Intersection of 2D planes in 4D space If I had a four dimensional space, in which I embedded two planes, what possible intersections could they have?
Constructing a Plane
To give this more context, consider the following. If I had a 4d tuple (w, x, y, z) with no restriction, it can assume any position in 4D space. 
By imposing one restriction, and then another, we can reduce the degrees of freedom by two, such that I end up with a 2D plane in 4D space, we can write this pair of restrictions in matrix form as:
$$ \begin{bmatrix} p_1 & p_2 & p_3 & p_4 \\ q_1 & q_2 & q_3 & q_4  \end{bmatrix}
\begin{bmatrix} w \\x \\ y\\ z  \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2  \end{bmatrix}
 $$
Now if this forms a plane, then the rank of the matrix should be two dimensional, as there will only be two linearly independent column vectors. I believe this is a condition for the plane to be two dimensional.
Intersecting Planes
Now I'd like to consider the result when I intersect that plane with the plane represented by:
$$ \begin{bmatrix} r_1 & r_2 & r_3 & r_4 \\ s_1 & s_2 & s_3 & s_4  \end{bmatrix}
\begin{bmatrix} w \\x \\ y\\ z  \end{bmatrix} = \begin{bmatrix} c_3 \\ c_4  \end{bmatrix}
 $$
I've already established that this will display weird behaviour. For example, if we take the xy plane and the wz planes and intersect them, there is only one solution (the zero vector), indicating that they intersect at a plane. 
Getting Insights
If we constructed this final matrix, what restrictions can we apply knowing that the original matrix had a rank of 2?
$$ \begin{bmatrix} p_1 & p_2 & p_3 & p_4 \\ q_1 & q_2 & q_3 & q_4 \\r_1 & r_2 & r_3 & r_4 \\ s_1 & s_2 & s_3 & s_4  \end{bmatrix}
\begin{bmatrix} w \\x \\ y\\ z  \end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\\ c_3 \\ c_4  \end{bmatrix}
 $$
 A: Let $\mathbf{n}, \mathbf{m}$ be two linearly independent vectors in $\mathbb{R}^4$. A plane $P$ is given by 
$$\mathbf{n}\cdot (\mathbf{r}-\mathbf{r}_0)=
\mathbf{m}\cdot (\mathbf{r}-\mathbf{r}_0)=0
$$
and the second plane $P'$ is given by
$$\mathbf{n}'\cdot (\mathbf{r}-\mathbf{r}_0')=
\mathbf{m}'\cdot (\mathbf{r}-\mathbf{r}_0')=0
$$
note that $\mathbf{r}_0$ and $\mathbf{r}'_0$ can be changed to any other point on the plane and the equations still describe the same plane. Hence if the intersection is non-empty we can take $\mathbf{r}_0=\mathbf{r}_0'$. Then we only need to find the solutions to
$$
A\cdot (\mathbf{r}-\mathbf{r}_0):=
\begin{pmatrix}
\mathbf{n}\\
\mathbf{m}\\
\mathbf{n}'\\
\mathbf{m}'
\end{pmatrix}
\cdot
(\mathbf{r}-\mathbf{r}_0)=0
$$
Note that without loss of generality we can choose $\mathbf{n}', \mathbf{m}'$ to either not lie in $V=\mathrm{span}(\mathbf{n},\mathbf{m})$, or only one of them lies there or both. 
Now if $\mathbf{n}', \mathbf{m}'$ do not lie in the subspace spanned by $\mathbf{n},\mathbf{m}$ then this matrix is invertible, meaning the intersection is $\mathbf{r}_0$ (a point). 
If one of $\mathbf{n}',\mathbf{m}'$ is in the span of $\mathbf{n},\mathbf{m}$ then the kernel of this linear transformation is 1-dimensional and the intersection becomes line.
If both $\mathbf{n}', \mathbf{m}'$ lie in $V$ since they also share a common point $\mathbf{r}_0$ by assumption then the two planes coincide. You should exclude this possibility if your original planes are distinct.
So finally we need to understand when the intersection is empty (we assumed it is non-empty above). This only happens if the planes are parallel. If they are not parallel then the rank of $A$ is at least 3. You can investigate that
$$
A\mathbf{r} = \begin{pmatrix}\mathbf{n}\cdot \mathbf{r}_0\\
\mathbf{m}\cdot \mathbf{r}_0
\\
\mathbf{n}'\cdot \mathbf{r}'_0\\
\mathbf{m}'\cdot \mathbf{r}'_0\end{pmatrix}
$$
has a solution if rank $A$ is at least 3. So we must have rank $A$ at most 2. But then $A$ comes from the equations of $P,P'$ so it has rank at least two. This means rank $A$ is necessarily $2$, and hence the planes are parallel.
A: I'll answer with what I think is a more intuitive solution.
All the possible options for two planes in R4:
I'll put examples where A and B (and C) are planes in R4 (x, y, z, t).
1. Non-parallel, with no intersection
Imagine two non-parallel planes in 3D, which would obviously intersect, and now fix the 4th dimension differently for each of them. Now they don't intersect anymore.
This is similar to what happens with non-parallel lines that do not intersect in 3D.
Example:A: {x=0; t=0}; B: {y=0; t=1}; C: {z=0; t=-1};
B and C are at a distance 1 from A, and of 2 between them, but none of them are parallel.
2. Parallel
Just as in 3D.  Displace plane A by any vector and you get a parallel plane B.  As a curiosity, you can now displace in two dimensions.
Example:A: {x=0; t=0}; B: {x=0; t=1}; C: {x=1; t=0};
Both B and C are parallel to A (and themselves too), and both are at a distance of 1 from A, and of sqrt(2) between them.
3. Intersection in a point
This would be the generic case of an intersection between two planes in 4D (and any higher D, actually).
Example:A: {z=0; t=0}; B: {x=0; y=0};
You can think of this example as:A: a plane that exists at a single instant in time.
B: a line that exists all the time.
Now you can visualize the intersection in your head, and it should become obvious that they intersect in a single point: (0, 0, 0, 0).
4. Intersection in a line
This is a special case of intersection, where planes intersect in a line, but still are different planes.
Example:A: {x=0; t=0}; B: {y=0; t=0};
This example is just a 3D intersection of planes, fixing the 4th D as 0.
5. Coincident
Two planes can be the same, as in 3D.
Example:A: {x=0; t=0}; B: {x=0; t=0};
