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Proof that VxU=(determinant) derives from VxU=|V||U|sinαe

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data
    I need to proof that VxU=(determinant) starting from VxU=|V||U|sinαe

    2. Relevant equations
    VxU=|V||U|sinαe and what I'm aiming to is VxU=(uy⋅uz - uz⋅vy)i - (ux⋅vz - uz⋅vx)j + (ux⋅vy - uy⋅vx)k

    3. The attempt at a solution
    U x V = |U||V|Sinαe
    (U x V)^2 = |U|^2|V|^2 cos^2α - 1 e
    (U x V)^2 = |U|^2|V|^2 cos^2α - |U|^2|V|^2
    U x V = (ux + uy + uz) (vx + vy + vz)cos^2α - (ux + uy + uz) (vx + vy + vz) e

    This is where I get stuck. Can someone point me on the right direction? that cosine and e are disorienting me. Thanks.
     
  2. jcsd
  3. Jun 11, 2015 #2

    Ray Vickson

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    The magnitude of ##\vec{A} \times \vec{B}## is the area of a certain parallogram with sides ##\vec{A}, \, \vec{B}##. Can you find another formula for that area?
     
  4. Jun 11, 2015 #3

    Mark44

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    What is e in the formula you show?

    Also, your formula is incorrect. V x U is a vector, but what you have on the right side is a scalar, unless e happens to be a vector (which isn't explained).
    What does (U x V)2 mean? With vectors there are three kinds of multiplication: cross product, dot product, multiplication by a scalar.
    Where did cos2α = 1e come from? I don't know what this means.
     
  5. Jun 11, 2015 #4
    Besides |U||V|Sinα wouldnt that be |U x V|? which confuses me too because it means that |U||V|Sinα = |U x V| = U x V

    I'm working on it and post if there is any or not much progress...

     
  6. Jun 11, 2015 #5
    e is a unit vector.
    (U x V)2 I squared both sides
    sin2α = cos2α - 1 and here I should have written (cos2α - 1)e
     
  7. Jun 11, 2015 #6
    This is probably wrong anyway
     
  8. Jun 11, 2015 #7

    Mark44

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    In what direction?
    How do you square a vector? By that, I mean, which type of multiplication do you mean? I listed three.
    ##\sin^2(\alpha) = 1 - \cos^2(\alpha)##. You have ##\cos^2(\alpha) = 1##, which has the opposite sign.
     
  9. Jun 11, 2015 #8

    Ray Vickson

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    Google is your friend.
     
  10. Jun 11, 2015 #9
    Ups..

    The unit vector is in the direction of the normal vector I supose.
     
  11. Jun 11, 2015 #10
    Not getting really far on this. I do not understand why this forum makes a point on being so cryptic with the help provided. I have no doubts on the good intentions of its members but after asking a couple of questions related with homework I couldn't help get frustrated for the meager hints (not just in this thread but in others that me and my friends have posted). I'm not trying to slack, I'm not asking you to do my homework, I don't want to copy and paste the answer from google, I need help when I get stuck. I agree you shouldn't give me the answer but c'mon, it is so impractical the way help is shared here. Basically I learned nothing but to copy the thing from google today.

    Thank you Ray Vickson and Mark44 anyway.
     
  12. Jun 11, 2015 #11

    SteamKing

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    Sorry you feel this way, but these are the rules governing HW help here at PF. You knew this going into the process when you signed up. If the HW Helpers don't follow the rules, they get awarded infraction points.

    In any event, I think Ray Vickson gave you an excellent hint in Post #2, which you didn't seem to follow up on in the succeeding posts.

    FWIW, I've never thought of the determinant form of the cross product as anything more than a handy mnemonic which can be used to find the components of the product. I've never thought that anyone would go to the trouble to construct a full mathematical proof.

    http://math.oregonstate.edu/home/pr...estStudyGuides/vcalc/crossprod/crossprod.html
     
  13. Jun 12, 2015 #12

    Fredrik

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    The problem is already confusing at this point. The standard definition of the cross product is ##A\times B=(A_2B_3-A_3B_2)i+(A_1B_3-B_1A_3)j+(A_1B_2-B_1A_2)k##, and if you use this, you only need to verify that this is exactly what you get from the determinant formula. This is trivial, so I guess you're not supposed to use the standard definition, and instead take ##V\times U=|V||U|(\sin\alpha) e## as the definition of the cross product. But what then is ##e##?

    I think the fact that you haven't properly answered that, is the main reason why there's been so little progress in this thread. You said nothing about it until Mark44 asked you. You gave the incomplete answer that it's a unit vector. So Mark44 had to ask the follow-up question "in what direction?". You gave an incomplete answer to that too. I assume you meant normal to the plane spanned by ##V## and ##U##, but there are two of those. I'm guessing that your answer to the obvious follow-up question would be "the one given by the right-hand rule".

    But we have little use for a wordy answer like that. We need a formula for ##e##. The problem is that the obvious formula is the equation ##V\times U=|V||U|(\sin\alpha) e##. This looks more like a definition of ##e## than a definition of ##V\times U##. But if we are to take it as the definition of ##e##, we need another definition of ##V\times U##.

    If the equation ##V\times U=|V||U|(\sin\alpha) e## is supposed to be the definition of ##e##, and we are allowed to use the standard definition of the cross product, the problem is trivial. And if that equation is supposed to be the definition of the cross product, the problem doesn't seem solvable at all without a proper definition of ##e##.
     
    Last edited: Jun 12, 2015
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