Question

The cartesian equation of a line is $\frac{x-4}{3}=\frac{y+1}{-2}=\frac{z+2}{5}$. Find the vector equation of line.

Answer 1

Comparing with standard form, $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
$\Rightarrow x_1=4,y_1=-1,z_1=-2$ and $a=3,b=-2,c=5$

Thus, the required line passes through the point $(4,-1,-2)$ and is parallel to the vector $(3\hat{i}-2\hat{j}+5\hat{k})$.

Let $\overrightarrow{r}$ be the position vector of any point on the line, then the vector equation of the line is given by
$\overrightarrow{r}=(4\hat{i}-\hat{j}-2\hat{k})+\lambda (3\hat{i}-2\hat{j}+5\hat{k})$