Angle between two planes in four dimensions Suppose I have two planes defined in 4D space, either in terms of
vectors spanning the planes, $X = t_1 A_1 + t_2 B_2$ and $X = t_3 A_3 + t_4 B_4$
(where $X$, $A$'s, and $B$'s are vectors with four elements and $t$'s are
scalars), or in terms of null space, $[C_1; D_1] X = 0$ and $[C_2; D_2] X = 0$
(where the matrices are $2$ x $4$ and $X$ has four elements). 
I understand
that these two planes generally intersect in just a single point
(unless the matrix $[C_1; D_1; C_2; D_2]$ is rank deficient). But is it
meaningful to ask what the angle is between the planes? If so, how
would it be computed? There is an explicit formula for the 3D case: simply the angle between the normals to the planes. Is there no equivalent explicit expression for the 4D case?
 A: Choose orthonormal bases $(E_1, E_2)$ and $(F_1, F_2)$ for the two planes. Then, the (hyper)volume of the parallelepiped spanned by the $E_1, E_2, F_1, F_2$ on the one hand is $\sin \theta$ where $\theta$ is the angle between the two planes, and on the other hand is (the absolute value of) the determinant of the matrix given by adjoining the four vectors:
$$\left\vert\det[E_1 \, E_2 \, F_1 \, F_2]\right\vert .$$
This quantity is independent of the choices of orthogonal bases, and in fact, we can even take the bases of the planes to be any that span parallelograms of unit area, so we don't need to produce an orthogonal basis.
If we start out with any bases $(E_i)$ or $(F_i)$, say ones that span areas $\lambda$ and $\mu$, we can always normalize them by rescaling vectors, but we might as well build this into our equation: The bases $(\lambda^{-1} E_1, E_2)$ and $(\mu^{-1} F_1, F_2)$ both span parallelograms of unit area, so for general bases the angle satisfies
$$\sin \theta = |\det[\lambda^{-1} E_1 \, E_2 \, \mu^{-1} F_1 \, F_2]| = \frac{|\det[E_1 \, E_2 \, F_1 \, F_2]|}{\lambda \mu}, $$
which we might write as
$$\color{#bf0000}{\boxed{\sin \theta = \dfrac{|\det[E_1 \, E_2 \, F_1 \, F_2]|}{|E_1 \wedge E_2| |F_1 \wedge F_2|}}} ,$$
where $|G_1 \wedge G_2|$ denotes the area of the parallelogram spanned by $G_1, G_2$.
This formula generalizes readily to formulas the angle between $k$-planes and $(n - k)$-planes in vector spaces of dimension $n$ (try this for the familiar situation $n = 2$, $k = 1$), and with just a little more work to finding the angle between $k$- and $l$- planes in vector spaces of dimension more than $k + l$.
Remark This leaves the matter of computing explicitly the areas $|G_1 \wedge G_2|$ of the parallelograms that the bases define. The area $A$ of the parallelogram defined by vectors $H_1 = (x_1, y_1), H_2 = (x_2, y_2)$ in the plane is $A = \left\vert\det \left(\begin{array}{cc} x_1 & x_2 \\ y_1 & y_2\end{array}\right)\right\vert = |x_1 y_2 - x_2 y_1|$, and its square is $A^2 = (x_1 y_2 - x_2 y_1)^2$, which we can rewrite as
$$A^2 = [(x, y) \cdot (x, y)] [(x', y') \cdot (x', y')] - [(x, y) \cdot (x', y')]^2 = (H_1 \cdot H_1) (H_2 \cdot H_2) - (H_1 \cdot H_2)^2 .$$
Now, the formula $A^2 = (H_1 \cdot H_1) (H_2 \cdot H_2) - (H_1 \cdot H_2)^2$ doesn't depend on coordinates, it just uses the Euclidean structure (namely the dot product $\cdot\,$), so it works just as well for computing the areas of the parallelograms in our original problem, that is, we may write
$$|G_1 \wedge G_2|^2 = (G_1 \cdot G_1) (G_2 \cdot G_2) - (G_1 \cdot G_2)^2$$
and then take square roots.
A: I wonder whether we should distinguish following two cases when defining angle between two planes (not intersecting) in dimension four.
Case one is when both planes $K$, $L$ belong to the same complex structure on $\mathbb R^4$. In such case for each vector $v$ in $K$ angle between $v$ and $L$ is the same. (The angle between vector $v$ and plane $L$ is defined as minimum angle between vectors $v$ and $w$ for $w$ in $L$). The examples are 
$ K=<[1,0,0,0], [0,1,0,0]>, L_1=<[1,1,1,1]/2, [-1,1,-1,1]/2>$
$    L_2:=<[t,0,1,-s]/2,[0,t,s,1]/2> $.
In this definition $t=\dfrac{\sqrt 5-1}{2}$ and $s=\dfrac{\sqrt 5+1}{2}$.
Case two is when both planes $K$, $L$ do not belong to the same complex structure. In this case for $v$ in $K$ the resulting angle between $v$ and $L$ can vary depending on choice of $v$. Examples are 
$K=<[1,0,0,0], [0,1,0,0]>, L=<[1,1,1,1]/2, [0,1,-1,0]/\sqrt 2 >$, 
$L_2=<[1,1,1,1]/2, [0,1,-s, t]/2>$. How to define angle in this case ? We may define it as minimum of angles between vectors $v$ and $w$ for $v$ in $K$ and $w$ in $L$. We may also define it as determinant of matrix $[v_1,v_2,w_1,w_2]$ where $v_1,v_2$ is orthonormal basis of $K$ and $w_1, w_2$ is orthonormal basis of $L$.
Interesting question is to ask about product of reflections in the planes. By reflection in the plane I mean matrix which reverse vectors in the plane and fix vectors perpendicular to the plane. If the angle between planes is $\dfrac{2 \pi}{k}$ then we expect that product of reflections has order $k$.
Of course I do not settle things in my answer. I rather asked another questions here in order to expand horizons $\begin {matrix}'''\\{..}\\{\smile} \end {matrix}$
A: $$cos\left (\theta \right ) = \widehat{A_1 \times B_2} \cdot \widehat{A_3 \times B_4}$$
Where cross product is extended through geometric algebra, and is computed as the vector whose elements are the coefficients to the linear combination
$$I'A\wedge B\wedge\cdots$$
which in the 4D vector case, becomes
$$
\begin{pmatrix}
A_w\\ 
A_x\\ 
A_y\\
A_z
\end{pmatrix}
\times
\begin{pmatrix}
B_w\\ 
B_x\\ 
B_y\\
B_z
\end{pmatrix} = \begin{pmatrix}
A_z B_y - A_y B_z\\
A_z B_x - A_x B_z\\ 
A_y B_x - A_x B_y\\ 
A_z B_w - A_w B_z\\ 
A_y B_w - A_w B_y\\ 
A_x B_w - A_w B_x 
\end{pmatrix}
\\
$$
A: Note that the angle between two planes in $\mathbb R^3$ is not the smallest angle between two vectors from either plane: that would always give an angle of $0$ for any two planes (since they intersect).  So that is not a good basis for a generalization.
The following alternative does generalise the concept of angle between planes in $\mathbb R^3$ to any two linear subspaces of $\mathbb R^n$ of equal dimension for $n \geq 1$. It also has the useful property that the angle is zero if and only if the subspaces are equal.
So let $A, B \subseteq \mathbb R^n$ be two non-zero linear subspaces of dimensions $0 < d \leq n$ and let $\pi_A, \pi_B$ be the orthogonal projections onto $A$ and $B$.  Define $\mu(A, B)$ as the smallest eigenvalue of the map $\pi_A \pi_B$ restricted to $A$. (Note that $\pi_A \pi_B$ is a symmetric positive semi-definite endomorphism of $A$.)  It turns out that
$$
\begin{eqnarray}
\mu(A, B) &\in& [0, 1] \\[1ex]
\mu(A, B) &=& \mu(B, A)
\end{eqnarray}
$$
for all pairs of subspaces.  Moreover, if we take $A, B$ to be lines (or planes) in $\mathbb R^3$ then $$\mu(A, B) = \cos^2(\alpha)$$ where $\alpha$ the angle between these lines (or planes).    So, in general, we could define the angle $\alpha$ between spaces $A, B$ in the same way: $\cos^2(\alpha) = \mu(A, B)$.
The function $\mu$ can be computed in the following way.  Let $M_A, M_B$ be $n \times d$ matrices such that their column spaces span $A$ and $B$ respectively and $$M_A^{\mathrm t} M_A = M_B^{\mathrm t} M_B = \mathbb 1_2.$$ (That is, their columns form orthonormal bases.) Then $\mu(A, B)$ is the smallest eigenvalue of the $d \times d$ matrix $$M_A^{\mathrm t} M_B M_B^{\mathrm t} M_A = (M_B^{\mathrm t} M_A)^{\mathrm t} M_B^{\mathrm t} M_A.$$
