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