How can I find the point on an ellipsoid whose normal passes through a given point?

  • Thread starter golmschenk
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  • #1
golmschenk
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Right now I'm running this with a brute force program which takes points on an ellipsoid and checks the distance to the point, slightly readjusts, and keeps moving toward the minimum, but it takes far to long for the mass amount of points I want to run through the program. Is there an equation I can use to find the line?

The question more specifically is the following. I have the parameteric equation for an ellipsoid. I have the location of a point. I want to find the shortest distance from the surface of the ellipsoid to the point. The point could be on the inside or the outside of the ellipsoid. I currently have a brute force program solving this but would like an equation or something else cleaner/faster to solve this. Is there a better way I could be doing this? Thanks for your time.
 

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  • #2
HallsofIvy
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Suppose the equation of the ellipsoid is
[tex]\frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{c^2}= 1[/tex]

Then the gradient at the point [itex](x_0,y_0, z_0)[/itex] is
[tex]\frac{2}{a^2}x_0\vec{i}+ \frac{2}{b^2}y_0\vec{j}+ \frac{2}{c^2}y_0\vec{k}[/tex]

The line through given point (u, v, w), parallel to that vector is
[tex]x= \frac{2}{a^2}x_0t+ u[/tex]
[tex]y= \frac{2}{b^2}y_0t+ v[/tex]
[tex]z= \frac{2}{c^2}z_0t+ w[/tex]

Solve the three equations
[tex]x_0= \frac{2}{a^2}x_0t+ u[/tex]
[tex]y_0= \frac{2}{b^2}y_0t+ v[/tex]
[tex]z_0= \frac{2}{c^2}z_0t+ w[/tex]
for [itex]x_0[/itex], [itex]y_0[/itex], and [itex]z_0[/itex] to find the point at which the shortest line (i.e. the perpendicular line) from the point intersects the ellipsoid, the find the distance between (u, v, w) and [itex](x_0, y_0, z_0)[/itex].
 
  • #3
golmschenk
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Fantastic! Thanks! I knew there had to be something like this, but wasn't finding it.
 

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