The 3D case

Now we try to describe the set of all 3D rotations by following the same general template as the 2D case. The matrix from (3.4) is extended from 2D to 3D, resulting in $ 9$ components:

$\displaystyle M = \begin{bmatrix}m_{11} & m_{12} & m_{13}  m_{21} & m_{22} & m_{23}  m_{31} & m_{32} & m_{33}  \end{bmatrix}.$ (3.14)

Thus, we start with $ 9$ DOFs and want to determine what matrices remain as valid rotations. Follow the same three rules from the 2D case. The columns must have unit length. For example, $ m_{11}^2 + m_{21}^2 + m_{31}^2 = 1$. This means that the components of each column must lie on a unit sphere. Thus, the unit-length rule reduces the DOFs from $ 9$ to $ 6$. By following the second rule to ensure perpendicular axes result, the pairwise inner products of the columns must be zero. For example, by choosing the first two columns, the constraint is

$\displaystyle m_{11} m_{12} + m_{21} m_{22} + m_{31} m_{32} = 0.$ (3.15)

We must also apply the rule to the remaining pairs: The second and third columns, and then the first and third columns. Each of these cases eliminates a DOF, resulting in only $ 3$ remaining DOFs. To avoid mirror images, the constraint $ \det M = 1$ is applied, which does not reduce the DOFs.

Finally, we arrive at a set of matrices that must satisfy the algebraic constraints; however, they unfortunately do not fall onto a nice circle or sphere. We only know that there are $ 3$ degrees of rotational freedom, which implies that it should be possible to pick three independent parameters for a 3D rotation, and then derive all $ 9$ elements of (3.14) from them.

Steven M LaValle 2020-01-06