For every convex 3-D polyhedron, there is a dual polyhedron with the roles of corners and faces swapped. Let us compute the dual polyhedron of the cube. Start with a representation of the cube with the origin as the center.

| (2.1) |

Now rewrite this in matrix form.

| (2.2) |

| (2.3) |

| (2.4) |

| (2.5) |

In general, if a bounded convex polyhedron is given by a set of inequalities of the form , then the vertices of its dual are exactly the columns of the matrix . The convex hull of these six vertices defines an octahedron. Using the same method as before, compute the bounding planes for this octahedron.

| (2.6) |

| (2.7) |

| (2.8) |

| (2.9) |

| (2.10) |

| (2.11) |

The first two inequalities represent that projection of the octahedron onto the axis:

| (2.12) |

Together with the following four inequalities, which involve only and but not , this defines the projection of the octahedron onto the -plane:

| (2.13) |

The following plot shows this projection (blue) together with the four boundary lines.

Finally, the last 8 inequalities, i.e., the ones containing , define the bounding planes of the octahedron.

| (2.14) |

The following 3-D plot displays the tetrahedron as well as the last of these 8 bounding planes. Rotate the plot with the mouse in order to see that the plane actually touches one of the faces.