Let $a,b,c,d$ be reals such that $a^2+b^2+c^2>d^2$.
How do I prove that $\{(x,y,z)\in S^2: ax+by+cz=d\}$ is infinite?
This is geometrically trivial, but I'm stuck at proving it rigorously..
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1$S^2 = {(x,y,z): x^2+y^2+z^2 = 1}$ – DeepSea Jul 03 '15 at 00:31
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1Hint: $(x,y,z) = s (a,b,c) + \bf v$ where ${\bf v} \cdot (a,b,c) = 0$, ... – Robert Israel Jul 03 '15 at 00:49
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2And please: there's a world of difference between "an infinite solution" and "infinitely many solutions". – Robert Israel Jul 03 '15 at 00:50
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A way to prove this can be found here – Winther Jul 03 '15 at 01:04
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Both $D^3$ and the plane are convex sets, hence their intersection is a convex set and its boundary is connected, hence if you prove that there are at least two different points in the intersection between the sphere and the plane, there are an infinite number of them, since the intersection has to be connected.
An alternative way is to use the implicit function theorem, or the fact that both the plane and the sphere are invariant with respect to a rotation around $(a,b,c)$, so their intersection has to be invariant, too.
Jack D'Aurizio
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