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Proof for dot products

  1. Sep 11, 2012 #1
    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.
  2. jcsd
  3. Sep 11, 2012 #2


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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.
  4. Sep 12, 2012 #3
    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.
    Last edited: Sep 12, 2012
  5. Sep 12, 2012 #4


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    I think he might mean how the definition in R^n is derived as opposed to something just being an inner product in general.
  6. Sep 12, 2012 #5
    what you on about?

    Cauchy-Schwarz inequality applies to any inner product space including ##\mathbb{R}^n##!
  7. Sep 12, 2012 #6


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    I mean that <x,y> = x1.y1 + x2.y2 + ... + xn.yn. (i.e. the actual definition not just an abstract one).
  8. Sep 12, 2012 #7
    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.
  9. Sep 12, 2012 #8


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    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)
  10. Sep 12, 2012 #9
    Don't you realize that Cauchy-Schwarz inequality is at the very root of that "cosine rule"?
  11. Sep 12, 2012 #10
    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|| = <cosm, sinm> and b / ||b|| = <cosn, sinn>. 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.
  12. Sep 13, 2012 #11


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    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.
  13. Sep 15, 2012 #12


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    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 [itex]u\cdot v[/itex].
  14. Sep 15, 2012 #13


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    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).

    I've already outlined this above.

    If the author doesn't want to consider a specific space (like R^n) then OK, but if they do then that's another thing.
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