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Find-shortest-distance-between-pairs-lines-whose-vector-are-vec-r-left-8-3-lambda-right-hat-i-left-9-16-lambda-right-hat-j-left-10-7-lambda-right-hat-k - Mathematics

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प्रश्न

Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i}  + 8 \hat{j} - 5 \hat{k}  \right)\]

योग
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उत्तर

The vector equations of the given lines can be re-written as 

\[\overrightarrow{r} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k}  + \lambda\left( 3 \hat{i}  - 16 \hat{j}  + 7 \hat{k}  \right)\] and \[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]

Comparing the given equations with the equations 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]

We get , 

\[\overrightarrow{a_1} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k}\]

\[\overrightarrow{b_1} = 3 \hat{i} - 16 \hat{j} + 7 \hat{k}\]

\[\overrightarrow{a_2} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k}\]

\[\overrightarrow{b_2} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k}\]

\[\therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \left( 15 \hat{i} + 29 \hat{j} + 5 \hat{k} \right) - \left( 8 \hat{i} - 9 \hat{j} + 10 \hat{k} \right) = 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \]

\[\overrightarrow{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i}  & \hat{j} & \hat{k} \\ 3 & - 16 & 7 \\ 3 & 8 & - 5\end{vmatrix} = 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \]

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{{24}^2 + {36}^2 + {72}^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84\]

Also,

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

\[ = \left( 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \right) . \left( 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \right)\]

\[ = 7 \times 24 + 38 \times 36 + \left( - 5 \right) \times 72\]

\[ = 168 + 1368 - 360\]

\[ = 1176\]

We know that 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

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

∴ Required shortest distance between the given pairs of lines,

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

\[ = \left| \frac{1176}{84} \right|\]

\[ = 14\]

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अध्याय 28: Straight Line in Space - Exercise 28.5 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.5 | Q 1.8 | पृष्ठ ३७

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