मराठी

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

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प्रश्न

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

बेरीज
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उत्तर

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

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

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

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

\[Here, \]

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

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

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

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

\[Here, \]

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

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

Now, 

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

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

\[ = 2 \hat{i} - 4 \hat{j} - 3 \hat{k} \]

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{2^2 + \left( - 4 \right)^2 + \left( - 3 \right)^2}\]

\[ = \sqrt{4 + 16 + 9}\]

\[ = \sqrt{29}\]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \vec{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{j} - 4 \hat{k} \right) . \left( 2 \hat{i} - 4 \hat{j} - 3 \hat{k} \right)\]

\[ = - 4 + 12\]

\[ = 8\]

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

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

\[ = \left| \frac{8}{\sqrt{29}} \right|\]

\[ = \frac{8}{\sqrt{29}}\]

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पाठ 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
पाठ 27 Straight Line in Space
Exercise 28.5 | Q 2.3 | पृष्ठ ३८

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