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Consider a given vector $a$ and scalar $d$. What is the set $X$ such that for any $x \in X$ their dot product equals $d$ : $\forall x \in X: x \cdot a = d$ ?

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    It's the unique hyperplane orthogonal to a and passing through a * d/|a|^2 (assuming your vectors are real).2011-01-27
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    @Qiaochu Yuan: What is meant by "a*d" ?2011-01-27
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    the scalar product of $a$ and the scalar $d/|a|^2$.2011-01-27
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    Yeah, got it. It describes the intersection point.2011-01-27

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It's easy to write down one point $p = \frac{a * d}{|a|^2}$ in this set. For any other point $q$, we have $(p - q) \cdot a = 0$, so the set of vectors $p - q$ is precisely the set of vectors orthogonal to $a$.