Question

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

Answer 1

$x+y-z=3$

$x+y-z-3=0$ ____ (1)

$2x+3y+z+4=0$ _____ (2)

$\left(1\right)+\left(2\right)$

$3x+4y+1=0$ ____ (3)

$3x=-4y-1=\lambda$

$x=\frac{\lambda }{3}$

$-4y=\lambda +1$

$y=-\frac{(\lambda +1)}{4}$

(1) Now equals

$\frac{\lambda }{3}-\frac{(\lambda +1)}{4}-z-3=0$

$z=\frac{\lambda }{3}-\frac{\lambda +1}{4}-3$

$z=\frac{4\lambda -3\lambda -3-36}{12}$

$z=\frac{\lambda -39}{12}$

$\lambda =12z+39$

$\left(3\right)3x=-4y-1=12z+39$

$\frac{x}{4}=\frac{y+\frac{1}{3}}{-3}=\frac{z+\frac{39}{12}}{1}$

Point $(0,\frac{-1}{3},\frac{-13}{4})$ air.Ratio $(4,-3,1)$