Proving Euler Angle Rotations of ZXZ spans SO(3)

Question

I am looking for a succinct proof that showcases a ZXZ combination of rotations can reach and orientation in SO(3). I know this combination is one of the 6 (or 12 if you include both intrinsic and extrinsic rotation), but the majority of proofs I find really get into the weeds. Is there something out there complete, yet brief?

Answer 1

There isn't much to prove. Each rotation in $SO(3)$ can be identified with a set of three mutually orthogonal unit vectors $\hat{\mathbf x}, \hat{\mathbf y}, \hat{\mathbf z}$, obtained by applying the rotation to the coordinate vectors $\hat{\mathbf i}, \hat{\mathbf j}, \hat{\mathbf k}$. Since the rotation must preserve orientation, $\hat{\mathbf x}, \hat{\mathbf y}, \hat{\mathbf z}$ follow the same relative relationships as $\hat{\mathbf i}, \hat{\mathbf j}, \hat{\mathbf k}$. In particular, $\hat{\mathbf y} = \hat{\mathbf z} \times \hat{\mathbf x}$.

1. A $Z$ rotation - that is, a rotation about the $z$ axis (the vector $\hat{\mathbf k}$) - can move any vector not on the $yz$-plane into that plane. Apply this to move $\hat{\mathbf z}$ into the $yz$-plane. Let $\hat{\mathbf x}_1, \hat{\mathbf y}_1, \hat{\mathbf z}_1$ be the results of applying this $Z$ rotation to the original vectors.
2. An $X$ rotation - about the $x$ axis, or vector $\hat{\mathbf i}$ - can rotate any non-zero vector in the $yz$-plane onto the positive $z$-axis. Do this for the vector $\hat{\mathbf z}_1$. Let $\hat{\mathbf x}_2, \hat{\mathbf y}_2, \hat{\mathbf z}_2$ be the results of applying this rotation to $\hat{\mathbf x}_1, \hat{\mathbf y}_1, \hat{\mathbf z}_1$. Now $\hat{\mathbf z}_2$ is a unit vector pointing in the direction of positive $z$-axis. This is the definition of $\hat{\mathbf k}$. So $\hat{\mathbf z}_2 = \hat{\mathbf k}$. Since $\hat{\mathbf x}_2, \hat{\mathbf y}_2$ are perpendicular to $\hat{\mathbf z}_2$, they must both lie in the $xy$-plane.
3. Another $Z$ rotation can rotate any non-zero vector in the $xy$-plane onto the positive $x$-axis. Apply this to $\hat{\mathbf x}_2$, and let $\hat{\mathbf x}_3, \hat{\mathbf y}_3, \hat{\mathbf z}_3$ be the results of applying this rotation to $\hat{\mathbf x}_2, \hat{\mathbf y}_2, \hat{\mathbf z}_2$. Since $Z$-rotations do not affect the $z$-axis, $\hat{\mathbf z}_3 = \hat{\mathbf z}_2 = \hat{\mathbf k}$. $\hat{\mathbf x}_3$ is a unit vector pointing along the positive $x$-axis, and thus must be $\hat{\mathbf i}$, and finally $\hat{\mathbf y}_3 = \hat{\mathbf z}_3 \times \hat{\mathbf x}_3 = \hat{\mathbf k} \times \hat{\mathbf i} = \hat{\mathbf j}$.

Combining these three rotations gives a result that rotates $\hat{\mathbf x}, \hat{\mathbf y}, \hat{\mathbf z}$ onto $\hat{\mathbf i}, \hat{\mathbf j}, \hat{\mathbf k}$. Just the opposite of what we need. But that is a simple matter of reversing the three component rotations and their order:

• Reverse the direction of the third rotation to get a $Z$ rotation taking $\hat{\mathbf i}, \hat{\mathbf j}, \hat{\mathbf k}$ to $\hat{\mathbf x}_2, \hat{\mathbf y}_2, \hat{\mathbf z}_2$.
• Reverse the direction of the second rotation to get an $X$ rotation taking $\hat{\mathbf x}_2, \hat{\mathbf y}_2, \hat{\mathbf z}_2$ to $\hat{\mathbf x}_1, \hat{\mathbf y}_1, \hat{\mathbf z}_1$.
• Reverse the direction of the first rotation to get a $Z$ rotation taking $\hat{\mathbf x}_1, \hat{\mathbf y}_1, \hat{\mathbf z}_1$ to $\hat{\mathbf x}, \hat{\mathbf y}, \hat{\mathbf z}$

The combination of those three rotations takes $\hat{\mathbf i}, \hat{\mathbf j}, \hat{\mathbf k}$ to $\hat{\mathbf x}, \hat{\mathbf y}, \hat{\mathbf z}$, and is thus the rotation in $SO(3)$ we needed.