Question

Find the vector equation of a line passing through the point $(-1,-1,2)$ and parallel to the line $2x-2=3y+1=6z-2$.

Answer 1

$2(x-1)=3(y+\frac{1}{3})=6(z-\frac{1}{3})$

$\frac{x-1}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{3}}{1}$.......(dividing by 6)

The direction ratio of the line are $3,2$and $1$ .

The equation of line passing through point $(-1,-1,2)$ and parallel to line with direction ratio as $3,2,1$ is

$\overrightarrow{r}=(-x-y+2z)+t(3x+2y+z)$where t is a parameter