= Orthogonal projection of a circle between planes
{title2=$b/a=|\boldsymbol n_1\cdot\boldsymbol n_2|$}
Under <orthogonal projection> from one plane onto another, a <circle> of <radius> $R$ becomes an <ellipse> with semiaxes $R$ and $R|\cos\alpha|$, where $\alpha$ is the <angle> between the planes' unit <normal vectors>. To prove this, choose one in-plane <unit vector> parallel to the planes' intersection and the other perpendicular to it. The first vector is unchanged by the projection; the second has projected length $|\cos\alpha|$ and remains perpendicular to the first. Parametrizing the circle in that <orthonormal basis> therefore gives projected coordinates $(R\cos t,R\cos\alpha\sin t)$. Parallel planes have a circular image, and perpendicular planes give the degenerate case of a line segment. An arbitrary center simply projects to the center of the image.
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