Is the intersection of two orthogonal planes a line, or the zero vector?

I am struggling to grasp a relatively simple concept in linear algebra:

I know that the intersection between two orthogonal subspaces is the zero vector.

But I also know that the intersection between two orthogonal planes is a line.

A plane is a subspace.

But....a line is not the zero vector.

Where did I go wrong in my logic?

• Depending on the context, a plane may not be a subspace. E.g. the plane $z=1$ in $\mathbb{R}^3$ is not a subspace. Aug 6, 2017 at 21:55
• You need a definition of "orthogonal subspaces" for your question to make sense. The answer would depend on the dimension of the larger space the two subspaces lie inside. Aug 6, 2017 at 21:56
• the intersection between two orthogonal planes is a line Geometrically orthogonal planes are not orthogonal as vector subspaces of $\mathbb{R}^3$. The orthogonal of a plane in $\mathbb{R}^3$ is its normal, which is a line, and their intersection is one single point.
– dxiv
Aug 6, 2017 at 21:57

edit, tl;dr: What usually is meant by two planes being orthogonal to one another in geometry is their normals being orthogonal to each other.

In other words: two one-dimensional subspaces being orthogonal to each other.

2 planes have their normals being orthogonal to each others are sometimes said to be orthogonal. There it is a specific geometric orthogonality pointing out that the normals of the planes are orthogonal to each other. When talking about subspaces orthogonal to each other what is usually meant is all their vectors are pairwise orthogonal. But you can verify for yourself that 2 2D subspaces can not have 0 vector intersection in $\mathbb R^3$.

But if you think about it closer, you will see that the geometric meaning of normals being orthogonal to each other actually means the complement of set 1 and the complement of set 2 are orthogonal to each other. So there is a connection to the same orthogonality concept, but the subspaces are 1 dimensional.

2 planes in $\mathbf {R}^3$ may be orthogonal, but they will never be orthogonal subspaces. An easy way to check this is the fact that a the dimension of a vector space is equal to the sum of the dimension of a subspace plus the dimension of the subspace's orthogonal space. In other words, a subspace orthogonal to a plane in $\mathbf {R}^3$ would necessarily be a line normal to the plane through the origin.

Every vector in an orthogonal subspace must be orthogonal to every vector in the subspace to which the orthogonal subspace is orthogonal. You can verify this is not the case for 2 planes in $\mathbf {R}^3$. Both of them contain vectors in their intersection, which is a line. You can see vectors in that line are not orthogonal to themselves, except for the 0 vector.

For manifolds $V,W$, or, here, subspaces, in general position, living in a space of dimension $n$, the dimension of the intersection $V \cap W$ is $n-(Dim(V)+Dim(W))$. Since you are working with planes, let's assume $n=3$. Then $Dim(U\cap W)=2+2-3=1$ EDIT: Of course we make "reasonable" adjustments when this value is negative.

We usually say that subspaces $U\subseteq V$ and $W\subseteq V$ of a Euclidean vector space $V$ are orthogonal if all vectors $u\in U$ and $w\in W$ are orthogonal to each other. Now, for $V=\Bbb R^3$ with $U$ and $W$ any two distinct planes, $U\cap W$ is a line containing some nonzero vector $v$. So $u:= v\in U$ and $w:=v\in W$ are not orthogonal to each other, because $v$ is not orthogonal to itself. So as pointed out in the comments, two planes in $\Bbb R^3$ are never orthogonal to each other as subspaces.

When you speak of orthogonal planes in $\Bbb R^3$, you might be thinking of planes whose normal vectors are orthogonal, but this property is unrelated to the one above, as we just saw that planes in $\Bbb R^3$ can never be orthogonal to one another anyway.

A (2-d) plane is only a subspace if it contains the zero vector $(0, \dots,0)$. The intersection of orthogonal subspaces is $(0,\dots ,0)$. It is conceivable that there could be two planes in $R^3$ that are subspaces with intersection $(0,0,0)$. However as pointed out above, there are no "orthogonal planes " in $R^3$... For if there were, then $dim (V+W)$ would be 4. In $R^n$, $n\gt 3$, there is enough room to do this, I think...

is simply false. In $\mathbb R^3$ there aren’t two orthogonal planes. The orthogonal complement of a plane is a line.
One can give the following definition: two vector subspaces $$V$$, $$W$$ of $$\mathbb{R}^n$$ are "perpendicular" if the two orthogonal complements of $$V\cap W$$ in $$V$$ and $$W$$ are orthogonal to each other. This definition reduces to the usual notion of orthogonality when $$V\cap W=0$$. Also, it can be extended to affine subspaces.
• Edit: to avoid weird cases, in the above definition one should assume that $V\cap W\neq V, W$ Dec 11, 2022 at 4:55