Generalized notion of perpendicularity, (not orthogonal)

Question

In 3 dimensions, we might call 2 planes perpendicular iff their normals are orthogonal.

But this does not coincide with the definition of orthogonal subspaces - the dot product of any pair of vectors is $0$.

Is there a notion of perpendicularity any text introduces, generalising in $\mathbb R^n$ the perpendicular planes in $\mathbb R^3$?

For example - I was thinking if we had two subspaces be "perpendicular" iff they are not parallel and are the sides of some $n$-cube.

Answer 1

The geometric algebra analyses this scenario. In the geometric algebra, two subspaces are orthogonal means there is a vector in one that is in the orthogonal complement of the other. This can be determined using the contraction operator, which generalizes the dot product, which is very nice.

So $V\perp W$ iff $V \rfloor W = 0$.

Answer 2

I think what you want when stated geometrically is that affine subspaces $V$ and $W$ are "perpendicular" in your new sense if when you project both spaces onto the orthogonal complement of their intersection the resulting affine subspaces, which will meet at just one point, are orthogonal.

That's equivalent to choosing the origin as that point, then an orthogonal basis for the intersection and then extending it to an orthogonal basis of the sum by choosing new basis vectors from $V$ and $W$. That basis will define the $n$ cube you have imagined.

It don't know whether this configuration has a special name.

Answer 3

Well, there is a quite natural generalisation of the very first phrase in your question to the case of arbitrary dimension $n$, that is: we might call 2 hyperplanes (subspace of dimension $n-1$) perpendicular iff their normals are orthogonal.