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Find the Shortest Distance Between the Lines X + 1 7 = Y + 1 − 6 = Z + 1 1 a N D X − 3 1 = Y − 5 − 2 = Z − 7 1

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Question

Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and }  \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]

Sum
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Solution

\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]

Since the first line passes through the point (-1,-1,-1) and has direction ratios proportional to 7, -6 , 1 , its vector equation is

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

\[\text{ Here }, \]

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

\[ \overrightarrow{b_1} = 7 \hat{i} - 6 \hat{j} + \hat{k} \]

Also, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1, -2,1.Its vector equation is

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

\[\text{ Here }, \]

\[ \overrightarrow{a_2} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \]

\[ \overrightarrow{b_2} = \hat{i} - 2 \hat{j} + \hat{k} \]

Now, 

\[\overrightarrow{a_2} - \overrightarrow{a_1} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \]
\[\text{ and  } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}  \\ 7 & - 6 & 1 \\ 1 & - 2 & 1\end{vmatrix}\]
\[ = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 4 \right)^2 + \left( - 6 \right)^2 + \left( - 8 \right)^2}\]
\[ = \sqrt{16 + 36 + 64}\]
\[ = \sqrt{116}\]
\[\text{ and }  \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 4 \hat{i}  + 6 \hat{j} + 8 \hat{k} \right) . \left( - 4 \hat{i} - 6 \hat{j} - 8 \hat{k}  \right)\]
\[ = - 16 - 36 - 64\]
\[ = - 116\]

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 

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

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

\[ = \sqrt{116}\]

\[ = 2\sqrt{29}\]

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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 7.2 | Page 38

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