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I am looking for a convenient parametrisation $\tau:I\to\mathbb{R}^3$ of the set (sphere/plane intersection) $$ C=\{(x,y,z)\in\mathbb{R}^3\,|\,x+y+z=1,x^2+y^2+z^2=1\}$$ where $I$ is an interval of $\mathbb{R}$. By convenient I should mean that with such $\tau$ the computation of the length of the curve $C$ is straightforward.

user431632
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  • What have you tried, effort, context? Rotate to a coordinate system where the plane is of the form $z=k$. – Ninad Munshi Dec 16 '21 at 21:17
  • So far, presumably in the direction of your suggestion, I was trying to describe the intersection (circle) with some angle from the centre of the sphere and along the normal to the plane passing through the sphere's centre. How is your rotation implemented? and what would then be the parametrisation? – user431632 Dec 16 '21 at 21:22
  • Isn't the intersection of a sphere and a plane always a circle or a point? (not sure if this helps, but just askin smth to addon) –  Dec 16 '21 at 21:32
  • Precisely. I'm trying to conveniently parametrise such a circle. – user431632 Dec 16 '21 at 21:34

1 Answers1

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Set

$$\begin{cases}u&=\dfrac{x+y+z}{\sqrt3},\\v&=\dfrac{x-y}{\sqrt2},\\w&=\dfrac{x+y-2z}{\sqrt{6}}.\end{cases}$$

As you can check, $$u^2+v^2+w^2=x^2+y^2+z^2=1$$ and you intersect a unit sphere and the plane

$$u=\frac1{\sqrt3}.$$

Hence a parameterization is

$$v=\sqrt\frac23\cos t,w=\sqrt\frac23\sin t$$ and $$ds=\sqrt\frac23dt.$$

As the first transformation is orthogonal, it is easy to invert.