Wikipedia:Reference desk/Archives/Mathematics/2021 May 14

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May 14

Angles of cube-plane intersections

How does one calculate the angle (on the plane) of the corners of the shape created by the intersection of a plane and a geometrical body (for example a cube). I have some ideas but I have no clue how to verify my result, other than constructing a model cube or something. 95.168.118.77 (talk) 16:14, 14 May 2021 (UTC)

The usual way to do this is to coordinatize, find the vectors in the directions of the two rays that form your angle, and use the dot product to find the angle between them. --JBL (talk) 17:36, 14 May 2021 (UTC)

Presumably, the body is a polyhedron, bounded by planar faces, and the shape of the intersection is then a polygon. Only orientations are important; if the body is translated by a small amount but not rotated, the directions of the sides of the polygon remain the same. As JBL indicates, if you have the directions of the sides of an angle, you can determine its size. But how to find these directions? The sides of the polygon lie in the intersection of two planes: the plane of the question, and the plane in which some face of the polyhedron is contained. The equation of a plane in space can be written in the form of a linear equation $ax+by+cz+d=0,$ and then the vector $(a,b,c)$ is a normal. All lines in the plane are perpendicular to this normal, so the intersection of two planes, which is a line lying in both planes, is perpendicular to each of the normals of the two planes. So we have two vectors $(a_{0},b_{0},c_{0})$ and $(a_{1},b_{1},c_{1}),$ and we need to find a third vector that is perpendicular to both. This is given by the cross product; the article gives several ways to compute it, which (if I have made no mistake) all result in $(b_{0}c_{1}{-}c_{0}b_{1},c_{0}a_{1}{-}a_{0}c_{1},a_{0}b_{1}{-}b_{0}a_{1}).$  --Lambiam 01:13, 15 May 2021 (UTC)