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Vector Proof

  1. Jun 2, 2005 #1
    For non-zero vectors v and w show that ||v||w + ||w||v is orthogonal to ||v||w - ||w||v

    I am baffled by this problem, I know that a way to solve it would be to say that the first dot the second = 0, but I am just unable to prove it for all cases from there. I know it is true though. Any ideas?

    btw... don't feel inclined to urgently help, this was a problem on an already turned in exam :yuck:
     
  2. jcsd
  3. Jun 2, 2005 #2

    Hurkyl

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    What cases?

    Anyways, if you show your work, we could find the step you're missing.
     
  4. Jun 2, 2005 #3
    For all cases.

    I will have to redo my work, since I turned it in (it was an exam problem). I will have it up in a bit.

    I basically said that vector v = <v1, v2, v3> and w = <w1, w2, w3> and ||v|| = a and ||w|| = b

    Then,


    ||v||w + ||w||v = a(<w1, w2, w3>) + b(<v1, v2, v3>)

    and

    ||v||w - ||w||v = a(<w1, w2, w3>) - b(<v1, v2, v3>)

    Then the dot product of the two will equal

    (aw1 + bv1)(aw1 - bv1) + (aw2 + bv2)(aw2 - bv2) + (aw3 + bv3)(aw3 - bv3) , which must equal 0. That means that aw1 = bv1 ... aw3 = bv3, and got stuck there. I will write out my work in a few, need to eat my pizza now though :smile:
     
  5. Jun 2, 2005 #4

    Hurkyl

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    No it doesn't!

    Just try expanding it. (Or just doing vector algebra without turning to coordinates!)
     
  6. Jun 2, 2005 #5
    Wait, sorry, I meant that if I can prove that aw1 = bv1, etc, then I would have the proof. Would I not?

    When I tried expanding it I got 0 = (something like) aw1^2 + aw2^2 .... Something that would only work if both of the vectors were 0 vectors, which does not satisfy the problem.

    What do you mean by vector algebra without turning to coordinates? How would that work? I talked to my teacher, and he said that there is a geometric proof of it, but I do not see it.
     
  7. Jun 2, 2005 #6

    Hurkyl

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    Nope -- reread what you're trying to prove.


    'a' isn't an arbitrary constant...


    The same way ordinary algebra works -- manipulate the equation using the algebraic properties of vectors. For instance, dot products are distributive. You also know how the dot product relates to the norm.
     
  8. Jun 2, 2005 #7
    Hmm, ok I will try it some other ways. I think I see what you are saying about my way not working, and about how to do it algebraically. No one post the answer :smile:
     
    Last edited: Jun 2, 2005
  9. Jun 2, 2005 #8
    "Dot products are distributive" Does this mean that [tex](\mathbf(a) + \mathbf(b)) \cdot (\mathbf(c) - \mathbf(d)) = \mathbf(a) \cdot \mathbf(c) - \mathbf(a) \cdot \mathbf(d) + \mathbf(b) \cdot \mathbf(c) - \mathbf(b) \cdot \mathbf(d) [/tex] ?
     
    Last edited: Jun 2, 2005
  10. Jun 2, 2005 #9

    Hurkyl

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    Yep.

    (Okay, technically, it means a(b+c) = ab + ac, but you can use this to prove the identity you quoted, in exactly the same way you do with ordinary multiplication)

    It might help to have the list of all allowed operations handy -- the axioms of a vector space, and the axioms of an inner product.
     
  11. Jun 2, 2005 #10
    Ok, I thought it worked due to a(b+c) = ab + ac, thanks.

    What are axioms of a vector space? and the axioms of an inner product?
     
  12. Jun 2, 2005 #11

    Hurkyl

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    They should be collected in your textbook. For example, in my book, the axioms of a vector space are in the section on "Abstract Vetor Spaces". I don't know your book, but they should be in there. :smile:

    wikipedia (en.wikipedia.org) probably has a nice table of them.
     
  13. Jun 2, 2005 #12
    Ok I figured out the entire proof algebraically, thanks for the help :smile:

    Hurkyl, my text does not having anything about axioms, nor has my teacher said anything about them. Note, though, I am not taking vector analysis, I am taking calc 3.
     
  14. Jun 2, 2005 #13

    Hurkyl

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    Aah, I see. I had assumed you were in a linear algebra course!

    This proof is a nice example of why vector algebra is so important -- proofs are shorter and cleaner. Also, many facts are naturally expressable as vectors.

    As such, since the algebra is much simpler, and coordinate free, it makes it easier to focus on the geometric aspects of the problems.
     
  15. Jun 3, 2005 #14

    mathwonk

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    even in cvalc 3 the book will tell you what proeprties of vector operations are allowed. but it is hard to believe anyone needs to read them as they are so obvious. i.e. multiplication is distributive over addition, addition is commutative, and inner products at least are commutative. addition and multiplication are associative where that makes sense.
     
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