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प्रश्न
Find the shortest distance between the lines \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
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उत्तर
\[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
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 ,
\[\overrightarrow{a_1} = \hat{i} + 2 \hat{j} + \hat{k} \]
\[ \overrightarrow{a_2} = 2 \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - \hat{j} + \hat{k} \]
\[ \overrightarrow{b_2} = 2 \hat{i} + \hat{j} + 2 \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} - 3 \hat{j} - 2 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 2 & 1 & 2\end{vmatrix}\]
\[ = - 3 \hat{i} + 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 3 \right)^2 + 3^2}\]
\[ = \sqrt{9 + 9}\]
\[ = 3\sqrt{2}\]
\[\text{ and }\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{i} - 3 \hat{j} - 2 \hat{k} \right) . \left( - 3 \hat{i} + 3 \hat{k} \right)\]
\[ = - 3 - 6\]
\[ = - 9\]
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
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{- 9}{3\sqrt{2}} \right|\]
\[ = \frac{3}{\sqrt{2}}\]
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