- #1

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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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- #1

- 166

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Given two vectors

Is there a way to prove that

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

- #2

chiro

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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.

- #3

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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.

##\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.

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- #4

chiro

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Cauchy-Schwarz inequality applies to any inner product space including ##\mathbb{R}^n##!

- #6

chiro

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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.

- #8

chiro

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Don't you realize that Cauchy-Schwarz inequality is at the very root of that "cosine rule"?

- #10

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Sorry for the readability.

- #11

chiro

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- #12

HallsofIvy

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What makes you think

- #13

chiro

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What makes you thinkthatdefinition 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].

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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