Convert spherical vector field to cartesian vector field I have a vector field defined in spherical coordinates as follows:
$$\vec{F}\big\langle\rho,\theta,\phi\big\rangle = \bigg\langle\rho  \sin\theta \cos\phi ,\rho^3 \cos \phi, \frac{\tan \theta}{\sqrt{r}}\bigg\rangle$$
And a point in the spherical coordinate system that lies on the surface of a sphere of radius 10, given by
$$ P(\rho,\theta,\phi) =\bigg (10, \frac{\pi}{6}, \frac{\pi}{3}\bigg)  $$
I need to find the perpendicular and parallel component of the force with respect to the surface at the point P. I was thinking of solving this by converting the vector field and the point to cartesian, find the normal vector to the sphere at point P and the field vector at point P, find the projection of the field vector on the normal vector, which gives the perpendicular component of F, and then subtract from F the perpendicular component to find the parallel component, then convert the perpendicular and parallel vector back to spherical. The issue is that I don't know how to convert the vector field (or vectors) from cartesian to spherical and vice versa. I know how to convert a point from cartesian to spherical or from spherical to cartesian, but it looks like the same process cannot be used when vectors are involved.
So my question is, how would I solve the problem using the above method. Furthermore, could someone show me how to solve it in spherical coordinates directly, if possible, with a step by step explanation. Assume I know very little about spherical coordinates and try to describe each step as clearly as possible. Thank you.
 A: Let's convert everything back to cartesian.
(1): The spherical point $P=\bigg(10,\frac{\pi}{6},\frac{\pi}{3}\bigg)$ converts to $P=\bigg(\frac{15}{2},\frac{5\sqrt{3}}{2},5\bigg)$ in cartesian.
(2): The vector $\vec{n}=\big<3,\sqrt{3},2\big>$ is orthogonal to the sphere $x^2+y^2+z^2=100$ at the point $P$. (This vector $\vec{n}$ is in cartesian form).
(3): The target vector $\vec{F}(P)=\Big<\frac{5}{2},500,\frac{1}{\sqrt{5}{\sqrt[\leftroot{-2}\uproot{2}4]{27}}}\Big>$ expressed in spherical coordinates converts to the cartesian vector $$\vec{F}(P)=\Bigg<\frac{5}{2}\sin\bigg(\frac{1}{\sqrt{5}{\sqrt[\leftroot{-2}\uproot{2}4]{27}}}\bigg)\cos(500),\frac{5}{2}\sin\bigg(\frac{1}{\sqrt{5}{\sqrt[\leftroot{-2}\uproot{2}4]{27}}}\bigg)\sin(500),\frac{5}{2}\cos\bigg(\frac{1}{\sqrt{5}{\sqrt[\leftroot{-2}\uproot{2}4]{27}}}\bigg)\Bigg>$$ Knowing all this, we can now write $$\vec{F}(P)=\Big[\vec{F}(P)\Big]_{||}+\Big[\vec{F}(P)\Big]_{\perp}$$ where $\Big[\vec{F}(P)\Big]_{\perp}=\bigg(\frac{\vec{F}(P)\cdot \vec{n}}{\vec{n}\cdot \vec{n}}\bigg)\vec{n}$ is orthogonal projection of $F(\vec{P})$ onto $\vec{n}$ and $$\Big[\vec{F}(P)\Big]_{||}=F(\vec{P})-\Big[\vec{F}(P)\Big]_{\perp}$$
