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Vector Proof

  • Thread starter Cake
  • Start date
  • #1
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Homework Statement


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.

Homework Equations


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

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(Edit: This is no longer valid)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:

Answers and Replies

  • #2
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Are you sure the problem doesn't say the sum of the squares of the cosines?
 
  • #3
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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
 
  • #4
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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?
 
  • #5
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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?
I don't quite follow. Provide more details, please (in the form of equations).

Chet
 
  • #6
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16
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?
 
  • #7
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4,096
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?
Good so far. what next?

Chet
 
  • #8
100
16
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.
 
  • #9
19,921
4,096
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.
The left side of this equation should be A2. Does this give you a hint as to what to do in 3D?

Chet
 
  • #10
100
16
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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