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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

enter image description here

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

enter image description here

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

Any help will be appreciated.

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  • $\begingroup$ 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 $a$... $\endgroup$
    – Jean Marie
    Nov 20, 2020 at 14:05
  • $\begingroup$ Besides, you have to define to what your "orthogonal plane" is orthogonal to ... $\endgroup$
    – Jean Marie
    Nov 20, 2020 at 14:07
  • $\begingroup$ 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). $\endgroup$ Nov 20, 2020 at 14:12
  • $\begingroup$ sorry I updated question a bit (: hope now it's clear ? $\endgroup$
    – GPrathap
    Nov 20, 2020 at 14:15
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    $\begingroup$ 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... $\endgroup$
    – Jean Marie
    Nov 20, 2020 at 14:50

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