Dot Product Derviation

The dot product or inner product in Euclidean Space $A\cdot B$ has two definitions:

1. Algebraically defined as: $$A \cdot B = \sum_{i=1}^{n}A_i \cdot B_i=A_1B_1 + A_2B_2 ... A_nB_n$$

2. Geometrically defined as: $$A \cdot B = \Vert A \Vert \Vert B \Vert \cos{\theta}$$

Is there a way to prove that they are equal other than simply saying they are defined so?

• Always when two definitions are around there ought to be a proof stating that the definitions are equivalent, when that's done you can use whatever definition that is most practical in your context. You shouldn't really have two definitions at the same time - one is a definition and the other should be presented as a theorem. – skyking Sep 3 '15 at 6:11
• So which one is a theorem? – ChaoSXDemon Sep 3 '15 at 6:13
• Depends on the choice of the author. One approach is to first put out (and prove) a theorem stating that always $\sum A_jB_j = |A||B|\cos\theta$ and then pick one as as the definition (or you could do it the other way around - it could be that the other is not needed or easily provable until far later). – skyking Sep 3 '15 at 6:16
• – Blue Sep 3 '15 at 6:41

Note that if $A$ and $B$ are vectors representing two sides of a triangle, then the third side is represented by vector $C=B-A$. Let the corresponding magnitudes of the sides be $a,b,c$
The cosine rule tells us that $c^2=a^2+b^2-2ab \cos \theta$
Computing the squares of the magnitudes from the components (using Pythagoras) we obtain $$2ab\cos \theta=\sum A_i^2+\sum B_i^2-\sum (B_i-A_i)^2=2\sum A_iB_i$$and the result follows.