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Find the Shortest Distance Between the Lines → R = ^ I + 2 ^ J + 3 ^ K + λ ( ^ I − 3 ^ J + 2 ^ K ) and → R = 4 ^ I + 5 ^ J + 6 ^ K + μ ( 2 ^ I + 3 ^ J + ^ K )

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Question

Find the shortest distance between the lines \[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and }  \overrightarrow{r} = 4 \hat{i} + 5 \hat{j}  + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]

Sum
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Solution

\[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and }  \overrightarrow{r} = 4 \hat{i} + 5 \hat{j}  + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]

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 ,

\[\overrightarrow{a_1} = \hat{i}  + 2 \hat{j}  + 3 \hat{k} \]
\[ \overrightarrow{a_2} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - 3 \hat{j} + 2 \hat{k} \]
\[ \vec{b_2} = 2 \hat{i} + 3 \hat{j} + \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = 3 \hat{i} + 3 \hat{j}  + 3 \hat{k}  \]
\[\text{ and  } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 1 & - 3 & 2 \\ 2 & 3 & 1\end{vmatrix}\]
\[ = - 9 \hat{i}  + 3 \hat{j}  + 9 \hat{k}  \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 9 \right)^2 + 3^2 + 9^2}\]
\[ = \sqrt{81 + 9 + 81}\]
\[ = \sqrt{171}\]
\[\text{ and }  \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \right) . \left( - 9 \hat{i} + 3 \hat{j} + 9 \hat{k}  \right)\]
\[ = - 27 + 9 + 27\]
\[ = 9\]

The shortest distance between the lines 

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

\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{9}{\sqrt{171}} \right|\]
\[ = \frac{3}{\sqrt{19}}\]

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Chapter 27: Straight Line in Space - Exercise 28.5 [Page 38]

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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 7.3 | Page 38

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