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Find the Shortest Distance Between the Following Pairs of Lines Whose Cartesian Equations Are: X − 1 2 = Y − 2 3 = Z − 3 4 a N D X − 2 3 = Y − 3 4 = Z − 5 5

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प्रश्न

Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\] 

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उत्तर

The equations of the given lines are

\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]     ...........(1)

\[\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\]     .............(2)

Since line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is  \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]

\[Here, \]

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

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

Also, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5.
Its vector equation is \[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]

\[\text{ Here } , \]

\[ \overrightarrow{a_2} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k} \]

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

Now, 

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

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

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

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

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

\[ = \sqrt{6}\]

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

\[ = - 1 + 2 - 2\]

\[ = - 1\]

The shortest distance between the lines  \[\vec{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and  } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] 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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अध्याय 27: Straight Line in Space - Exercise 28.5 [पृष्ठ ३८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
Exercise 28.5 | Q 2.1 | पृष्ठ ३८

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