Geometric Transformation for Convex Hull Reorientation

Problem Statement:

Reorient a convex hull (specifically a 3D rectangular prism defined by its vertices) in a global coordinate system such that:

1. The origin of the local coordinate system remains invariant (no translation of the origin).
2. A specific vector in the local frame, representing the recurrence mapping from Cin(0,0) to Cout(0,0) and defined as $\begin{bmatrix} 0 \\ 0 \\ k \end{bmatrix}$, is mapped to the vector $\begin{bmatrix} 0 \\ k \\ 0 \end{bmatrix}$ in the global frame.
3. All coordinates of the transformed vertices of the convex hull are non-negative.

The original rectangular prism is defined by the vertices: v0: (0, 0, 0) v1: (m, 0, 0) v2: (m, 0, k) v3: (0, 0, k) v4: (0, n, k) v5: (0, n, 0) v6: (m, n, 0) v7: (m, n, k)

Solution:

The required transformation can be achieved by a single rotation matrix that is a result of two sequential rotations: a -90 degree rotation around the x-axis followed by a -90 degree rotation around the y-axis.

1. Rotation around the x-axis by -90 degrees ($R_x(-90^\circ)$):

The rotation matrix is:

$$R_x(-90^\circ) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(-90^\circ) & -\sin(-90^\circ) \\ 0 & \sin(-90^\circ) & \cos(-90^\circ) \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}$$

2. Rotation around the y-axis by -90 degrees ($R_y(-90^\circ)$):

The rotation matrix is:

$$R_y(-90^\circ) = \begin{bmatrix} \cos(-90^\circ) & 0 & \sin(-90^\circ) \\ 0 & 1 & 0 \\ -\sin(-90^\circ) & 0 & \cos(-90^\circ) \end{bmatrix} = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$

3. Combined Rotation Matrix ($R$):

The combined rotation is obtained by multiplying the second rotation matrix by the first:

$$R = R_y(-90^\circ) \cdot R_x(-90^\circ) = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$

4. Verification of Synthesis Vector Mapping:

Applying the rotation matrix $R$ to the synthesis vector $\begin{bmatrix} 0 \\ 0 \\ k \end{bmatrix}$:

$$R \begin{bmatrix} 0 \\ 0 \\ k \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ k \end{bmatrix} = \begin{bmatrix} 0 \cdot 0 + 1 \cdot 0 + 0 \cdot k \\ 0 \cdot 0 + 0 \cdot 0 + 1 \cdot k \\ 1 \cdot 0 + 0 \cdot 0 + 0 \cdot k \end{bmatrix} = \begin{bmatrix} 0 \\ k \\ 0 \end{bmatrix}$$

The synthesis vector is mapped as required.

5. Transformation of Vertices:

Applying the rotation matrix $R$ to each vertex of the original prism:

v0: (0, 0, 0) -> v0’ = $(0, 0, 0)$ v1: (m, 0, 0) -> v1’ = $(0, 0, m)$ v2: (m, 0, k) -> v2’ = $(0, k, m)$ v3: (0, 0, k) -> v3’ = $(0, k, 0)$ v4: (0, n, k) -> v4’ = $(n, k, 0)$ v5: (0, n, 0) -> v5’ = $(n, 0, 0)$ v6: (m, n, 0) -> v6’ = $(n, 0, m)$ v7: (m, n, k) -> v7’ = $(n, k, m)$

6. Positive Coordinate Constraint:

Assuming the dimensions $m, n, k$ are positive, all the transformed vertices have non-negative coordinates.

Conclusion:

The single rotation matrix that achieves the desired reorientation, maps the synthesis vector $(0, 0, k)$ to $(0, k, 0)$, keeps the origin invariant, and results in all non-negative coordinates for the transformed rectangular prism is:

$$R = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$

This rotation corresponds to a specific reorientation of the coordinate axes. The original x-axis becomes the new z-axis, the original y-axis becomes the new x-axis, and the original z-axis becomes the new y-axis.