# how to prove that spherical coordinates are orthogonal using cross product in cartesian?

The principle which I have been taught is taking $$d\hat{\phi}$$,$$d\hat{\theta}$$ and $$d\hat{r}$$, and the cross product of any two for example $$d\hat{\phi}$$ and $$d\hat{\theta}$$ should give me vector which is in direction with the $$d\hat{r}$$ vector. That is the cross product between the initial cross product and $$d\hat{r}$$ is $$0$$. I tried it but I am not getting as $$0$$. Following is my attempt.
Spherical Coordinates:
$$x=r\sin{\theta} \cos{\phi}$$
$$y=r\sin{\theta} \sin{\phi}$$
$$z=r\cos{\theta}$$
$$d\vec{l}_\phi=r\sin{\theta} d\phi\ \hat{\phi}$$
$$d\vec{l}_\theta=rd\theta\ \hat{\theta}$$
$$d\vec{r}=dr\ \hat{r}$$
To calculate $$d\vec{\phi}=|{d\phi}|\hat{\phi}$$ and $$d\vec{\theta}=|{d\theta}|\hat{\theta}$$, making use of the spherical coordinates in terms of cartesian, I get $$d\vec{\phi}=\frac{-ydx}{x^2+y^2}\hat{i}+\frac{xdy}{x^2+y^2}\hat{j}$$ $$d\vec{\theta}=\frac{xzdx}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}\hat{i}+\frac{yzdy}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}\hat{j}-\frac{\sqrt{x^2+y^2}dz}{(x^2+y^2+z^2)}\hat{k}$$
then I put them in the $$d\vec{l}_\phi$$ and $$d\vec{l}_\theta$$ which I am getting as
$$\frac{xdydz}{\sqrt{x^2+y^2+z^2}}\hat{i}-\frac{ydxdz}{\sqrt{x^2+y^2+z^2}}\hat{j}-\frac{zdxdy}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}\hat{k}$$ further I cross product this with $$\vec{r}$$ I am supposed to get each term as $$0$$ but I don't.

• Why not compute pairwise dot products instead? You method only tells you whether or not $d\phi$ and $d\theta$ are both orthogonal to $dr$. It says nothing about the orthogonality of the first two vectors, so you still have to check that some other way.
– amd
May 1, 2020 at 16:30
• yes that is a way, dot products didnt come to my mind. Still, how are the dot products $0$?how do the ${dx}^2$ and ${dy}^2$ terms vanish? May 1, 2020 at 16:39
• Crossposted to physics.stackexchange.com/q/550643/2451 May 9, 2020 at 16:13
• sorry for that, i guess i was too impatient. i wont repeat it. May 9, 2020 at 16:52

I'm here from your question in the Physics StackExchange. (Please keep questions such as these on this site.) So, let's get started. I'll use the $$(r,\theta,\varphi)$$ convention. We know that $$\begin{array}{l} \hat{\mathbf{r}} =\sin( \varphi )\cos( \theta )\hat{\mathbf{i}} +\sin( \varphi )\sin( \theta )\hat{\mathbf{j}} +\cos( \varphi )\hat{\mathbf{k}}\\ \hat{\boldsymbol{\theta}} =-\sin( \theta )\hat{\mathbf{i}} +\cos( \theta )\hat{\mathbf{j}}\\ \hat{\boldsymbol{\varphi }} =\cos( \varphi )\cos( \theta )\hat{\mathbf{i}} +\cos( \varphi )\sin( \theta )\hat{\mathbf{j}} -\sin( \varphi )\hat{\mathbf{k}} \end{array}$$ You can take the dot products from here.