Find the intersection of two planes.

Question

Find the intersection of the planes $x+(y-1)+z=0$ and $-x+(y+1)-z=0$. These two planes are 3-dimensional and I am confused on how to solve it.

Answer 1

Hint : You have to solve the system :

$$\left\{\begin{array}{l}x+(y-1)+z=0\\-x+(y+1)-z=0 \end{array} \right. $$

that means expressing two of the variables in terms of the third one. Imagine you get something like :

$$\left\{\begin{array}{l}x=1+3z\\y=-2+z \end{array} \right. $$

then adding $z=t$ you get a system of parametric equation of a line :

$$\left\{\begin{array}{l}x=1+3t\\ y=-2+t\\z=t \end{array} \right., t\in\mathbb{R}.$$

in my example this would be the line passing through $(1;-2;0)$ (constant values in the equations) with direction $\overrightarrow{u}(3;1;1)$ (coefficients multiplying $t$ in the equations)

Your example is a bit particular as it yields $y=0$ quickly. You can write it as $y=0z$ and continue as planed.