Question

The vector equation of the line through the points $(3,4,-7)\text{and}(1,-1,6)$ is

Answer 1

Given points are $(3,4,-7)\text{and}(1,-1,6)$
Position vector of the point are $\overrightarrow{a}=3\hat{i}+4\hat{j}-7\hat{k}\text{and}\overrightarrow{b}=\hat{i}-\hat{j}+6\hat{k}$

Vector equation of aline passing through the point$\overrightarrow{a}\text{and}\overrightarrow{b}$ is given by:
$\overrightarrow{r}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a})$
$\Rightarrow \overrightarrow{r}=3\hat{i}+4\hat{j}-7\hat{k}+\lambda \left[\begin{array}{c}(\hat{i}-\hat{j}+6\hat{k})\\ -(3\hat{i}+4\hat{j}-7\hat{k})\end{array}\right]$

Substituing $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=3\hat{i}+4\hat{j}-7\hat{k}+\lambda (-2\hat{i}-5\hat{j}+13\hat{k})$
$\therefore (x-3)\hat{i}+(y-4)\hat{j}+(z+7)\hat{k}=\lambda (-2\hat{i}-5\hat{j}+13\hat{k})$

Hence , the equation of line in vector form is
$\therefore (x-3)\hat{i}+(y-4)\hat{j}+(z+7)\hat{k}=\lambda (-2\hat{i}-5\hat{j}+13\hat{k})$