Orthogonal planes in n-dimensions
In 3D two planes are orthogonal when their normal vectors are orthogonal (their inner product is zero). For example, planes $xy$ and $xz$ are orthogonal because their normal vectors $\hat{z}$ and $\hat{y}$ respectively are orthogonal, i.e $\hat{z}\cdot \hat{y}=0$.
How we define orthogonality of planes in $n$ dimensions? I am talking about 2d planes through the origin, in n-dimensional Euclidean space, that are specified by orthonormal vectors $\hat{x}_1, \hat{x}_2,.., \hat{x}_n$.
In 4D we have four orthogonal axes x,y,z,w defined by normal vectors $\hat{x}, \hat{y}, \hat{z}, \hat{w}$. However these axes make six planes: $xy, xz, xw, yz, yw, zw$. Are these planes orthogonal to each other? For example, the normal vectors $\hat{z}$ and $\hat{w}$ are perpendicular to the plane $xy$, but they are orthogonal, i.e $\hat{z}\cdot \hat{w}=0$, not parallel. How it is possible that they are not parallel when they are perpendicular to the same plane and how we check if the plane $xy$ is orthogonal to the plane $wz$?
It depends on what you want "orthogonal" to mean, of course. Typically, this means as subspaces, which does not accord with your meaning in 3-d. There are no other definitions in widespread use. There are a few different ways we can characterize your 3-d meaning:
- The normal unit vectors are perpendicular.
- In the plane perpendicular to the line of intersection, each plane's intersection is perpendicular to the other.
- Normal unit vectors of one are contained in the other.
As mentioned, in higher dimensions, there are multiple normal unit vectors, even beyond the sign ambiguity in 3-d. Similarly, in higher dimensions, planes can intersect in just a point, rather than a line.
Definition 3 seems promising though. Given the existence of multiple normal vectors We can generalize it in at least two ways: (a) all normal vectors must be contained in the other, or (b) at least one normal vector must be contained in the other. I assume that we want the planes that were considered orthogonal in 3-d to also be considered orthogonal in 4-d. That rules out "all normal vectors", but still allows "at least one normal vector".
Is this a useful or interesting definition? I don't know, but it seems consistent. All principle planes are perpendicular to all others, which seems right.
Generally, two linear subspaces are considered orthogonal if every pair of vectors from them are perpendicular to each other. This doesn't wok in three dimensions: two planes are either parallel or they share a common line, hence in the latter case two vectors can be chosen both from the shared line and these are not orthogonal.
What you're talking about, however, is a scenario where the two planes' normal vectors are orthogonal. This does, in fact, naturally generalize to any number of dimensions (as well as hyperplanes of any dimension): let two linear subspaces $V$ and $W$ be dubbed converse - I'm making that word up - if $V^\perp\perp W^\perp$. That is, for any vectors $\vec{n}$ orthogonal to $V$ and $\vec{m}$ orthogonal to $W$, $\vec{n}$ and $\vec{m}$ will themselves be orthogonal.
Another meaning of orthogonal is "intersecting at right angles". This fits the $3D$ case with the $xy$ and $xz$ planes. In $3D$, planes have to have a line in common unless they are parallel. In $4D$, planes can miss entirely. Think of the planes $x=0, y=0$ and $x=5, z=9$. You can also have planes that intersect in only one point. So what you mean by orthogonal planes needs definition. Of course, you can insist they intersect in a line and measure the angle, but then they must be within a $3D$ subspace of your $4D$ space.
In more than 3 dimensions, planes are orthogonal if and only if when one vector is taken parallel to one plane and another vector is taken parallel to the other plane, those vectors must be orthogonal.
The idea comes from http://mathworld.wolfram.com/OrthogonalSubspaces.html