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Two impossible crossproduct problems

  1. Sep 6, 2011 #1
    Hi, I really need some help here.. Vectors remind me why I hate geomtery.

    problem 1: Prove that |UxV|2 = |U|2+|V|2 - (U*V)2

    How can I prove that these two are equal without spending 1 hour using algebra? Maybe there is some geometry quirk that I'm not seeing?

    problem 2: We have two vectors A=[1,1,1] and B=[1,2,3]

    Find a vector C so that AxB = AxC, where C =! B.

    I tried using algebra on this but I just ended up with crazy expressions for Cx, Cy and Cz where each of them were dependent on the others.

    So.. Is there some other way? All help is appreciated =)
     
    Last edited: Sep 6, 2011
  2. jcsd
  3. Sep 6, 2011 #2
    I would also like to point out that I haven't learned yet about things like the ratio between sine(x) and cosine(x), so pls if possible tell me this can be solved by simple algebra or by using geometry?
     
    Last edited: Sep 6, 2011
  4. Sep 6, 2011 #3

    Mute

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    For the second one, since it just asks for a vector C not equal to B, the easiest thing to do would be to choose C = B + V, where V is some vector such that A x V = 0. Do you know what kinds of vectors have that property?
     
  5. Sep 6, 2011 #4

    Fredrik

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    Gold Member

    Do you know how to rewrite U·V and |U×V| in terms of |U|, |V| and the angle between U and V. You need to use those identities, and also [itex]\cos^2x+\sin^2x=1[/itex].
     
    Last edited: Sep 6, 2011
  6. Sep 6, 2011 #5
    ahh, so AxC = Ax(B + V) = AxB + AxV where AxV=0.

    Very cunning. Yes, V equals any vector which is pararell with A. Thank you 4 the help!

    Sure, |U|2*|V|2 - |U|2*|V|2*cos(x)2= |U|2*|V|2(1-cos(x)2)=|U|2*|V|2(sine(x)2)=|UxV|2

    correct? tho we haven't learned about the 1=cos(x)^2 + sine(x)^2 trick in my maths class so maybe there is another way to prove it?

    thank u very much 4 the help anyways
     
    Last edited: Sep 6, 2011
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