Cos^2(φ_1) +cos^2 (φ_2) + cos^2(φ_3) = 1 in a three dimensional cartesian system

[karkas]

Homework Statement

I seem to be stuck for an assignment that I have for one of my classes, in which we are asked to prove that cos^2(φ_1) +cos^2 (φ_2) + cos^2(φ_3) = 1 in a three dimensional cartesian system, where φ_1 ,φ_2, φ_3 are the angles that a random vector r (x,y,z) is to the x,y and z axxi respectively.

Homework Equations

Prove that cos^2(φ_1) +cos^2 (φ_2) + cos^2(φ_3) = 1.

The Attempt at a Solution

I have made various attempts at linking the angles together and forming some kind of equation but none of them lead to the solution. It just seems really random to me, maybe I'm wrong because it's so early in the morning...

 

[Dick]

If phi_1 is the angle (x,y,x) makes with the x axis, then x=sqrt(x^2+y^2+z^2)*cos(phi_1), yes? That's just trig. What are the other two coordinates?

 

[karkas]

If phi_1 is the angle (x,y,x) makes with the x axis, then x=sqrt(x^2+y^2+z^2)*cos(phi_1), yes? That's just trig. What are the other two coordinates?

I'm guessing you mean r(x,y,z). So its got to be y=sqrt(x^2+y^2+z^2)*sin(phi_2) and z=sqrt(x^2+y^2+z^2)*cos(phi_3).

 

[SammyS]

y has cosine like the others, not sine.

 

[Dick]

I'm guessing you mean r(x,y,z). So its got to be y=sqrt(x^2+y^2+z^2)*sin(phi_2) and z=sqrt(x^2+y^2+z^2)*cos(phi_3).

Well, I meant (x,y,z) to be the coordinates of the point. Why did you put sin in the y coordinate? sqrt(x^2+y^2+z^2) is the length of the vector. cos is the ratio between the hypotenuse and the coordinate, yeah?

 

[karkas]

y has cosine like the others, not sine.

Oh yeah my bad. I'm then guessing that it's wrong to say that cos(phi_2)=sin(phi_1).I'm not entirely sure which angles our teacher wanted us to use, therefore I'm confused. The fact is that I have formed these equations, but messing with them led me to the beggining which generally means I'm missing something.

 

[Dick]

y has cosine like the others, not sine.

Ah, ok. Then do you understand why those things are true? If so, then compute x^2+y^2+z^2 using x=cos(phi_1)*sqrt(x^2+y^2+z^2), etc.

 

[Dick]

y has cosine like the others, not sine.

Oh, that was you SammyS. Want to take it from here??

 

[karkas]

Ah, ok. Then do you understand why those things are true? If so, then compute x^2+y^2+z^2 using x=cos(phi_1)*sqrt(x^2+y^2+z^2), etc.

Ah I think I got it and I guess it's just a matter of not spotting the answer, my everlasting doom. x^2+y^2+z^2 = |z|^2
and
x^2+y^2+z^2 = [ cos^2(phi_1) + cos^2(phi_2) + cos^2(phi_3) ] * ( x^2+y^2+z^2)
= [ cos^2(phi_1) + cos^2(phi_2) + cos^2(phi_3) ] * |z|^2,
therefore we have proven it?

 

[Dick]

Ah I think I got it and I guess it's just a matter of not spotting the answer, my everlasting doom.


x^2+y^2+z^2 = |z|^2
and
x^2+y^2+z^2 = [ cos^2(phi_1) + cos^2(phi_2) + cos^2(phi_3) ] * ( x^2+y^2+z^2)
= [ cos^2(phi_1) + cos^2(phi_2) + cos^2(phi_3) ] * |z|^2,
therefore we have proven it?

Yes.