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By-computing-shortest-distance-determine-whether-following-pairs-lines-intersect-or-not-r-i-j-k-3-i-j-r-4-i-k-2-i-3-k - Mathematics

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प्रश्न

By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \overrightarrow{r} = \left( 4 \hat{i} - \hat{k}  \right) + \mu\left( 2 \hat{i}  + 3 \hat{k} \right)\] 

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उत्तर

\[\vec{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and }  \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k}  \right)\] 

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 ,

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

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

\[ \overrightarrow{b_1} = 3 \hat{i} - \hat{j} \]

\[ \overrightarrow{b_2} = 2 \hat{i} + 3 \hat{k} \]

\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = 3 \hat{i} - \hat{j} \]

\[\text{ and }  \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 3 & - 1 & 0 \\ 2 & 0 & 3\end{vmatrix}\]

\[ = - 3 \hat{i}  - 9 \hat{j}  + 2 \hat{k} \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \vec{b_1} \times \overrightarrow{b_2} \right) = \left( 3 \hat{i}  - \hat{j}  \right) . \left( - 3 \hat{i}  - 9 \hat{j}  + 2 \hat{k} \right)\]

\[ = - 9 + 9\]

\[ = 0\]

\[\text{ We observe } \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = 0\]

\[\text{ Thus, the given lines intersect  } .\]

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अध्याय 28: Straight Line in Space - Exercise 28.5 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.5 | Q 3.2 | पृष्ठ ३८

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