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

  • #1
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Homework Statement


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.

Homework Equations




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.
 

Answers and Replies

  • #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.
 
  • #3
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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.
(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.
 
  • #4
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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.
 
  • #5
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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
 
  • #6
Orodruin
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I believe the OP has already completed thid problem.
 
  • #7
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how can we (a) Geometrically and (b) Physically(in real life application) interpret the equation ???
 

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