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Find the Shortest Distance Between the Following Pairs of Lines Whose Vector Are: → R = ( 2 ^ I − ^ J − ^ K ) + λ ( 2 ^ I − 5 ^ J + 2 ^ K ) and , → R = ( ^ I + 2 ^ J + ^ K ) + μ ( ^ I − ^ J + ^ K )

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Question

Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k}  \right) + \lambda\left( 2 \hat{i}  - 5 \hat{j} + 2 \hat{k}  \right) \text{ and }, \overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - \hat{j}  + \hat{k}  \right)\]

Sum
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Solution

\[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k}  \right) + \lambda\left( 2 \hat{i}  - 5 \hat{j} + 2 \hat{k}  \right) \text{ and }, \overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - \hat{j}  + \hat{k}  \right)\] 

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 , 

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

\[ \overrightarrow{a_2} = \hat{i} + 2 \hat{j} + \hat{k} \]

\[ \overrightarrow{b_1} = 2 \hat{i}  - 5 \hat{j}  + 2 \hat{k}  \]

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

\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = - \hat{i} + 3 \hat{j} + 2 \hat{k} \]

\[\text{ and }  \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j} & \hat{k}  \\ 2 & - 5 & 2 \\ 1 & - 1 & 1\end{vmatrix}\]

\[ = - 3 \hat{i} + 3 \hat{k} \]

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

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

\[ = 3\sqrt{2}\]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \vec{b_1} \times \overrightarrow{b_2} \right) = \left( - \hat{i} + 3 \hat{j} + 2 \hat{k} \right) . \left( - 3 \hat{i}  + 3 \hat{k} \right)\]

\[ = 3 + 6\]

\[ = 9\]

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 

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

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

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

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Chapter 27: Straight Line in Space - Exercise 28.5 [Page 37]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Exercise 28.5 | Q 1.6 | Page 37

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