Question

The cartesian equation of a line is $6x-2=3y+2=2z-2$. Find its direction ratios & also find vector equation of the line.

Answer 1

The cartesian equation of the line is

$6x-2=3y+2=2z-2$

$\begin{array}{c}6\left(x-\frac{1}{3}\right)=3\left(y+\frac{2}{3}\right)=2\left(z-1\right)\\ \frac{x-\frac{1}{3}}{1}=\frac{y+\frac{2}{3}}{2}=\frac{z-1}{3}\end{array}$

p($\frac{1}{3},\frac{-2}{3},1$)

Direction ratios are ($1,2,3$)

Vector equation of the line is

$\overrightarrow{r}=\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\hat{k}+\lambda (\hat{i}+2\hat{j}+3\hat{k})$