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Find the Shortest Distance Between the Lines → R = 6 ^ I + 2 ^ J + 2 ^ K + λ ( ^ I − 2 ^ J + 2 ^ K ) and → R = − 4 ^ I − ^ K + μ ( 3 ^ I − 2 ^ J − 2 ^ K )

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प्रश्न

Find the shortest distance between the lines \[\overrightarrow{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) \text{ and }  \overrightarrow{r} = - 4 \hat{i}  - \hat{k}  + \mu\left( 3 \hat{i} - 2 \hat{j} - 2 \hat{k}  \right)\]

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उत्तर

\[\overrightarrow{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) \text{ and }  \overrightarrow{r} = - 4 \hat{i}  - \hat{k}  + \mu\left( 3 \hat{i} - 2 \hat{j} - 2 \hat{k}  \right)\]

Comparing the given equations with the equations

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \vec{b_2}\]

we get , 

\[\overrightarrow{a_1} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{a_2} = - 4 \hat{i} - \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - 2 \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} - 2 \hat{j} - 2 \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = - 10 \hat{i} - 2 \hat{j}  - 3 \hat{k} \]
\[\text { and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2\end{vmatrix}\]
\[ = 8 \hat{i}  + 8 \hat{j}  + 4 \hat{k}  \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{8^2 + 8^2 + 4^2}\]
\[ = \sqrt{64 + 64 + 16}\]
\[ = \sqrt{144}\]
\[ = 12\]
\[\text { and }  \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 10 \hat{i} - 2 \hat{j} - 3 \hat{k} \right) . \left( 8 \hat{i} + 8 \hat{j} + 4 \hat{k}  \right)\]
\[ = - 80 - 16 - 12\]
\[ = - 108\]

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 

\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{- 108}{12} \right|\]
\[ = 9\]

 

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अध्याय 27: Straight Line in Space - Exercise 28.5 [पृष्ठ ३८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
Exercise 28.5 | Q 7.4 | पृष्ठ ३८

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