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Finding the angle between two vectors

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    A force F =( 6 i - 2 j ) N acts on a particle that under
    goes a displacement D r = ( 3 i + j )m. Find (a) the work done
    by the force on the particle and (b) the angle between F
    and D r .



    2. Relevant equations

    I've found the work to be about 16N. My problem is finding the angle.

    The equation thebook gives is cos(inv)* (Products of vectors A and B) / (A)(i^2+j^2)*(B)(i^2+j^2)

    3. The attempt at a solution

    I used cas(inv)*((VectorA * VectorB) / sqrt(6^2-2^2)(3^2+1^2))

    Which came out to 26 degrees. The back of the book says 36.9 degrees for the answer. I don't think I'm missing anything. All your help is appreciated. Thank you.
     
  2. jcsd
  3. Sep 23, 2011 #2
    The easiest way that i can think of to find the angle between two vectors is the dot product.

    Remember [itex]\vec{A}\cdot\vec{B}=|A||B|Cos(\theta)[/itex]
     
  4. Sep 23, 2011 #3
    That's what I used, but I still got 26.5 degrees instead of the 36.9.

    I got part a correct (finding the Force on the object) so I don't think I did any previous calculations incorrectly for the numbers I'm using now.
     
  5. Sep 23, 2011 #4

    SammyS

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    The magnitude of vector A is √( 62 + (-2)2 ) = √( 36 + 4 ) .
     
  6. Sep 24, 2011 #5
    Holy crap.. thank you.
     
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