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I have $r, d \in \mathbb{R}^n$. I want to find all $y \in \mathbb{R}^n$ such that $r \cdot y = 0$ and $d \cdot \langle y_1^2, \dots, y_n^2 \rangle = 0$.

The trivial $y = 0$ solution is not terribly interesting for my purposes.

Ideally, I'd like to be able to do this for all $r$ and $d$. In a pinch, though: $r$ and $d$ are sort of randomly generated, so we can assume $r$ and $d$ have no $0$ elements (which might make their intersection look weird) and be right with probability $1$.

Extra credit if it's doable in polynomial time.

Any advice?

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What does "find all $y$" mean, when there are infinitely many of them? –  Gerry Myerson Sep 24 '12 at 1:27
    
Good point. In the sense that if $y$ is a solution then $\lambda y$ is a solution, I only need one such $y$ per family. Do you expect infinite such families? If so, I'm interested in why - that might be helpful. –  GMB Sep 24 '12 at 1:34
    
Suppose you intersect $x=0$ with $x^2+y^2+z^2-w^2=0$. You get all solutions of $y^2+z^2=w^2$. For any fixed $w$, you'll have infinitely many $y,z$, e.g., infinitely many solutions $(0,y,z,1)$, all from different families, no? –  Gerry Myerson Sep 24 '12 at 1:43
    
Ah, yes, thanks. Along that vein, perhaps rotating the coordinate system until $r \cdot x = 0$ becomes $x_j = 0$ would be helpful, if there's no simpler way to solve this. –  GMB Sep 24 '12 at 1:47

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