Calculate normal vector to $2$-face of polytope in $\Bbb R^n$ I am trying to work through a divergence theorem application for a function integrated over an $n$-dimensional convex polytope, but I can't seem to figure out how to properly calculate the normal vectors to its surface. 
As an example, suppose my polytope lives in $\Bbb R^4$, and is defined as the hypersimplex: $$\left[0,1\right]^4 \cap \left\{ \boldsymbol{x}: \sum x_i = k \right\}$$ 
If $k \in (2,3)$, this shape is a tetrahedron with the vertices cut off, so now the vertices are replaced with triangular $2$-faces, and original triangular $2$-faces of the tetrahedron are now hexagons. I want to use the divergence theorem to integrate over the boundary (the surface of the hypersimplex), recognizing that this integral will be the sum of integrals over the $2$-faces.
What I am struggling with is how to find the normal unit vector to the $2$-faces in more than $3$ dimensions, and how to ensure that it points "outward" from the convex shape. 
Let me define one of the $2$-faces by its vertices so you can answer a specific question. Suppose $k = 2.5$. Then one of the triangular $2$-faces is defined by:
$$(0, 1, 1, 0.5),\ (0, 1, 0.5, 1),\ (0, 0.5, 1, 1)$$
How do I find the normal vector and how do I ensure it points outward?
 A: First off, you should be careful about applying the divergence theorem to a 3-dimensional object in $\mathbb{R}^4$.  You will have to make sure to project the vector field onto the hyperplane containing the object and then calculate its divergence.  Otherwise the vector field might diverge in the normal direction instead of out of the boundary.
It is also true that a 2-dimensional face has infinitely many normal vectors in $\mathbb{R}^4$, the same way that an edge has infinitely many normal vectors in $\mathbb{R}^3$.
What you want to do is find the vector which is normal to the face, and is also parallel to the hyperplane that contains the object.  As you mention in the comments, the vectors along two of the edges are 
$$
w_1 = v_2 - v_1 = (0,0,-0.5,0.5),\qquad w_2=v_3-v-1 = (0,-0.5,0,0.5)
$$
In addition, the normal vector to the whole polytope is
$$
n = (1,1,1,1).
$$
Thus, you want to find a vector in $\mathbb{R}^4$ which is perpendicular to all three of these.
There is something like cross product in $\mathbb{R}^4$ for solving this problem.  Given the three vectors, we can compute the "determinant"
$$
\left|\begin{matrix}\textbf{e}_1 & \textbf{e}_2 & \textbf{e}_3 & \textbf{e}_4 \\[3pt] 0 & 0 & -0.5 & 0.5 \\[3pt] 0 & -0.5 & 0 & 0.5 \\[3pt] 1 & 1 & 1 & 1\end{matrix}\right| \;=\; (-0.75,0.25,0.25,0.25).
$$
(This operation is actually a special case of the exterior product.)  As you can see, the resulting vector is perpendicular to $w_1$, $w_2$, and $n$, so this is the normal vector you are looking for.
