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Find the Shortest Distance Between the Following Pairs of Lines Whose Vector Are: → R = ( 1 − T ) ^ I + ( T − 2 ) ^ J + ( 3 − T ) ^ K and → R = ( S + 1 ) ^ I + ( 2 S − 1 ) ^ J − ( 2 S + 1 ) ^ K

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प्रश्न

Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k}  \text{ and }  \overrightarrow{r} = \left( s + 1 \right) \hat{i}  + \left( 2s - 1 \right) \hat{j}  - \left( 2s + 1 \right) \hat{k} \]

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उत्तर

\[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k}  \text{ and }  \vec{r} = \left( s + 1 \right) \hat{i}  + \left( 2s - 1 \right) \hat{j}  - \left( 2s + 1 \right) \hat{k} \]

The vector equations of the given lines can be re-written as

\[\overrightarrow{r} = \hat{i}  - 2 \hat{j} + 3 \hat{k} + t\left( - \hat{i} + \hat{j} - \hat{k}\right) \text{ and }  \overrightarrow{r} = \hat{i}  - \hat{j} - \hat{k}  + s\left( \hat{i}  + 2 \hat{j}  - 2 \hat{k}  \right)\]

Comparing the given equations with the equations  \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]

we get , 

\[\overrightarrow{a_1} = \hat{i} - 2 \hat{j} + 3 \hat{k} \]

\[ \overrightarrow{a_2} = \hat{i} - \hat{j} - \hat{k} \]

\[ \overrightarrow{b_1} = - \hat{i} + \hat{j} - \hat{k} \]

\[ \overrightarrow{b_2} = \hat{i} + 2 \hat{j} - 2 \hat{k} \]

\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \hat{j} - 4 \hat{k} \]

\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j} & \hat{k}  \\ - 1 & 1 & - 1 \\ 1 & 2 & - 2\end{vmatrix}\]

\[ = - 3 \hat{j} - 3 \hat{k}  \]

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 3 \right)^2 + \left( - 3 \right)^2}\]

\[ = \sqrt{9 + 9}\]

\[ = 3\sqrt{2}\]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{j} - 4 \hat{k}  \right) . \left( - 3 \hat{j} - 3 \hat{k} \right)\]

\[ = - 3 + 12\]

\[ = 9\]

The shortest distance between the line 

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

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

\[ = \left| \frac{9}{3\sqrt{2}} \right|\]

\[ = \frac{3}{\sqrt{2}}\]

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अध्याय 27: Straight Line in Space - Exercise 28.5 [पृष्ठ ३७]

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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 1.4 | पृष्ठ ३७

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