<s>[INST] Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
 [/INST] With spherical coordinate system, $r=\sqrt{6}$ and
$$z=-x-y=-\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi$$
so your parametric equation will be
$$
\begin{aligned}x&=\sqrt{6}\,\sin \theta \,\cos \varphi \\y&=\sqrt{6}\,\sin \theta \,\sin \varphi \\z&=\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi \end{aligned}$$
where $ θ ∈ [0, π], φ ∈ [0, 2π)$. 
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