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I know that the intersection of an ellipsoid and a plane is an ellipse. However, how can I derive the conditions for an ellipsoid and a plane not to intersect? Suppose I have a plane $$Ax+By+Cz+D=0$$ and I have an ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2} +\frac{z^2}{c^2}=1$$ How I can write down the conditions that they won't intersect? I was thinking that they will not intersect if for any given point $(x,y,z)$ the plane is parallel to the tangent parallel to the ellipsoid at $(x,y,z)$. However, how will $D$ come into play? Do we use $D$ to make sure that the tangent hyperplane and the ellipsoid don't coincide with one another? If so, how can I impose this condition?

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  • $\begingroup$ The ellipsoid is bounded, so as you translate a tangent plane, it will intersect the ellipsoid for a while, and then it won't. Think of the tangent plane at one of the poles, and the plane at the opposite pole. Parallel planes between will intersect the ellipsoid, but other parallel planes won't. $\endgroup$ – saulspatz Feb 14 '18 at 3:12
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The plane does not intersect the ellipsoid iff its pole is in the interior of the ellipsoid. In homogeneous coordinates, this point is $[Aa^2:Bb^2:Cc^2:-D]$. Plugging this into the equation of the ellipse leads to the inequality $$(Aa)^2+(Bb)^2+(Cc)^2 \lt D^2.$$


In a comment to another answer, you ask about the distance from this plane to the ellipsoid. This is equal to the distance between the plane and a parallel tangent plane on the same side of the ellipse. This is fairly easily computed by using the ellipsoid’s dual: a plane with equation $Px+Qy+Rz+S = 0$ is tangent to the ellipsoid iff the coefficients of the equation satisfy the dual conic equation $$P^2a^2+Q^2b^2+R^2c^2-S^2=0.$$ So, the two tangent planes parallel to $Ax+By+Cz+D=0$ have a constant term of $\pm\sqrt{(Aa)^2+(Bb)^2+(Cc)^2}$ instead of $D$. The distances between these tangent planes and the reference plane are then $${\left|D\pm\sqrt{(Aa)^2+(Bb)^2+(Cc)^2}\right| \over \sqrt{A^2+B^2+C^2}}.$$ Choose the least of the two values that result. Since this ellipsoid is centered on the origin, that will be given by the solution for $S$ that has the same sign as $D$.

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  • $\begingroup$ Can you explain what you mean by "its pole is in the interior of the ellipsoid"? $\endgroup$ – user1691278 Feb 14 '18 at 5:21
  • $\begingroup$ @user1691278 See en.m.wikipedia.org/wiki/Pole_and_polar. $\endgroup$ – amd Feb 14 '18 at 5:48
  • $\begingroup$ How did you get the distance formula? I'm not sure what a dual is. $\endgroup$ – user1691278 Feb 14 '18 at 7:25
  • $\begingroup$ @user1691278 The constant tern in the equation $Ax+By+Cz+D=0$ of a plane is the (signed) distance of the plane from the origin times the length of the normal vector $(A,B,C)$. The distance between two planes is then just the absolute value of the difference of their distances to the origin, i.e., the difference of the normalized constant terms in their equations. $\endgroup$ – amd Feb 14 '18 at 7:47
  • $\begingroup$ @user1691278 As for the dual conic, there’s a brief description here. For more detail, see www-m10.ma.tum.de/foswiki/pub/Lehre/ProjektiveGeometrieWS0607/…. The latter also covers pole/polar relationships. $\endgroup$ – amd Feb 14 '18 at 7:49
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The idea is to perform a linear scaling of the space such that the ellipsoid becomes a unit sphere $x^2+y^2+z^2=1$. The two objects intersect before this scaling iff they do after it. $x$ becomes $ax$, $y$ becomes $by$ and $z$ becomes $cz$: $$Aax+Bby+Ccz+D=0$$ The (new) plane and sphere do not intersect iff the origin's distance from the plane is more than 1: $$\left|\frac D{\sqrt{(Aa)^2+(Bb)^2+(Cc)^2}}\right|>1$$ $$D^2>(Aa)^2+(Bb)^2+(Cc)^2$$ They have a tangential intersection iff this value is 1 ($=$ instead of $>$ in the second equation) and a normal elliptical intersection iff this value is less than 1 ($<$).

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  • $\begingroup$ Is there a way to derive conditions with equalities? Because after defining these conditions I need to minimize the distance between such a plane and the ellipsoid. I'm not sure how to tackle it with inequalities. $\endgroup$ – user1691278 Feb 14 '18 at 3:32
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    $\begingroup$ @user1691278 See here for calculating the minimal distance. In particular, with the original plane and ellipsoid you work out the point on the ellipsoid that has normal vector parallel to the plane's normal. There are two points; pick the one that gives the shortest distance. $\endgroup$ – Parcly Taxel Feb 14 '18 at 3:38

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