Consider a given vector $a$ and scalar $d$. What is the set $X$ such that for any $x \in X$ their dot product equals $d$ : $\forall x \in X: x \cdot a = d$ ?
What is the locus such that any vector from it has a given dot product with the given vector?
2
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linear-algebra
analytic-geometry
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4It'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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0@Qiaochu Yuan: What is meant by "a*d" ? – 2011-01-27
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0the scalar product of $a$ and the scalar $d/|a|^2$. – 2011-01-27
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0Yeah, got it. It describes the intersection point. – 2011-01-27
1 Answers
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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 $a$.