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 $wz$?
Best Answer
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:
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.