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Vectors Dot Product

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that if u and v are nonzero vectors, and theta is the angle between them then u dot product v = ||u|| ||v|| cos (theta). Consider the triangle with sides u ,v , and u-v. The Law of Cosines implies that ||u-v||^2 = ||u||^2 + ||v||^2 - 2||u|| ||v|| cos(theta). On the other hand, ||u-v||^2 = (u-v) dot product (u-v)... Simplify this last expression, then equate the two formulas for ||u-v||^2.


    2. Relevant equations



    3. The attempt at a solution

    I have no idea how to start! I thought of doing
    cos(theta)= uxvx + uyvy / sqrt[(uxvy- vxuy)^2] + (uxvx + uyvy)^2] When x and y are subscripts, because
    u= uxi + uyj
    v= vxi + vyj
     
  2. jcsd
  3. Oct 21, 2013 #2

    CAF123

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    Expand the dot product ##(u-v) \cdot (u-v)## using distributivity.
     
  4. Oct 21, 2013 #3
    Yes I did that, and I get u^2 - 2uv + v^2.
    But as for the left side ||u-v||^2, I get [sqrt(u^2 + v^2)]^2 = u^2 + v^2.
    That isn't equal though.
     
  5. Oct 21, 2013 #4

    CAF123

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    Since u and v are vectors, you should write that as||u||2-2(u . v) + ||v||2. Now compare this with the law of cosines.
     
  6. Oct 21, 2013 #5
    So how is -2 ( u dot v) = -2 ||u|| ||v|| cos theta?
    I don't understand. I thought once you take the norm of u and v you'll be left with -2 u times v cos theta?
     
  7. Oct 21, 2013 #6

    CAF123

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    The full expression before simplification is ##u \cdot u -2(u \cdot v) + v \cdot v##. The middle term is a projection, while the first and third terms are the norm of the vectors squared (which is also the projection of a vector onto itself)
     
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