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General Proof of Cross/Dot Product

  1. Mar 3, 2013 #1
    how do you prove the distributive quality of the cross and dot products?
    thanks very much!
     
  2. jcsd
  3. Mar 3, 2013 #2

    SteamKing

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    Set em up and expand em out.
     
  4. Mar 3, 2013 #3
    geometry derived

    how would you derive the concept of geometry from first principles?
     
  5. Mar 3, 2013 #4

    micromass

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    What do you mean?? What does this have to do with the OP?
     
  6. Mar 3, 2013 #5

    Fredrik

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    Is that your way of saying that the answer you got is inadequate? It is not. What he's suggesting is just that you use the definitions of the dot product and the cross product to rewrite both the left-hand side and the right-hand side of the equality you want to prove. Then you just compare the results. Are they equal?

    The way this usually goes is that we give you a hint like that, and then you either use it to solve the problem, or show us your attempt up to the point where you get stuck. Then we can give you a new hint on how to get past that point.
     
  7. Mar 4, 2013 #6
    |A||B+C|costheta3=|A|B|costheta1+|A||C|costheta2

    |A||B+C|sintheta3n=|A||B|sintheta1n+|A||C|sintheta2n


    I was referring to a different problem.... I mean Valenzia refers to deriving geometry from general concepts and it was fascinating to see an exposition of that
     
  8. Mar 4, 2013 #7

    micromass

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    If you want to be able to derive geometry from basic axioms, then I highly recommend the book "Geometry: Euclid and beyond" by Hartshorne. See: https://www.amazon.com/Geometry-Euclid-Beyond-Robin-Hartshorne/dp/0387986502

    In that book, he starts from basic axioms, and he derives all of geometry and trigonometry (or he makes it clear how to do it). Furthermore, he constructs [itex]\mathbb{R}[/itex] from purely geometric axioms. And later on he deals with notions such as areas.
     
    Last edited by a moderator: May 6, 2017
  9. Mar 4, 2013 #8
    thanks!
     
  10. Mar 4, 2013 #9

    micromass

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    Yes, but that's because "deriving geometry from first principles" is far away from modern mathematics. Most mathematicians consider these kind of questions "solved" and do other things.
     
  11. Mar 4, 2013 #10
    Euclid set the standard for mathematical exposition that is still followed today (definition theorem proof). But, otherwise, what micromass said.
     
  12. Mar 4, 2013 #11

    Fredrik

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    I was thinking that you should use the definition $$A\cdot B=\sum_{i=1}^3 A_i B_i$$ of the dot product, and a similar definition of the cross product. This seems easier to me than to use the formulas ##A\cdot B=|A||B|\cos\theta## and ##|A\times B|=|A||B|\sin\theta##. If you want to continue with the approach you have started, you will have to figure out how the angles you called ##\theta_1,\theta_2,\theta_3## are related. In the case of the cross product, you have the additional problem that it's not sufficient to prove that ##\left|A\times(B+C)\right| =\left|A\times B+A\times C\right|##. You also need to verify that ##A\times(B+C)## is in the same direction as ##A\times B+A\times C##.

    So I suggest using the definitions directly instead of using these two theorems about the angle between the vectors. How does your book define the cross product?
     
  13. Mar 7, 2013 #12
    as you stated in component form, AXB=(AyBz-AzBy)xhat....it seems worth memorizing since the matrix notation is a bit hard to read.... there's supposed to be a more general proof
    http://en.wikipedia.org/wiki/Cross_product
    by the way how do I use the nice latex notation?
     
  14. Mar 7, 2013 #13

    Fredrik

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    If you use that definition to rewrite ##A\times (B+C)## and ##A\times B+A\times C##, you can easily see that the two are equal.

    Yes, it's worth the effort to commit some version of the definition to memory. If you understand how to compute a determinant by cofactor expansion along the first row, then that determinant gives you an easy way to remember the formula that you partially typed. I like to use the Levi-Civita symbol ##\varepsilon_{ijk}## myself.

    The notation ##\varepsilon_{ijk}## is defined for all ##i,j,k\in\{1,2,3\}## by saying that ##\varepsilon_{123}=1## and that the sign changes when you swap two of the indices. This means e.g. that ##\varepsilon_{312}=-\varepsilon_{132}=\varepsilon_{123}=1##, and that ##\varepsilon_{ijk}=0## when two (or three) of the indices are equal. (For example, we have ##\varepsilon_{113}=-\varepsilon_{113}## and this implies that ##\varepsilon_{113}=0##).

    It seems a bit awkward at first, but once you get used to this notation, it makes everything much easier. The definition of the cross product can be written as
    $$A\times B=\sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 e_i \varepsilon_{ijk} B_j C_k,$$ but no one actually writes it like that. People who use this symbol also use the Einstein summation convention, which is to not write any summation sigmas when we're working with expressions where there is always a sum over those indices that appear twice. So I would write the above as
    $$A\times B=e_i \varepsilon_{ijk} B_j C_k.$$ This notation makes it very easy to remember the definition of the cross product. It also makes many proofs trivial. For example,
    $$A\times (B+C) =e_i \varepsilon_{ijk}A_j(B+C)_k =e_i \varepsilon_{ijk}A_j (B_k+C_k) =e_i \varepsilon_{ijk}A_j B_k +e_i \varepsilon_{ijk}A_j C_k =A\times B+A\times C.$$ If you learn a few basic results about the properties of the Levi-Civita symbol, then all the basic properties of the cross product are very easy to prove. Unfortunately, results like $$\varepsilon_{ijk}\varepsilon_{ilm} =\delta_{jl}\delta_{km} -\delta_{jm}\delta_{lk}$$ are a bit tricky to prove. So this approach may not save you time now, but it will in the long run.

    More general than what? If you use the definition (any version of it), you can prove that for all ##A,B,C\in\mathbb R^3##, we have ##A\times (B+C)=A\times B + A\times C##.

    If you click on the quote button next to one of my posts, you can see how I did it. If you want to know more, check out this FAQ post about it.
     
  15. Mar 7, 2013 #14
    thanks very much!

    in G.E. Hays Vector and Tensor Analysis there are proofs of the general case, chapter 1, section 7 and 8
     
  16. Mar 7, 2013 #15

    Fredrik

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    I had a quick look at these proofs. They prove the same statements that we're discussing here.

    I'm not a fan of the book's definitions:
    1. Definitions. Quantities which have magnitude only are called scalars. The following are examples: mass, distance, area, volume. A scalar can be represented by a number with an associated sign, which indicates its magnitude to some convenient scale. There are quantities which have not only magnitude but also direction. The following are examples: force, displacement of a point, velocity of a point, acceleration of a point. Such quantities are called vectors if they obey a certain law of addition set forth in § 2 below.​
    These are old and weird definitions. In this century, a vector is a member of a vector space. There's a field associated with each vector space, and the members of that field are called scalars.
     
  17. Mar 9, 2013 #16
    Thanks so much! Sorry I was just asking a question about the general proof in Hay, not one about the easy one. perhaps the proofs in Rudin Real Analysis aren't great?

    as you know the theta proof is with 3 triangles- construct an obtuse triangle with B+C, B, C and a larger right triangle with A as the bottom side, then carve out 2 smaller triangles next to B+C with a rectangle next to them and 3 theta angles to the left bottom corner of the 3 triangles. |B+C|costheta3=|B|costheta1+|C|costheta2
     
    Last edited: Mar 9, 2013
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