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Find the Vector Equation of the Line Passing Through the Points (−1, 0, 2) and (3, 4, 6).

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Question

Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).

Sum
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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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Chapter 27: Straight Line in Space - Exercise 28.1 [Page 9]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Exercise 28.1 | Q 2 | Page 9

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