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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

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Question

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

Sum
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Solution 1

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

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 , 

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

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

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

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

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

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

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

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

\[ = - 1\]

\[\text{ We observe } \]

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

\[\text{ Thus, the given lines do not intersect } .\]

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Solution 2

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

Comparing the given equations with the equations 

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

\[\vec{a_1} = \hat{i} - \hat{j} \]

\[ \vec{a_2} = 2 \hat{i} - \hat{j} \]

\[ \vec{b_1} = 2 \hat{i} + \hat{k} \]

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

\[ \therefore \vec{a_2} - \vec{a_1} = \hat{i} \]

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

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

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

\[ = - 1\]

\[\text{ We observe } \]

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

\[\text{ Thus, the given lines do not intersect } .\]

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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 3.1 | Page 38

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