How can we plot four dimensional planes and how the planes intersect? I am well concerned about the three dimensional planes but don't know how to plot four dimensional planes. So help me plotting $3$ four dimensional planes $u+v+w+z=6,u+w+z=4,u+w=2$. Regards
 A: Take one coordinate as time, let's say $t = v$. 
The planes $u+w+z = 4$, $u+w=2$ are fixed in $\mathbb{R}^3$, independent of time.
While as time progress, the plane $u + w + z = 6 - t$ moves till it intersects (at time instant $t = 2$).
A: Let the last variable be represented by time.

A: Trick:
All three equations contain the term $u+v$. Let us call it $t$.
Then
$$\begin{align}&t+z+v&=6\\
&t+z&=4\\
&t&=2\end{align}$$
is an ordinary 3D problem, giving the single point $t=2,z=2,v=2$, which is actually the line $u+v=2,z=2,v=2$.
(In fact, we were projecting 4D onto 3D, i.e. looking "in the direction of $u+v$".)
A: Illustrations of three-dimensional planes and their intersections
are actually only two-dimensional drawings.
The image you see (collected by the retina of your eye)
also is two-dimensional.
You accept the idea that the third dimension was projected onto the image
and infer the structure it must have provided before
the projection.
You can also project four dimensions onto two, like this:

This figure depicts the hyperplane $u+w+z=4$ and the triangle
$(4,0,0,0)$, $(0,0,4,0)$, $(0,0,0,4)$ where it intersects the
$uvw$-, $vwz$-, and $uvz$-hyperplanes.
It depicts the hyperplane $u+w+z=6$ and the tetrahedron
$(6,0,0,0)$, $(0,6,0,0)$, $(0,0,6,0)$, $(0,0,0,6)$
where it intersects the 
$uvw$-, $vwz$-, $uwz$-, and $uvz$-hyperplanes.
It also depicts the triangle $(4,2,0,0)$, $(0,2,4,0)$, $(0,2,0,4)$
in the plane of intersection of those two hyperplanes.
The shaded regions of course are only parts of each hyperplane;
in drawings of three-dimensional planes, if we colored in the
entire plane it would fill the picture with a
background color without adding any information, and the
same is true for these hyperplanes.
If we add the third hyperplane to the picture, the common
intersection of all three is the line containing the points
$(2,2,0,2)$ and $(0,2,2,2)$.
But it's better to start a new drawing for that intersection;
this one already has more going on than one can easily keep track of
(except possibly for some individuals who are
exceptionally good at four-dimensional visualization).
