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Find the Value os Cos Vectors

  1. Sep 2, 2011 #1
    The points P, Q and R have position vectors 2i + 5j -3k, i + 4j - 2k and 3i + 3j - 2k, respectively. Given that the angle between PQ and PR is theta, find the value of cos of theta:

    So, the first thing I did was to find PQ and PR. I got:

    PQ = -i -j + k
    PR = i + 2j + k

    Then what I did was I found the dot product between the two of those and then found the magnitude of each and multiplied them. I thought I would get the value of cos of theta by dividing the dot product by the the product of the magnitude of each but the answer in the book does not agree with mine. What am I doing wrong?

    Thanks,
    Peter
     
  2. jcsd
  3. Sep 2, 2011 #2
    PQ dot PR = |PR| |PQ| cos theta

    (PQ dot PR) / (|PR| |PQ|) = cos theta

    should be right

    post what you got for PQ dot PR and the magnitudes
     
  4. Sep 2, 2011 #3
    Ok, for the dot product: 2
    Product of Magnitudes: √18
     
  5. Sep 2, 2011 #4
    -1*1 + -1*2 + 1*1 =

    -1 + -2 + 1

    = -2
     
  6. Sep 2, 2011 #5
    Isn't the j component -1 * -2?
     
  7. Sep 2, 2011 #6
    you wrote +2j and -j
     
  8. Sep 2, 2011 #7
    Oh, I'm so sorry... Ok, if we look back at the first post, R - P will yield i - 2j + k. So the answer is in fact 2/√18?
     
  9. Sep 2, 2011 #8
    seems like it
     
  10. Sep 2, 2011 #9
    the angle should be about 61 degrees you can graph it to verify on a graphing system if you want
     
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