Question

Find the vector equation of the line joining the points $(2,1,3)$ and $(-4,3,-1)$.

Answer 1

Step $1$: Evaluate the normal vector for the given points

Given, $(2,1,3)$ and $(-4,3,-1)$

Here, the position vectors are

$\overrightarrow{a}=2\hat{i}+\hat{j}+3\hat{k}$ and

$\overrightarrow{b}=-4\hat{i}+3\hat{j}-\hat{k}$

So, the normal vector can be found as

$\overrightarrow{n}=\overrightarrow{b}-\overrightarrow{a}$

$\Rightarrow \overrightarrow{n}=\left(-4\hat{i}+3\hat{j}-\hat{k}\right)-\left(2\hat{i}+\hat{j}+3\hat{k}\right)$

$\Rightarrow \overrightarrow{n}=-4\hat{i}+3\hat{j}-\hat{k}-2\hat{i}-\hat{j}-3\hat{k}$

$\Rightarrow \overrightarrow{n}=-6\hat{i}+2\hat{j}-4\hat{k}$

Step $2$: Form the equation of the line

We know that, $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{n}$

So,

$\Rightarrow \overrightarrow{r}=\left(2\hat{i}+\hat{j}+3\hat{k}\right)+\lambda \left(-6\hat{i}+2\hat{j}-4\hat{k}\right)$

$\Rightarrow \overrightarrow{r}=2\hat{i}+\hat{j}+3\hat{k}+\lambda -6\hat{i}+\lambda 2\hat{j}-\lambda 4\hat{k}$

$\Rightarrow \overrightarrow{r}=\left(2-6\lambda \right)\hat{i}+\left(1+2\lambda \right)\hat{j}+\left(3-4\lambda \right)\hat{k}$

$\Rightarrow \overrightarrow{r}=2\left(1-3\lambda \right)\hat{i}+\left(1+2\lambda \right)\hat{j}+\left(3-4\lambda \right)\hat{k}$

Hence, the equation of the line is $\overrightarrow{r}=2\left(1-3\lambda \right)\hat{i}+\left(1+2\lambda \right)\hat{j}+\left(3-4\lambda \right)\hat{k}$.