Is this method of determining if 4 points are coplanar, valid? I had a test today in school and a question asked me to prove that four points A,B,C, and D were coplanar. I made vector equations for lines AB and CD respectively and showed, by the simulataneous equation method, that these two lines intersect. I then went on to explain that this means that the lines are coplanar and therefore, the points A,B,C, and D are coplanar. I haven't seen this method being used, but it occurred to me during the exam. Is this method valid?
 A: Yes it is. Nicely done coming up with that yourself.
If you have two intersecting lines, you always have a plane where both of them lie in. Then all your four points will lie on that plane too.
But beware: If you find that the lines $AB$ and $CD$ do not intersect, this does not automatically mean that the points are not coplanar.
A: Yes, this will work providing the lines aren't parallel (in which case just choose different pairs: if all such lines are parallel, the points are all collinear). It turns out it's equivalent to showing that the minimum distance between a plane and a point is zero.

Start with $A,B,C$. Any three non-collinear points determine a plane. If $A,B,C$ are collinear, then $A,B,D$ determine a plane (or a line if they are all collinear) and $C$ lies in it automatically.
So suppose that $A,B,C$ determine a plane. We want to show that $D$ lies in this plane. A plane through three points is given by the points
$$ sA + (1-s-t)B +tC $$
for all real $s,t$ (this is a two-dimensional linear space that contains $A,B,C$, so it must be the plane). To check that $D$ is in the plane is to find $s,t$ so that
$$ D = sA+(1-s-t)B+tC. $$
On the other hand, the lines $AB$ and $CD$ are given by
$$ \lambda A + (1-\lambda)B, \\
\mu C + (1-\mu) D $$
for real $\lambda $ and $\mu$ (for the same reasons: these are one-dimensional linear spaces that contain $A,B$ and $C,D$). To say these intersect means that there are $\lambda,\mu$ so that
$$ \lambda A + (1-\lambda)B = \mu C + (1-\mu) D. $$
If $\mu=1$, $C$ lies on $AB$ anyway, and we're back in the collinear case, so suppose that $\mu \neq 1$. Then
$$ D = \frac{\lambda}{1-\mu} A + \frac{1-\lambda}{1-\mu} B + \frac{-\mu}{1-\mu} C = \frac{\lambda}{1-\mu} A + \left( 1 - \frac{-\mu}{1-\mu} -\frac{\lambda}{1-\mu} \right) B + \frac{-\mu}{1-\mu} C, $$
which is in the form we found above. Therefore if there is a solution to these equations, it shows that $D$ lies in the plane $ABC$.
What if there is no solution? There are two possibilities: either the lines are parallel (in which case $D$ still lies in the plane $ABC$) or the lines are skew (so $D$ does not lie in the plane $ABC$). One checks for parallelism using the cross product: $(A-B) \times (C-D)=0$ if and only if $A=B$ or $C=D$ or $A-B \parallel C-D$.
A: I would figure out the distances between all six pairs of points and, treating these as the edges of a trerahedron, use the Cayley-Menger determinant (http://mathworld.wolfram.com/Tetrahedron.html) to get the volume.  Zero volume = coplanar.
