Question

Find the vector equation of a line passing through a point with position vector $2\hat{i}-\hat{j}+\hat{k}$ and parallel to the line joining the points $-\hat{i}+4\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$. Also find the cartesian equivalent of this equation.

Answer 1

We know that vector equation of a line passing through a vector $\overline{a}$ and its parallel vector $\overline{b}$ is

$\overline{r}=\overline{a}+t\overline{b}$

Here line passing through the vector $2\hat{i}-\hat{j}+\hat{k}$ and its parallel vector is

$\hat{i}+2\hat{j}+\hat{k}-(-\hat{i}+4\hat{j}+\hat{k})$

$=2\hat{i}-2\hat{j}$

$\therefore$ vector equation is,$(2\hat{i}-\hat{j}+\hat{k})+\lambda (2\hat{i}-2\hat{j})$

and

$\therefore$ the cartesion equivalent is

$\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-1}{0}$.