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A plane can be defined by a point and the normal vector, but these have to be reworked before feeding to InfinitePlane

enter image description here

I could find two vectors perpendicular to the given normal, but they would be arbitrary and I think I should be able to just use the point and normal vector that I have.

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  • $\begingroup$ You may want to use Hyperplane instead. $\endgroup$ – user1747134 Mar 22 '19 at 13:24
  • $\begingroup$ @user, Hyperplane[] is mentioned in the OP's own answer. $\endgroup$ – J. M.'s technical difficulties Mar 22 '19 at 13:42
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A plane can be defined by a point and the normal vector

This is only true for 3-dimensional space. InfinitePlane is meant to give a 2-dimensional plane no matter the embedding dimension. In higher dimensions, giving a point and normal is not sufficient to define a 2D plane.

Use Hyperplane,

graphic = EntityValue[
    Entity["Chemical", "1,2Dichlorobenzene"], "StickMoleculePlot"]
plane = Hyperplane[{15.78`, 67.73`, -71.86`}, {-106.51`, -21.78`, -43.46`}]

Show[ graphic, Graphics3D @ plane]

enter image description here

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  • $\begingroup$ Given any plane is there a way to extract its normal vector? $\endgroup$ – user13892 Feb 17 at 22:05
  • $\begingroup$ @user, that depends on how the plane is specified, but usually, yes. $\endgroup$ – J. M.'s technical difficulties May 18 at 16:56

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