Question

Find vector equation of line passing through the point whose position vector is
$3\hat{i}-4\hat{j}+\hat{k}$ and parallel to the vector $2\hat{i}+\hat{j}-3\hat{k}$. Also write the equation in Cartesian form.

Answer 1

Here, $\overrightarrow{a}=3\hat{i}-4\hat{j}+\hat{k}$
and $\overrightarrow{b}=2\hat{i}+\hat{j}-3\hat{k}$
$\therefore$ Vector equation of the line passing through the point $\overrightarrow{a}$ and parallel to the vector $\overrightarrow{b}$ is given by,
$\overrightarrow{r}=\overrightarrow{a}+t\overrightarrow{b}$ where $t$ is a scalar
$\overrightarrow{r}=(3\hat{i}-4\hat{j}+\hat{k})+t(2\hat{i}+\hat{j}-3\hat{k})$
Let $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$, then

$x\hat{i}+y\hat{j}+z\hat{k}=(3\hat{i}-4\hat{j}+\hat{k})+t(2\hat{i}+\hat{j}-3\hat{k})$
$x\hat{i}+y\hat{j}+z\hat{k}=(3+2t)\hat{i}+(-4+t)\hat{j}+(1-3t)\hat{k}$
Equating the coeffeicients of $\hat{i},\hat{j}$ and $\hat{k}$
$x=3+2t$
$y=-4+t$
$z=1-3t$
$\Rightarrow \frac{x-3}{2}=\frac{y+4}{1}=\frac{z-1}{-3}=t$
These are the equation of the line in cartesian form.