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Vectors: why is the cosine of this angle always -1/2?

  1. Nov 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Pick any numbers that add to x + y + z = 0. Find the angle between your vector v = (x, y, z) and the vector w = ( z, x, y). Challenge question: explain why v.w/|v||w| is always -1/2.

    2. Relevant equations


    3. The attempt at a solution
    I chose (1, -2, 1) for the first part, which is straightforward. The second bit has me somewhat flummoxed. Using x, y, and z, I get:
    (xz + xy + yz)/(x^2 + y^2 + z^2)
    I can see that the numerator will always be less than the denominator because either one or two of the components must be negative. The explanation proper eludes me though, and I feel missing something simple here. Please assist me in understanding this one.
     
  2. jcsd
  3. Nov 24, 2014 #2

    Orodruin

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    What is (x+y+z)^2?

    Edit: There is also a geometrical interpretation, but we can leave that until you understand why the equality holds.
     
  4. Nov 24, 2014 #3
    (xz + xy + yz)/(x^2 + y^2 + z^2) = -1/2 as a result of multiplying out the expression, which is equal to zero. I knew it was staring me in the face.
     
  5. Nov 26, 2014 #4
    V scalar w /|v| ×| w |= |v| ×| w | cos ß / |v| ×|w| = cos ß
    so you should find some kind of relationship between those vector in order to get ß = π -60° / π + 60°
    Thats all i can think of.
     
  6. Nov 26, 2014 #5
    You need to answer Orodruin's question in post #2. Once you see the answer to this question, you will know what to do next.

    chet
     
  7. Nov 26, 2014 #6

    Orodruin

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    I believe the OP has already completed thid problem.
     
  8. Aug 19, 2016 #7
    how can we (a) Geometrically and (b) Physically(in real life application) interpret the equation ???
     
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