Show that these vectors lie in a common two-dimensional plane, How could I show that 4 vectors $v_1, ... v_4$ lie in a common two-dimensional plane in $R^4$?
This plane $P$ does not contain the origin.
The vectors are:
$v_1=(3,0,1,4)$
$v_2=(3, 2, -1, 6)$
$v_3=(3,1,0,5)$
$v_4=(2,0, -1,1)$
I also want to find a description of $P$ using two linear equalities:
$$P = \{\vec x: f(\vec x) =a ,g(\vec x)=b\}$$
where $f$ and $g$ are linear functionals $f(\vec x) = f_1x_1 + ... f_4x_4$ and so on.
I have tried putting the 4 vectors into a 4x4 matrix and row-reduced it a few times, both times confirming that this matrix has rank 3, so that doesn't seem to help me.
Any ideas are welcome.
Thanks,
 A: You can treat one of the vectors like the origin and show that the differences span a two-dimensional subspace.
So, we can arbitrarily pick $v_4$ as our frame of reference. Now we get three difference vectors 
\begin{align*}
d_1 &= v_1 - v_4 \\
d_2 &= v_2 - v_4 \\
d_3 &= v_3 - v_4, \\
\end{align*}
and want to show that they span a two-dimensional subspace. You can use whatever technique you like, at this point: Row reducing a matrix, or by expressing $d_3$ as a linear combination of $d_1$ and $d_2$, probably among other possibilities.
If you can express this translated subspace as $\{d : f^*(d) = 0 = g^*(d)\}$ using linear functionals $f^*$ and $g^*$, I believe the original plane will be given by $\{v : f^*(v) = f^*(v_4),\ \ g^*(v) = g^*(v_4)\}$, although I haven't thought about that before.
A: It suffices to show that the three vectors $v_2-v_1, v_3-v_1,v_4-v_1$ lie in the some two-dimensional plane (containing the origin). This is equivalent to the three vectors being linearly dependent:
$$a(v_2-v_1)+b(v_3-v_1)+c(v_4-v_1)=0.$$
Supposing that this translated plane is not one-dimensional, we may WLOG assume $v_2-v_1$ and $v_3-v_1$ are not scalar multiples of each other. Using Gram-Schmidt, we can find unit vectors $w$ and $w'$ that are orthogonal to each other and to everything in our plane. Then, our translated plane can be written as $\{v \mid w^\top v = w'^\top v= 0\}$. Then, translating back, our original plane is $\{v \mid w^\top v = w'^\top v = w^\top v_1\}$.

In your case, $v_2-v_1 = (0,2,0,2)$, $v_3-v_1=(0,1,-1,1)$, and $v_4-v_1 = (-1,0,-2,-3)$, which spans three dimensions, so unfortunately your original vectors do not lie in the same two-dimensional plane.
A: Here's another approach. Find the plane $P$ containing $v_1, v_2, v_3$ (this is guaranteed to exist); let $w$ be its normal vector. Now look at $(v_1-v_4)\cdot w$. If this equals zero, then $(v_1-v_4)\perp w$;  if you draw a picture, you can see why this means $v_4$ lies in $P$.

Basically, what we're doing is using the geometric fact that $v_1, . . . , v_4$ are coplanar iff there is a vector $w$ which is perpendicular to each of the pairwise differences $v_i-v_j$. Note that this means it's actually equivalent, mathematically, to the other answers; but you might find it a little more concrete (or not; YMMV).
