OK, I screwed up in class explaining PLANES, so I will now do
it right in this web page!  It is really very easy...
my brain was clearly on drugs to screw this up.
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First, the set of (x,y,z) forming a PLANE is the set of x,y,z
satisfying an equation of the form
    a x + b y + c z  = 1                           (*)
or as vectors (a,b,c) . (x,y,z) = 1.
Here a,b,c are 3 constants that determine that plane, different a,b,c
give different planes.  Example:  2x+3y+4z=1 is a plane.

Second, the x-intercept of this plane is
the x such that
    (a,b,c) . (x,0,0) = 1
that is
      a x = 1,  that is   x = 1/a.
Similarly the y-intercept is  yi = 1/b  and the z-intercept is zi=1/c.

AXIS INTERCEPTS:   xi=1/a,  yi=1/b,  zi=1/c.
Example 2x+3y+4z=1 is a plane with axis-intercepts 1/2, 1/3, 1/4.

Finally, the direction
(a,b,c)
is PERPENDICULAR (also called "normal") to this plane,
and in fact the shortest vector from the origin (0,0,0) to the plane
is                    2
   (a,b,c) / |(a,b,c)|  .

Proof: Clearly this vector (regarded as a point)
is on the plane because if you substitute
(x,y,z)=this
into (*), you get 1 because of the length rescaling term.

Now clearly it is (regarded as a direction)
perpendicular to the plane because you can check that 
      (1/a, -1/b, 0)
and
      (1/a, 0,   -1/c)
and
      (0,   1/b,  -1/c)
(all of which are vectors lying within the plane - since they
are the x-intercept minus the y-intercept location, etc.)
all are perpendiacular to (a,b,c)  by performing the dot product and
getting 1-1=0 in each case.  

Finally, due to the previous 2 claims it must be the shortest vector
from (0,0,0) to the plane.
QED.

Example: (2,3,4) is normal to the plane 2x+3y+4z=1.

To find angles between planes: find the angle between their normal
vectors.  If these two vectors are A, B, then
                 A . B
   cos(angle) = --------
                |A|  |B|

To find the angle between a plane P and a vector V:
90 degrees minus the angle between the normal-vector-to-P
and V.