The vector equation of the line passing through the points $ (3,2,1) $ and $ (-2,1,3) $ is
- $ \vec{r}=3\hat{i}+2\hat{j}+\hat{k}+\lambda (-5\hat{i}-\hat{j}+2\hat{k}) $
- $ \vec{r}=3\hat{i}+2\hat{j}+\hat{k}+\lambda (-5\hat{i}+\hat{j}+\hat{k}) $
- $ \vec{r}=-2\hat{i}+\hat{j}+3\hat{k}+\lambda (5\hat{i}+\hat{j}+2\hat{k}) $
- $ \vec{r}=-2\hat{i}+\hat{j}+\hat{k}+\lambda (5\hat{i}+\hat{j}+2\hat{k}) $
The Correct Option is A
Solution and Explanation
The correct option is(A): \(\vec{r}=3\hat{i}+2\hat{j}+\hat{k}+\lambda (-5\hat{i}-\hat{j}+2\hat{k})\)
The vector equation of a line passing through \((3,2,1)\) and \((-2,1,3)\) is \(\vec{r}=3\hat{i}+2\hat{j}+\hat{k}+\lambda [(-2-3)\hat{i}+(1-2)\hat{j}\)
\(+(3-1)\hat{k}]\) \(=3\hat{i}+2\hat{j}+\hat{k}+\lambda (-5\hat{i}-\hat{j}+2\hat{k})\)
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Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)
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