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Find the Shortest Distance Between the Following Pairs of Lines Whose Cartesian Equations Are: X − 3 1 = Y − 5 − 2 = Z − 7 1 a N D X + 1 7 = Y + 1 − 6 = Z + 1 1 - Mathematics

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Question

Find the shortest distance between the following pairs of lines whose cartesian equations are:  \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]

Sum
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Solution

\[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} . . . (1) \]

\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} . . . (2)\] 

Since line (1) passes through the point (3, 5, 7) and has direction ratios proportional to 1, -2,1 its vector equation is 

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

\[Here, \]

\[ \overrightarrow{a_1} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \]

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

Also, line (2) passes through the point ( -1,-1,-1) and has direction ratios proportional to 7,-6,1 Its vector equation is

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

\[Here, \]

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

\[ \overrightarrow{b_2} = 7 \hat{i} - 6 \hat{j}  + \hat{k} \]

Now,

\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \]

\[\text{ and }\overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 7 & - 6 & 1\end{vmatrix}\]

\[ = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \]

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{4^2 + 6^2 + 8^2}\]

\[ = \sqrt{16 + 36 + 64}\]

\[ = \sqrt{116}\]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \vec{b_1} \times \overrightarrow{b_2} \right) = \left( - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \right) . \left( 4 \hat{i} + 6 \hat{j} + 8 \hat{k } \right)\]

\[ = - 16 - 36 - 64\]

\[ = - 116\]

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 

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

\[ = \left| \frac{- 116}{\sqrt{116}} \right|\]

\[ = \sqrt{116}\]

\[ = 2\sqrt{29}\]

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Chapter 28: Straight Line in Space - Exercise 28.5 [Page 38]

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RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.5 | Q 2.4 | Page 38

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