How to find the displacement vector in Spherical coordinate with out any coordinate convertion

Is there a way of subtracting two vectors in spherical coordinate system without first having to convert them to cartesian or other forms?

Since I have already searched and found the difference between Two Vectors in Spherical Coordinates as,

$$|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'[\cos(\theta-\theta')+\sin\theta\sin\theta'(cos(\phi-\phi')-1))])^{\frac{1}{2}}$$

which I believe the radius of displaced vector. I still didn't get any way to find the theta (angle from positive z axis)and psi(angle from positive x axis). Please help.

The given expression is the distance between two points whose spherical coordinates are given. It is derived on Cartesian coordinate basis considering differences of $$r,\theta,\phi$$ which is the most convenient method.
If $$\rho,\rho^{'},\phi, \phi^{'}, \theta$$ are given two values of $$\theta ^{'}$$ can be calculated.
• Thank You for your information. But here I need to find the $r,\theta,\phi$ of displaced point. Currently I have $\rho,\rho^{'},\phi, \phi^{'}, \theta, \theta'$ using these is it possible to find the new $r,\theta,\phi$. I also know we cannot simply subtract the final from the original. – Aswin Jagadeesh A Nov 9 '18 at 5:22
• If distance $R,\phi, \theta ,\rho$ are given then the center and radius of a sphere locus of the distance $R$ from that given point are fixed, Next if either of $\theta$ or $\phi$ are given then two values of $\phi. \theta$ can be respectively determined. – Narasimham Nov 9 '18 at 9:14
• Apologies, I didn't fully understand what you have said. My motive is to find the displacement vector, that means right now i have two points one original and the other final, which are of course in spherical coordinate. Now I want the vector difference so that I hope I'll get the $\delta r,\delta \phi,\delta \theta$. Don't consider that the two points are in the same sphere. – Aswin Jagadeesh A Nov 9 '18 at 9:47