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What are the possible $v\in R^3$ such that a plane consisting of $x\in R^3$ satisfying $x\cdot v = 0$, intersects the hyperboloid given by $\{(x,y,z)\in R^3|x^2+y^2-z^2=-1, z>0\}$? 

I think $x\cdot v=0$ describes a plane through the origin...

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The solid is an "upward" bowl shape whose vertex is on the $z$-axis at $z=1$. The cross-section in the $z$-$y$ plane of the solid is the upper branch of the hyperbola $z^2-y^2=1$. The set of lines in the $z$-$y$ plane passing through the origin that intersect the solid are precisely those that make an angle $\theta$ with respect to the positive $z$-axis with measure less than $45^\circ$. This is because the upper branch of the hyberbola $z^2-y^2=1$ is asymptotic to, but always above, the line $z=y$.

By symmetry (the solid is the upper branch of the hyperbola in the $z$-$y$ plane revolved about the $z$-axis), the set of lines passing through the origin that intersect the solid are those lying in the interior of the cone centered about the $z$-axis with vertex at the origin that makes an angle of $45^\circ$ with respect to the positive $z$-axis.

You can now determine the set of admissible vectors $v$ from this (see the comments below).

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Since the OP is asking for a normal vector, the admissible $v$ lies outside that cone. In other words, $v = (x,y,z)$ is admissible precisely if $x^2+y^2 > z^2$. – WimC Feb 26 '12 at 17:58
All $v$ outside the double-napped cone (the cone above and its reflection through the $x$-$y$ plane), I think. – David Mitra Feb 26 '12 at 18:13
Thank you, David. – Jeremy Feb 26 '12 at 19:31

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