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Vector Proof

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that sum of the cosines squared of the angles between a vector and the x, y, and z axes equals 1. Prove using either geometry or vector algebra.

    2. Relevant equations
    [itex]cos^2(\theta_x)+cos^2(\theta_y)+cos^2(\theta_z)=1[/itex]

    3. The attempt at a solution
    I started by trying to pull out the angles with
    [itex]\cos^{-1}[/itex]
    Getting
    [itex]\theta_x+\theta_y+\theta_z=0+2k\pi[/itex]
    (Edit: This is no longer valid)
    This doesn't make sense to me since I know that the maximum sum of the angles can only be around 180 degrees (I think). So I thought maybe the geometric approach would have something to do with triangles. But I don't know how to show that analytically.
    (Edit: Still clueless)

    I'm also lost on how to start with the vector algebra. I'm looking over the identities of vectors now but to no avail.

    Thoughts?
     
    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2
    Are you sure the problem doesn't say the sum of the squares of the cosines?
     
  4. Jan 10, 2015 #3
    Aaaah, edited :D

    Edit: I sort of haven't looked at the book in a while and have been banging my head so hard I must have forgotten that important part :D
     
  5. Jan 10, 2015 #4
    Hmm, so let's see, if there's 3 triangles between the vector and the axes, you can use Pythagorean theorem to get the cosine squared of each component right?
     
  6. Jan 10, 2015 #5
    I don't quite follow. Provide more details, please (in the form of equations).

    Chet
     
  7. Jan 10, 2015 #6
    What I'm saying is if you treat the components of the vector in relation to each axis, you can break the components into [itex]A cos(\theta_x), A cos(\theta_y), A cos(\theta_z)[/itex] right? That's how you would relate the vectors position in relation to the axes to cosine?
     
  8. Jan 10, 2015 #7
    Good so far. what next?

    Chet
     
  9. Jan 10, 2015 #8
    Can I imagine that this is 2 dimensional instead of 3 and put the vector on a 2D plane to use pythagorean theorem to get them to be squared?
    [itex]c^2=A^2 cos^2(\theta_x)+A^2 cos^2(\theta_y)[/itex]

    That's how I was visualizing earlier too it, but it gets rid of the z component, so I'm not sure how I can bring the z axis back into the equation.
     
  10. Jan 10, 2015 #9
    The left side of this equation should be A2. Does this give you a hint as to what to do in 3D?

    Chet
     
  11. Jan 10, 2015 #10
    The Pythagorean identity for 3D vectors is [itex]A^2 = A_x^2+A_y^2+A_z^2[/itex] but I didn't think it would be that easy that I could just use that willy nilly for this problem. But it makes sense. So:

    [itex]A^2 = A^2 cos^2(\theta_x)+A^2 cos^2(\theta_y)+A^2cos^2(\theta_z)[/itex]

    Divide by [itex]A^2[/itex]

    Aaaaand finished.

    Still not sure why I didn't get that originally.
     
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