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Homework Help: Orthogonal vector question

  1. Mar 10, 2014 #1
    1. The problem statement, all variables and given/known data

    vector A = 3U-V
    vector B = U+2V
    U and V are vectors
    |U| = 3|V|
    Given that vector A and vector B are perpendicular vectors, find the angle between vector U and vector V.

    2. Relevant equations

    A*B = |A||B|cos(∠AB)
    A*A = |A|^2

    3. The attempt at a solution

    Since A and B are perpendicular to each other that means that the dot product will equate zero because cos 90 deg = 0.
    So substituting in the vectors I end up with something like

    (3U-V)*(U+2V) = 0 = 3U*U + 5U*V - 2V*V

    Given that any vector dot product itself gives you the magnitude of the vector squared and that we are trying to figure out the angle UV:
    U*U = |U|^2
    U*V = |U||V|cosθ
    V*V = |V|^2
    3U*U + 5U*V - 2V*V = 3|U|^2 + 5|U||V|cosθ - 2|V|^2

    Rearrange: cosθ = (2|V|^2 - 3|U|^2)/(5|U||V|)

    Substitute in |U| = 3|V| and you get -25/15. I can't get the inverse cos of a number greater than 1 and I can't figure out where I went wrong. Any help would be greatly appreciated.
  2. jcsd
  3. Mar 10, 2014 #2


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    Deleted - suggestion was wrong.
    Last edited: Mar 10, 2014
  4. Mar 10, 2014 #3
    Oh I was trying to figure out what you meant by that, but I see you changed your suggestion. It's a real head scratcher, of all the stuff I've done with vectors, this question should work out the way I did it but nothing seems to yield a realistic result.
  5. Mar 10, 2014 #4
    Of course, ##U## and ##V## could also be the zero vector...
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