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.

enter image description here

Another view:

enter image description here

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?

  • $\begingroup$ You say that the six points whose coordinates are given all lie on the $x$-$y$-plane, where the $z$-coordinate is zero. But you do not specify the location of G. From your diagram, it seems to have $y$-coordinate zero. Without any further information about G, I don’t see what can be done. $\endgroup$
    – Lubin
    Jun 28, 2012 at 22:34
  • $\begingroup$ @Lubin I messed up and forgot to include the coordinates of G. I'm sorry about that. I have now edited my post to include G = {0, 0.705, 1.2211}. Thank you! $\endgroup$
    – Andrew
    Jun 28, 2012 at 22:57
  • 2
    $\begingroup$ For the first angle, it looks like you want the dihedral angle between AOG and OGC. You can find the normals of the planes by calculating the cross products OA $\times$ OG and GC $\times$ GO. The second angle is calculated the same way. $\endgroup$
    – Théophile
    Jun 28, 2012 at 23:55
  • 1
    $\begingroup$ @Théophile, I think that should be an answer, not a comment. $\endgroup$
    – user856
    Jun 29, 2012 at 0:32
  • $\begingroup$ @RahulNarain: Done! I just didn't have time to write out the calculations yesterday. Thanks for the reminder. $\endgroup$
    – Théophile
    Jun 30, 2012 at 17:08

1 Answer 1


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.

  • $\begingroup$ Maybe the coefficient should be $\frac{1}{\sqrt{5}}$, not $\frac{1}{5}$? $\endgroup$
    – Andrew
    Jul 2, 2012 at 0:28
  • $\begingroup$ Also, how do you know that $\textbf{n}_{CGO} = (1, 0, 0)$ instead of $(-1, 0, 0)$? It depends on the order in which you take the cross product. $\endgroup$
    – Andrew
    Jul 2, 2012 at 0:48
  • 1
    $\begingroup$ @Andrew: Thanks, I've fixed the norm; it should indeed be $\frac{1}{\sqrt{5}}$. As for the direction of the normal, the nice thing is that it doesn't matter: it suffices to find a normal of the plane. If we had used $(-1,0,0)$ instead of $(1,0,0)$, then the final calculation would have given $\phi = 116.56^\circ = -63.43^\circ$. This is the supplementary angle of $63.43^\circ$ and gives the same information. $\endgroup$
    – Théophile
    Jul 2, 2012 at 1:02
  • $\begingroup$ Thank you! But if I look down the G-O bond (see my edit to my question above), the angle between the planes CGO and GOA looks greater than $90^{\circ}$, not less than $90^{\circ}$. Why does it seem like $\phi = 116.56^{\circ}$ makes more physical sense? Maybe I am completely missing the point, though. Thanks for your time and help. $\endgroup$
    – Andrew
    Jul 2, 2012 at 1:18
  • 1
    $\begingroup$ @Andrew: Indeed, the angle you're looking for will be $\phi = 116.56^\circ$; you must choose from the calculation that I gave whether to take the smaller or the larger of the two angles. If you have a drawing already, as in your example, it should be clear which to take. Or are you writing a computer program of some kind? $\endgroup$
    – Théophile
    Jul 5, 2012 at 21:29

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