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Finding angle

  1. Jan 30, 2004 #1
    ive got two vectors and i need to find the angle between them.
    AB = -1.95i + 2.4j +.3k
    AC = 2.4j+1.8k
    i looked up the answer and found it to be 46.8 deg.
    i used the equation (AB)(AC) cos theta = ACxABx +ACyABy + ACzABz
    the dot product is 6.84. but the equation ends up like this
    cos theta = (ACxABx +ACyABy + ACzABz)/(AB)(AC)

    this is a dot product on top and on bottom right? i dont quite understand how this works. according to the answer, ABAC should be around 10.
     
  2. jcsd
  3. Jan 30, 2004 #2
    I get a slightly different answer. (I hope I haven't forgotten everything from last semester.)

    ACx is 0 so the dot product is 0 + (2.4)(2.4) + (.3)(1.8) = 6.3

    and, the denominator of your fraction is not a dot product. It is the product of the MAGNITUDES of the two vectors.

    Do you know how to get those?

    Based on this I got θ = 47.5o (approx)
     
  4. Jan 30, 2004 #3

    NateTG

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    Science Advisor
    Homework Helper

    If you have
    [tex]\vec{a}=<a_1,a_2,a_3>[/tex]
    and
    [tex]\vec{b}=<b_1,b_2,b_3>[/tex]

    then the angle is
    [tex]\cos^{-1}\frac{a_1b_1+a_2b_2+a_3b_3}{ \sqrt{a_1^2+a_2^2+a_3^2} \sqrt{b_1^2+b_2^2+b_3^2}}[/tex]
     
  5. Feb 2, 2004 #4
    argh, wish i had that this morning. we never went over that, he always gave us the angle already.
     
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