# Vector Proof

## 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(Edit: This is no longer valid)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?

Last edited:

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

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

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
Mentor
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

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
Mentor
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

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
Mentor
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

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.