# How to intersect arbitrary 3d plane (with form ax+by+cz + d = 0) with the XY plane?

So I know there have been other stack exchange questions around intersecting two arbitrary 3d planes like here (Plane-plane intersection in python) however I have a special case where I want to intersect two planes, where one is arbitrary and another is guaranteed to be the XY plane.

For the other questions I've seen, since there were two arbitrary planes, the final line equation was that for a line in 3D space (typically in the parameterized form L = O + t*D, where O is a point on the line and D is a direction vector) but since I know I'm going to be intersecting with the XY plane, I know my intersection line is going to lie on the XY plane, and so I want to express the final answer in the standard form Ax + By + C = 0.

How would I go about doing this?

• A point $(x,y,z)$ is in the xy-plane if and only if $z=0$. So a point$(x,y,z)$ is in your arbitrary 3D plane and the xy-plane if and only if it satisfies both $ax+by+cz=0$ and $z=0$. In other words, if and only if $ax +by + d=0$.
– MPW
Commented May 13, 2019 at 19:22
• Ax+By+C=$0$ represents a plane; z could be anything. Its intersection with z$=0$ (XY plane) is a line. Commented May 13, 2019 at 19:24

The equation $$ax+by+cz+d=0$$ defines a plane in the sense that $$P_1=\{(x,y,z)\in \mathbb{R}^3: ax+by+cz+d=0\}$$ is considered a plane (for good reason). One the other hand, the $$(x,y)-$$plane $$P_2$$ is defined by the equation $$z=0$$. So, to find the intersection $$P_1\cap P_2$$, you should find the set of points satisfying both sets of equations. That is, you should find the solution set in $$\mathbb{R}^3$$ of the system $$\begin{cases} ax+by+cz+d=0\\ z=0. \end{cases}$$ Unsurprisingly, it is given by simply setting $$z=0$$ in the first equation: $$P_1\cap P_2=\{(x,y,0): ax+by+d=0\}.$$ Notice that this is a line in the plane $$P_2$$.