Is there a command (or a combination of commands) that takes three points of a plane and outputs the normal vector of that plane? I'm told there is a ready way of automating this, but have been unable to find it.

  • $\begingroup$ Hyperplane maybe work. $\endgroup$
    – cvgmt
    Dec 22 '21 at 9:21
  • 1
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    – Syed
    Dec 22 '21 at 9:22
  • $\begingroup$ Similar to this, this and perhaps this. $\endgroup$
    – Syed
    Dec 22 '21 at 9:45

In 3D, "Cross" can be used to get a vector perpendicular to 2 given vectors.

Call the given vectors p1,p2,p3, then Cross[p2-p1,p3-p1] is a vector perpendicular to the plane through p1,p2,p3.

Here is an example:

{p1, p2, p3} = RandomReal[{-1, 1}, {3, 3}];
p4 = Cross[p2 - p1, p3 - p1];
  Arrow[{{0, 0, 0}, #}] & /@ {p1, p2, p3}, Red, Arrow[{{0, 0, 0}, p4}],
  Green, Opacity[0.5], InfinitePlane[{p1, p2, p3}]

enter image description here

  • $\begingroup$ I will usually recommend Daniel's method, but the method hinted in Roman's comment can be made to work: Normal[Last[CoefficientArrays[Det[PadRight[Prepend[{p1, p2, p3}, {x, y, z}], {4, 4}, 1]], {x, y, z}]]] $\endgroup$ Dec 22 '21 at 13:18

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