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Angle Between Vectors

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    If A = 1i + 2j + 3k and B = 1i + 2k, and if C = A X B, then the angle between the vector A and the vector C is:


    2. Relevant equations
    AxB = ((a2b3-a3b2)i + (a3b1-a1b3)j + (a1b2-a2b3)k)
    A·B = ABcosθ = AiBi + AjBj + AkBk


    3. The attempt at a solution
    I got C = 4i + j -2k and A = 1i + 2j + 3k
    A·C = 4i + 2j - 6k with a magnitude of 7.48
    then I used the magnitude of A and C individually to get 3.74 and 4.58 respectively.
    The I used the equation cos-1(7.48/(3.74 x 4.58)) = θ
    θ = 64.11 degrees. The actual answer is 90 degrees. What am I doing wrong?
     
  2. jcsd
  3. Feb 3, 2009 #2

    Nabeshin

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    Science Advisor

    This is not meant to be a numerical question! Do you know what physical (spacial) property the cross product of a two vectors has with respect to the original vectors?
     
  4. Feb 3, 2009 #3
    I suppose that the cross product is perpendicular to the plane of the initial vectors considering that the answer is 90 degrees. I was not fully aware of this, but if this is the case then I'll keep it in mind.
     
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