
\[ \overrightarrow{b_2} = - \hat{i} + 2 \hat{j} + \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = 2 \hat{i} - 2 \hat{j} + 3 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & - 1 \\ - 1 & 2 & 1\end{vmatrix}\]
\[ = 3 \hat{i} + 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{3^2 + 3^2}\]
\[ = \sqrt{9 + 9}\]
\[ = 3\sqrt{2}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 2 \hat{i} - 2 \hat{j} + 3 \hat{k} \right) . \left( 3 \hat{i} + 3 \hat{k} \right)\]
\[ = 6 + 9\]
\[ = 15\]
The shortest distance between the lines ,
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{15}{3\sqrt{2}} \right|\]
\[ = \frac{5}{\sqrt{2}}\]
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