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Find the Shortest Distance Between the Following Pairs of Lines Whose Cartesian Equations Are : X − 1 2 = Y + 1 3 = Z and X + 1 3 = Y − 2 1 ; Z = 2

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Question

Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]

Sum
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Solution

The equations of the given lines are 

\[\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 0}{1} . . . (1) \]

\[\frac{x + 1}{3} = \frac{y - 2}{1} = \frac{z - 2}{0} . . . (2)\]

Since line (1) passes through the point (1,-1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]

\[Here, \]

\[ \overrightarrow{a_1} = \hat{i}  - \hat{j}  + 0 \hat{k}  \]

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


Also, line (2) passes through the point ( -1, 2, 2) and has direction ratios proportional to 3, 1, 0. Its vector equation is

\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]

\[Here, \]

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

\[ \overrightarrow{b_2} = 3 \hat{i}  + \hat{j}  + 0 \hat{k} \]

Now, 

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

\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j} & \hat{k}  \\ 2 & 3 & 1 \\ 3 & 1 & 0\end{vmatrix}\]

\[ = - \hat{i} + 3 \hat{j} - 7 \hat{k} \]

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 1 \right)^2 + 3^2 + \left( - 7 \right)^2}\]

\[ = \sqrt{1 + 9 + 49}\]

\[ = \sqrt{59}\]

\[\text{ and }  \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \right) . \left( - \hat{i} + 3 \hat{j} - 7 \hat{k}  \right)\]

\[ = 2 + 9 - 14\]

\[ = - 3\] 

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 

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

\[ = \left| \frac{- 3}{\sqrt{59}} \right|\]

\[ = \frac{3}{\sqrt{59}}\] 

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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 2.2 | Page 38

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