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Find the Shortest Distance Between Following Pairs of Lines Whose Vector Are: → R =( ^ I + 2 ^ J + 3 ^ K ) + λ ( 2 ^ I + 3 ^ J + 4 ^ K ) And→ R = ( 2 ^ I + 4 ^ J + 5 ^ K ) + μ ( 3 ^ I + 4 ^ J + 5 ^ K

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Question

Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j}  + 4 \hat{k}  \right) \text{ and }  \overrightarrow{r} = \left( 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \right) + \mu\left( 3 \hat{i}  + 4 \hat{j}  + 5 \hat{k} \right)\]

Sum
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Solution

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

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 , 

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

\[ \overrightarrow{a_2} = 2 \hat{i} + 4 \hat{j} + 5 \hat{k}  \]

\[ \overrightarrow{b_1} = 2 \hat{i}  + 3 \hat{j}  + 4 \hat{k} \]

\[ \overrightarrow{b_2} = 3 \hat{i}  + 4 \hat{j}  + 5 \hat{k}  \]

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

\[\text{and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{vmatrix}\]

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

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 1 \right)^2 + 2^2 + \left( - 1 \right)^2}\]

\[ = \sqrt{1 + 4 + 1}\]

\[ = \sqrt{6}\]

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

\[ = - 1 + 4 - 2\]

\[ = 1\]

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 

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

\[ = \left| \frac{1}{\sqrt{6}} \right|\]

\[ = \frac{1}{\sqrt{6}}\]

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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 1.3 | Page 37

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