# Intersection of a Convex Polyhedron and a Plane

Let $$X= \{x|a_j^Tx\leq b_j, j=1,...,m \}$$ be a polyhedron, and p is a point inside a polyhedron as shown below. I want to derive an expression for calculating an orthogonal plane from point p which intersect the polyhedron. The orthogonal plane is constructed perpendicular to the trajectory at the considered point, i.e., p

In other words, this is what I want "Intersection of a Convex Polyhedron and a Plane"

This figure was taken from here: https://demonstrations.wolfram.com/IntersectionOfAConvexPolyhedronAndAPlane/

Any help will be appreciated.

• I don't understand where you find a polyhedron:$\left \| C(x-a) \right \|^2=(x-a)^T(C^TC)(x-a)\le1$ defines in general the interior of an ellipsoid centered in $a$... Nov 20, 2020 at 14:05
• Besides, you have to define to what your "orthogonal plane" is orthogonal to ... Nov 20, 2020 at 14:07
• I don't know what is an orthogonal plane, but that doesn't matter. Any plane through that point will intersect that polyhedron (or ellipsoid, or whatever it is that you have). Nov 20, 2020 at 14:12
• sorry I updated question a bit (: hope now it's clear ? Nov 20, 2020 at 14:15
• You haven't reacted to my remark: it is definitely not a polyhedron unless you mean $C(x−a)\le 1$ without the squared norm... Nov 20, 2020 at 14:50