Question

The Cartesian equations of a line are $3x+1=6y-2=1-z$. Find the fixed point through which it passes, its direction ratios and also its vector equation.

Answer 1

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

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

forced it is $(\frac{-1}{3},\frac{1}{3},1)$

Direction radius are $(2,1,-6)$ or

$(\frac{2}{\sqrt{41}},\frac{1}{\sqrt{41}},\frac{-6}{\sqrt{41}})$

vector equation of line $\Rightarrow$

$\overline{r}=(\frac{-1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k})+\lambda (2\hat{i}+\hat{j}-6\hat{k})$