## Proof for dot products

This is something that has been bothering me...

Given two vectors A and B

Is there a way to prove that A dot B = ABcosθ ?

I'm concerned with WHY this is the case... If anyone has a good proof that would be great.

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 Hey cytochrome. The proof for this is based on the cosine rule for triangles. Let A and B be the vectors you are considering. Now in vector terms we know A + C = B (following from head to tail of both vectors) which means that C = B - A and this means the length is the length of B - A (which you can use pythagoras rule for in n-dimensions). Now your cosine rule is C^2 = A^2 + B^2 - 2ABcos(theta). You know how to calculate lengths of all the vectors (using Pythagoras') so know collect the terms together and see what you get.
 It is a consequence of Cauchy-Schwarz inequality: ##\left |\left \langle a,b \right \rangle \right | \leq \left\|a\right\|\left\|b\right\|## Hence the ratio: ##cos\theta = \frac{\left \langle a,b \right \rangle}{\left\|a\right\|\left\|b\right\|}## Dot product is a inner product.

## Proof for dot products

I think he might mean how the definition in R^n is derived as opposed to something just being an inner product in general.

 what you on about? Cauchy-Schwarz inequality applies to any inner product space including ##\mathbb{R}^n##!
 I mean that = x1.y1 + x2.y2 + ... + xn.yn. (i.e. the actual definition not just an abstract one).
 Didn't I say dot product is inner product already? (edit: note I didn't say inner product is restricted to dot product only) The point here is not the dot product but rather the Cauchy-Schwarz inequality itself which applies to R^n if you take the inner product to be dot product. Besides, using what you mentioned as "cosine rule for triangle" is confusing for high dimension spaces.
 The length is known for arbitrary finite n through Pythagoras' theorem and the proof using lengths works in any dimension for R^n: it's a very simple proof since you only care about lengths of the triangle and it's very easy to understand (length is an invariant concept)
 Don't you realize that Cauchy-Schwarz inequality is at the very root of that "cosine rule"?
 This works for R2... I think it's a little more intuitive than the other proofs I've seen. Let a, b be two vectors. Then a / ||a||, b / ||b|| are to unit vectors. We can let a / ||a|| = and b / ||b|| = . then (a / ||a||) * (b / ||b||) = (cosm)(cosn)+(sinm)(sinn) = cos(m-n). The angle c between the vectors is m-n. So (a / ||a||) * (b / ||b||) = cos(c) and a*b= ||a|| ||b|| cos(c). Sorry for the readability.
 I know about the inequality, but I thought I made it very clear that I was talking about the actual specific definition (i.e. the formula to compute said quantity): I've said this quite a few times.

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 Quote by chiro I mean that = x1.y1 + x2.y2 + ... + xn.yn. (i.e. the actual definition not just an abstract one).
What makes you think that definition is any more "actual" than another? I've always though of "the length of the projection of u on v" as the basic definition of $u\cdot v$.

 Quote by HallsofIvy What makes you think that definition is any more "actual" than another? I've always though of "the length of the projection of u on v" as the basic definition of $u\cdot v$.
Well it is specific to defining A.B in R^n and the author wanted to prove the formula for R^n, then the above is a good on doing that.

I understand inner products are very general that follow specific axioms: I was talking about a very specific space (i.e. R^n).