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Question
Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).
Solution
We know that the vector equation of a line passing through the points with position vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] is \[\overrightarrow{r} = \overrightarrow{a} + \lambda \left( \overrightarrow{b} - \overrightarrow{a} \right)\] where \[\lambda\] is a scalar.
Here,
\[\overrightarrow{a} = - 1 \hat{i}+ 0 \hat{j}+ 2 \hat{k} \]
\[ \overrightarrow{b} = 3 \hat{i} + 4 \hat{j} + 6 \hat{k} \]
Vector equation of the required line is
\[\overrightarrow{r} = \left( - 1 \hat{i} + 0 \hat{j}+ 2 \hat{k} \right) + \lambda \left\{ \left( 3 \hat{i} + 4 \hat{j} + 6 \hat{k} \right) - \left( - 1 \hat{i} + 0 \hat{j} + 2 \hat{k} \right) \right\}\]
\[ \Rightarrow \overrightarrow{r} = \left( - 1 \hat{i} + 0 \hat{j} + 2 \hat{k} \right) + \lambda \left( 4 \hat{i} + 4 \hat{j} + 4 \hat{k} \right)\]
\[\text { Here }, \lambda \text { is a parameter } . \]
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