Proving cos^2(\theta_x)+cos^2(\theta_y)+cos^2(\theta_z)=1

[Cake]

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

$cos^2(\theta_x)+cos^2(\theta_y)+cos^2(\theta_z)=1$

The Attempt at a Solution

I started by trying to pull out the angles with
$\cos^{-1}$
Getting
$\theta_x+\theta_y+\theta_z=0+2k\pi$
(Edit: This is no longer valid)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?

 

[Chestermiller]

Are you sure the problem doesn't say the sum of the squares of the cosines?

 

[Cake]

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

 

[Cake]

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?

 

[Chestermiller]

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

 

[Cake]

What I'm saying is if you treat the components of the vector in relation to each axis, you can break the components into $A cos(\theta_x), A cos(\theta_y), A cos(\theta_z)$ right? That's how you would relate the vectors position in relation to the axes to cosine?

 

[Chestermiller]

What I'm saying is if you treat the components of the vector in relation to each axis, you can break the components into $A cos(\theta_x), A cos(\theta_y), A cos(\theta_z)$ right? That's how you would relate the vectors position in relation to the axes to cosine?

Good so far. what next?

Chet

 

[Cake]

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?
$c^2=A^2 cos^2(\theta_x)+A^2 cos^2(\theta_y)$

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.

 

[Chestermiller]

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?
$c^2=A^2 cos^2(\theta_x)+A^2 cos^2(\theta_y)$

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

 

[Cake]

The Pythagorean identity for 3D vectors is $A^2 = A_x^2+A_y^2+A_z^2$ 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:

$A^2 = A^2 cos^2(\theta_x)+A^2 cos^2(\theta_y)+A^2cos^2(\theta_z)$

Divide by $A^2$

Aaaaand finished.

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