How do I determine the dihedral angle of two (bond-defined) planes? I'm not sure if this question would be more appropriate for Chemistry or Physics SE, but if so please forgive me. 
I have drawn the following picture; the spheres represent atoms and the lines connecting them represent chemical bonds.  

Another view:

I would like to determine two dihedral angles:


*

*the dihedral angle comprised of the vertices C-G-O-A.  (Or maybe, although I am not sure, I should state this as the dihedral angle between the bonds C-G and O-A.)

*the dihedral angle comprised of the vertices D-A-O-G.  (Or perhaps this should be the dihedral angle between D-A and O-G.


O is the origin.  The Cartesian coordinates of the vertices are:


*

*O = {0, 0, 0}

*A = {-1.2211, -0.705, 0}

*B = {1.2211, -0.705, 0}

*C = {0, 1.41, 0}

*D = {-1.2211, -2.115, 0}

*E = {1.2211, -2.115, 0}

*G = {0, 0.705, 1.2211}


One problem that I am encountering is that it is not clear to me how a bond between two atoms can define a plane, since I think that a plane is only uniquely defined by three noncollinear points.  Wikipedia says that the dihedral angle $\varphi_{AB}$ between two planes $A$ and $B$ is simply $\cos \varphi_{AB} = \textbf{n}_A \cdot \textbf{n}_B$ where $\textbf{n}_A$ and $\textbf{n}_B$ are the unit vectors normal to planes $A$ and $B$.  But it is not clear to me how to actually define $A$ and $B$ given only two vertices (i.e., two atoms) defining each plane.
Do you have any suggestions?
 A: Let's find the dihedral angle between the planes by taking the dot product of normal unit vectors Three points are needed to determine a plane, so in the chain of bonds C-G-O-A we'll take the elements three at a time to get the two planes: C-G-O and G-O-A.
Let us now calculate unit vectors $\mathbf{n}_{CGO}$ and $\mathbf{n}_{GOA}$ normal to these planes. First note that C, G, and O all have $x$-component $0$, so C-G-O lies in the $yz$-plane and we can take $\mathbf{n}_{CGO} = (1,0,0)$. For the other plane, we have:
$$\begin{align}
OG \times OA &= \big((0, 0, 0) - (0, 0.705, 1.2211)\big) \times \big((0, 0, 0) - (-1.2211, -0.705, 0)\big)\\
&=(0, -0.705, -1.2211) \times (1.2211, 0.705, 0)\\
&=(0.8609,-1.4911,0.8609)
\end{align}$$
Before making this a unit vector, let's divide by $0.8609$ to clean things up: this last vector is parallel to $(1,-1.73206,1)$, which I take to be $(1,-\sqrt{3},1)$. Interesting. Shrinking this to a unit vector:
$$\begin{align}
\mathbf{n}_{GOA} &= \frac{OG \times OA}{|OG \times OA|}\\
&=\frac{(1,-\sqrt{3},1)}{||(1,-\sqrt{3},1)||}\\
&=\frac{1}{\sqrt{5}}(1,-\sqrt{3},1)\\
\end{align}$$
Finally, the dihedral angle $\phi$ between our two planes satisfies $\cos{\phi} = \mathbf{n}_{CGO} \cdot \mathbf{n}_{GOA} = \frac{1}{\sqrt{5}}$. In other words, $\phi = \arccos(\frac{1}{\sqrt{5}}) = 1.1071$, or about $63.43^\circ$.
The dihedral angle for D-A-O-G can be calculated the same way.
