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I need to find some "nice" parametrization of intersection of sphere $x^2 + y^2 + z^2 = 1$ and a plane $Ax + By + Cz = 0$. I know that the curve we get is an ellipse, but have no idea how to parametrize it

My attempt: Well if we assume $C \not = 0$, then we can see that $z = -\left(\frac{A}{C}x + \frac{B}{C}y \right)$, by pluging that in the first equation, we get $$\left(1+ \frac{A^2}{C^2}\right)x^2 + \left(1+ \frac{B^2}{C^2}\right)y^2 + \frac{2AB}{C^2}xy = 1$$ Which i know is an ellipse, but not sure how to parametrize it.

José Carlos Santos
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Nasal
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  • https://math.stackexchange.com/questions/305894/parametrization-for-intersection-of-sphere-and-plane - is a similar question to this! Check it out. – Taylor Rendon Jun 24 '20 at 16:28

1 Answers1

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Yes, it is an ellipse. But you can say more: it is a circle.

Take a vector $v$ of the plane whose norm is $1$. Now, let$$w=v\times\left(\frac A{\sqrt{A^2+B^2+C^2}},\frac B{\sqrt{A^2+B^2+C^2}},\frac C{\sqrt{A^2+B^2+C^2}}\right).$$Then $w$ also belongs to the plane and its norm is also equal to $1$. So, both $v$ and $w$ belong to the unit sphere. Now, take the parametrization$$\begin{array}{ccc}[0,2\pi]&\longrightarrow&\Bbb R^3\\\theta&\mapsto&\cos(\theta)v+\sin(\theta)w.\end{array}$$

José Carlos Santos
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