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The question really is in the title. I know what it means if the dot product equals 0 but I find it interesting thinking what it means when it equals exactly 1 and can't seem to find anything online to enlighten me.

Thanks

moony
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    Nothing special, really. You get some relation between their lengths and to what degree they point in the same direction, but it's not anything really firm. – Arthur Mar 20 '17 at 23:15
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    I don't think it means anything in particular, but check this post for an intuition about the dot product. – Bobson Dugnutt Mar 20 '17 at 23:15
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    The value of the dot product has dimensions square length, so it means nothing without a reference pair of lengths to compare it to (namely the lengths of the original vectors) unless it is zero, because this statement does not depend on a choice of units. – Qiaochu Yuan Mar 20 '17 at 23:20
  • Thanks very much. I wish I could accept one of these as answers as you have answered my question. Thanks – moony Mar 20 '17 at 23:47

1 Answers1

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If you already know the vectors are both normalized (of length one), then the dot product equaling one means that the vectors are pointing in the same direction (which also means they're equal).

If you already know the vectors are pointing in the same direction, then the dot product equaling one means that the vector lengths are reciprocals of each other (vector b has its length as 1 divided by a's length). For example, 2D vectors of (2, 0) and (0.5, 0) have a dot product of 2 * 0.5 + 0 * 0 which is 1. Also, (1, 1) has a length of sqrt(2), and (0.5, 0.5) has a length of 1/sqrt(2), and the dot product is also 1.

If you don't already know anything about the vectors, you can't concretely say anything about this.

Aaron Franke
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    Your second statement is false, take the vectors $[1,1]$ and $[1/2,1/2]$; the dot product is $1$ but they are not "reciprocals" – Alex Mathers Oct 02 '19 at 01:21
  • @AlexMathers Good catch! That's what I get for not thinking up counterexamples... I think the actual rule is the lengths are reciprocal, I've edited my answer to reflect this. – Aaron Franke Oct 02 '19 at 01:36