# Intersection of Two Planes

I keep hearing different answers for what the intersection of two planes is. I believe it is a line, but it can also be a plane IF the two planes are not distinct. However other sources are saying that the intersection of two planes can also be a point or an empty set. So what is the intersection of two distinct, nonparallel planes? A point, line, and/or a empty set?

• It can never be a point if it is an infinite plane. However, parallel planes can have no intersection. Non-parallel infinite planes will always yeild a line. If one is not infinte, it can give a point if its vertex just touches the other plane. – AvZ Jan 26 '15 at 14:25
• The intersection of two planes is a line. In order to explicitly find it, you need a point on the line and the direction of it. To find the direction, you determine the cross product of the two normals of the two planes (since the line must be perpendicular to both normals). – Autolatry Jan 26 '15 at 14:28
• the intersection of two (twodimensional) planes can also be a single point, but only if the ambient space is at least of dimension 4. – Thomas Jan 26 '15 at 14:49

$\newcommand{\Reals}{\mathbf{R}}$For definiteness, I'll assume you're asking about planes in Euclidean space, either $\Reals^{3}$, or $\Reals^{n}$ with $n \geq 4$.

The intersection of two planes in $\Reals^{3}$ can be:

• Empty (if the planes are parallel and distinct);
• A line (the "generic" case of non-parallel planes); or
• A plane (if the planes coincide).

The tools needed for a proof are normally developed in a first linear algebra course. The key points are that non-parallel planes in $\Reals^{3}$ intersect; the intersection is an "affine subspace" (a translate of a vector subspace); and if $k \leq 2$ denotes the dimension of a non-empty intersection, then the planes span an affine subspace of dimension $4 - k \leq 3 = \dim(\Reals^{3})$. That's why the intersection of two planes in $\Reals^{3}$ cannot be a point ($k = 0$).

Any of the preceding can happen in $\Reals^{n}$ with $n \geq 4$, since $\Reals^{3}$ be be embedded as an affine subspace. But now there are additional possibilities:

• The planes $$P_{1} = \{(x_{1}, x_{2}, 0, 0) : x_{1}, x_{2} \text{ real}\},\qquad P_{2} = \{(0, 0, x_{3}, x_{4}): x_{3}, x_{4} \text{ real}\}$$ intersect at the origin, and nowhere else.
• The planes $P_{1}$ and $$P_{3} = \{(0, x_{2}, 1, x_{4}): x_{2}, x_{4} \text{ real}\}$$ are not parallel (in the sense that neither is a translate of the other), but they do not intersect.

The planes $P_{1}$ and $P_{3}$ are "partially parallel" in the sense that there exist parallel lines $\ell_{1} \subset P_{1}$ and $\ell_{3} \subset P_{3}$. This turns out to be true for every pair of disjoint planes in $\Reals^{4}$.

In $\Reals^{5}$, there exist "totally skew" planes, such as $$P_{4} = \{(x_{1}, x_{2}, 0, 0, 0)\},\qquad P_{5} = (0, 0, 1, x_{4}, x_{5})\}.$$

(The terms "partially parallel" and "totally skew" are not standard as far as I know.)