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.
Since the other (very nice) answers here have not satisfied you, let me take a different route. I will illustrate what is going on in a simpler situation, and hopefully it will shed light on your question by analogy. Also, since the algebraic formalism doesn't seem to be to your liking, we'll think very geometrically.
Let's consider a 1-dimensional subspace $S$ of $\mathbf{R}^3$. Such a subspace is simply a line through the origin. In fact, for concreteness, let's take $S$ to be the familiar $x$-axis in $\mathbf{R}^3$.
What are all the possible dimensions of orthogonal subspaces to $S$?
Well, the empty-set is orthogonal to $S$ for trivial reasons, and that provides the unique zero-dimensional example. This is also the case in your problem.
Next, we have one-dimensional subspaces which are orthogonal to $S$. These are lines through the origin spanned by vectors making a "right angle" with the $x$-axis. There are many such lines. The $y$-axis and the $z$-axis provide the most familiar examples. So $1$-dimensional orthogonal subspaces are also possible, there are lots of them, and if you take $S$ together with a one-dimensional orthogonal subspace, there's still plenty of $\mathbf{R}^3$ that remains "untouched".
Now, take any two distinct one-dimensional orthogonal subspaces from the previous example. These lines are spanned by two linearly independent vectors. Thus, taken together they span a $2$-dimensional subspace of $\mathbf{R}^3$, and since both of the basis vectors are orthogonal to $S$, the entire space is orthogonal to $S$. A $2$-dimensional subspace of $\mathbf{R}^3$ is a plane. This plane is going to be the $y-z$-plane, which is probably familiar to you from multivariable calculus. You can probably visualize how the $x-$axis "points out of the $y-z$-plane orthogonally".
But now we're done, because the next dimension to consider would be dimension $3$, but the only $3$-dimensional subspace of $\mathbf{R}^3$ is all of $\mathbf{R}^3$ itself, which would certainly overlap with $S$, and hence not be orthogonal to it.
So the possible dimensions for orthogonal subspaces were 0, 1, and 2. Now, a subspace and its orthogonal complement, taken together, "fill up" the entire ambient vector space. What we noticed in our example was that the $x-$axis and the $y-z$-plane are not only orthogonal, but taken together they fill up $\mathbf{R}^3$. So the orthogonal complement of our one-dimensional subspace of $\mathbf{R}^3$ was two-dimensional.
Now, you can play the same sort of game with the problem in your question...it just won't be so easy to visualize what's going on. If $S$ is six-dimensional in $\mathbf{R}^9$, then you can find orthogonal subspaces of dimensions 0, 1, 2, or 3, and if you take a 3-dimensional orthogonal subspace, then you've filled up all of $\mathbf{R}^9$, so you've found the orthogonal complement.
Best Answer
In the case of orthogonal planes, the requirement is that each plane has a basis of two orthogonal vectors and that one of the vectors of each pair is shared between the two planes and the other vector of each pair form a pair orthogonal to each other. Here we have three mutually orthogonal vectors in three dimensional space.
In the case of orthogonal planes as subspaces, we require that each vector of each pair is orthogonal to each vector of the other pair. This requires the dimension of the ambient space to be at least $\,2+2=4.$