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For arbitrary $d \in \mathbb{R}^n$, I am interested in the set of $x \in \mathbb{R}^n$ such that $d \cdot \langle x_1^2, \dots, x_n^2\rangle = 0$. If $d > 0$, then $x = 0$ is the unique solution. If $d \ge 0$, then the solution set is a $k$-dimensional flat where $x_i$ can be anything iff $d_i = 0$ (otherwise, $x_i = 0$).

What about when $d$ contains both positive and negative elements? I suspect it's a hyperplane, on the observation that if $x$ is a solution, then $\lambda x$ is a solution for any $\lambda \in \mathbb{R}$. But does the figure always have enough dimension to be a hyperplane, or is it sometimes lower-dimensional?

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What you have in general is a type of cone, not a hyperplane. Consider the case $n=3$ with $d=\langle 1,1,-1\rangle$ where you have the cone $x^2 + y^2 = z^2$. – Robert Israel Sep 23 '12 at 19:00

If $d$ has both positive and negative elements, this variety always contains a manifold of dimension $n-1$, because the Jacobian of $F(x_1,\ldots,x_n) = \sum_j d_j x_j^2$ is $(2 d_1 x_1, \ldots, 2 d_n x_n)$, which is nonzero when some $d_i x_i \ne 0$. Use the Implicit Function Theorem.
There are always solutions with some $d_i x_i \ne 0$, e.g. if $d_1 > 0 > d_2$ take $x_1 = \sqrt{-d_2}$, $x_2 = \sqrt{d_1}$, $x_i = 0$ otherwise.

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