I have a vector in the $xy$ plane, $\vec{A}$, defined by $\|\vec{A}\|=8$, and $\theta=130^\circ$ with the positive $x$ axis. This gives the component form $\langle 8\cos130^\circ, 8\sin130^\circ, 0\rangle$. Now I need to find the spherical coordinates for $\vec{A}+3\hat{k}$, which are $\langle\|\vec{A}+3\hat{k}\|, \theta,\phi\rangle$. In order to evaluate $\theta$, I used the formula $$\theta=\cos^{-1}\left(\frac{A_x}{\|\vec{A}+3\hat{k}\|}\right)=\cos^{-1}\left(\frac{8\cos130^\circ}{\sqrt{(8\cos130^\circ)^2+(8\sin130^\circ)^2+3^2}}\right)\approx 127^\circ.$$ Why is it that I do not get the original $\theta=130^\circ$? Does the angle made with the positive $x$ axis change with the addition of a $z$ component?

  • $\begingroup$ because what you calculated is the angle in the plane made by axis $x$ and the "3D" vector, which is obviously different from the angle that its projection on $xy$ plane makes with $x$ axis. $\endgroup$ – G Cab Sep 24 '16 at 18:52
  • $\begingroup$ @GCab For spherical coordinates, which angle is the correct $\theta$? $\endgroup$ – Db3010 Sep 24 '16 at 19:01
  • $\begingroup$ In spherical coordinates, as I know for standard, you have the angle with $x$ axis of the projection of the vector on the $x,y$ plane, the $\theta$ in your case, and then the angle that the vector makes with the $z$ axis (or depending on conventions, with its projection on the $x,y$ plane, in your case $arctan(3/8)$). $\endgroup$ – G Cab Sep 24 '16 at 23:18

If $\theta$ in your spherical system is going to have the same meaning as $\theta$ in the polar coordinates, then it should be $130$. The change formula is $$ \theta_{\rm spherical}=\arctan\left(\frac yx\right)=\arctan\left(\frac{8\sin 130}{8\cos 130}\right)=\arctan(\tan(130))=130. $$


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