Question

If the equation of line $x+y-z-3=0=2x+3y+z+4$ is symmetric form. Find the direction ratios of the line

Answer 1

$x+y-z-3=\mathrm{0........}\left(1\right)$

$2x+3y+z+4=\mathrm{0......}\left(2\right)$

Eliminating $z$ from $\left(1\right)$ and $\left(2\right)$

$⟹3x+4y+1=0$

$x=t⟹y=\frac{-1-3t}{4}$

Eliminating $y$ from $\left(1\right)$ and $\left(2\right)$

$z=\frac{t-13}{4}⟹t=\frac{z+\frac{13}{4}}{\frac{1}{4}}$

The equation of line is $\frac{x}{1}=\frac{4y+1}{-3}=\frac{4z+13}{1}$

d.r.'s of the line are $(1,-\frac{3}{4},\frac{1}{4})$