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 $wz$?
 A: 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.
A: 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.
A: 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.
A: 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
A: To expound upon the definition of orthogonal spaces, you can prove that planes are orthogonal by using their basis elements.  Each (2d) plane has two basis elements and everything in the plane is a linear combination of them, so it suffices to show that both basis elements of one plane are orthogonal to both basis elements for another plane.  Say, in $\mathbb{R}^4$, you have the $xy$ plane with basis $\{\hat x,\hat y\}$ and the $zw$ plane with basis $\{\hat z, \hat w\}$.  Since all the dot products are zero, these two planes are orthogonal.
Now computing the dot product for arbitrary vectors in each plane, we get: $(a\hat x + b\hat y)\cdot (c\hat z + d\hat w)=0$.
Essentially, there are not enough dimensions to have orthogonal planes in $\mathbb{R}^3$.
Apologies for my previous, incorrect answer.
A: $\newcommand{\Brak}[1]{\langle#1\rangle}$Since this has gotten bumped: As wnoise says, orthogonality usually refers to vector subspaces $W_{1}$ and $W_{2}$ of an inner product space $(V, +, \cdot, \Brak{\ ,\ })$, and means $\Brak{w_{1}, w_{2}} = 0$ for all $w_{1}$ in $W_{1}$ and $w_{2}$ in $W_{2}$. This definition extends naturally to affine subspaces (which may fail to intersect at all.)
In this sense (as Eric Naslund, anon, and Ricky Demer point out), there are no "orthogonal planes" in Euclidean $3$-space. The six coordinate planes in Euclidean $4$-space partition into three pairs of mutually-orthogonal planes: $(w, x)$ and $(y, z)$; $(w, y)$ and $(x, z)$; and $(w, z)$ and $(x, y)$.

The only definition of angle between subspaces I've seen in the wild comes from Kobayashi-Nomizu, in their discussion of holomorphic sectional curvature: The angle between $W_{1}$ and $W_{2}$ is the infimum of the angle between $w_{1}$ and $w_{2}$, taken over all $w_{1}$ in $W_{1}$ and $w_{2}$ in $W_{2}$.
Kobayashi-Nomizu apply this, with $J$ denoting the complex structure of an Hermitian manifold, to pairs of real $2$-planes $P$ and $J(P)$. If $v \in P \cap J(P)$ for some non-zero $v$, then $Jv \in P \cap J(P)$, which implies $P = J(P)$ is spanned over the reals by $v$ and $Jv$. Contrapositively, if $P$ is not a complex line, then $P \cap J(P) = \{0\}$, and the angle is non-zero. (Generally, distinct subspaces can have non-zero intersection and therefore have an angle of zero.)
In case it matters:


*

*The condition "all normal vectors of one are contained in the other" does not capture the same meaning: Two planes in Euclidean $n$-space ($n \geq 5$) fail to span, so there exist unit vectors orthogonal to both planes no matter how they're situated.

*The condition "some normal vector of each is contained in the other" does not capture the same meaning, as is seen already in Euclidean $3$-space.
