Find equation of intersection of $2x + y - z = -1$ and $x + y -z = 1$

Question

$2x + y - z = -1$, and $x + y -z = 1$. I'm trying to find the parametric form.

When I solve these, I get $x = -2$, but not sure where to go from there.

Answer 1

As you said, $x=-2$ and so $z = y-3$. To find the parametric equation of a line, you just need two points on the line. We can take $P=(-2,0,-3)$ and $Q=(-2,3,0)$, for instance. Then a vector directing the line is given by $v=Q-P = (0,3,3)$, and a general parametric equation for the line is $P +tv = (-2,3t,-3+3t)$.

Answer 2

The respective normal vectors to the two planes are $u=(2,1,-1)$ and $v=(1,2,-2)$

the cross product gives the vector director of the intersection line.

we get $w=(0,1,1)$

the parametric equations of the line have following form $x=x_0$, $y=y_0+t$, and $z=z_0+t$.

on the other hand the point $(x_0,y_0,z_0)$ is in the intersection if $x_0=-2$ and $y_0-z_0=3$. so, let us take for example $z_0=0$. this gives $y_0=3$.

finally, the parametric equations are

$x=-2$

$y=3+t$

$z=t$

$t$ is the paremeter.