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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). –  Qiaochu Yuan Jan 27 '11 at 15:01
    
@Qiaochu Yuan: What is meant by "a*d" ? –  mbaitoff Jan 27 '11 at 16:12
    
the scalar product of $a$ and the scalar $d/|a|^2$. –  Arturo Magidin Jan 27 '11 at 16:13
    
Yeah, got it. It describes the intersection point. –  mbaitoff Jan 27 '11 at 16:15
    
@Qiaochu Yuan: Can you post your comment as an answer, so I can accept it? –  mbaitoff Jan 27 '11 at 16:17

1 Answer 1

up vote 2 down vote accepted

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 $a$.

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