Converting Cartesian Components of a Vector on a Sphere to Tangential-Normal Components I am dealing with a mechanics problem on a unit sphere where all the equations are written in terms of Cartesian components and wish to make a conversion to a system $n, t, s$ where $n$ is the unit normal to the surface, $t$ is tangential to the surface and $s$ is also tangential to the surface and orthogonal to $n$ and $t$.
In theory, this should be easy for a sphere because the normal vector is the same as the radial vector, then you can take the cross product of the normal vector and the unit vector in the $z$ direction to get the tangent vector.
$\textbf{t}     = \textbf{n} \: \times \: \hat{\textbf{k}}. $
The second tangential vector is then obtained via
$ \textbf{s}     = \textbf{n} \: \times \: \textbf{t}.  $
This seems fine, but then say, I have a radial vector with Cartesian components $[r_x \: r_y \: r_z]^T$, but now I want it to have the equivalent tangential-normal components $[r_n \: r_t \: r_s]^T$, so what would the formula be to take the $x$, $y$ and $z$ components and convert them, as there will no longer be a formula to convert as there would be with Cartesian to spherical polar coordinates, for example.
Edit: Added example of conversion.
$\begin{bmatrix}
    r_{x}       \\
    r_{y}      \\
    r_{z}    
\end{bmatrix} = \begin{bmatrix}
  n_x    &  t_x  & s_x \\
    n_y      &  t_y & s_y \\
   n_z     & t_z  & s_z
\end{bmatrix} \begin{bmatrix}
    r_{n}       \\
    r_{t}      \\
    r_{s}    
\end{bmatrix}=\begin{bmatrix}
  \cos \theta    &  -\sin \theta & 0 \\
    \sin \theta     &  \cos \theta & 0 \\
  0   & 0 & 1
\end{bmatrix}\begin{bmatrix}
    r_{n}       \\
    r_{t}      \\
    r_{s}    
\end{bmatrix}$
Edit: Should the matrix giving the conversion actually be:
$\begin{bmatrix}
    r_{x}       \\
    r_{y}      \\
    r_{z}    
\end{bmatrix} = \begin{bmatrix}
 \sin \theta \cos \phi   & \sin \phi &  \cos \theta \cos \phi \\
 \sin \theta \sin \phi      & - \cos \phi &  \cos \theta \sin \phi\\
 \cos \phi    & 0  &- \sin \theta
\end{bmatrix} \begin{bmatrix}
    r_{n}       \\
    r_{t}      \\
    r_{s}   
    \end{bmatrix} $

Let $(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ be the coordinates of the reference point on the sphere. From your definitions it follows that the components of $\hat n$, $\hat t$, $\hat s$ in cartesian coordinates are given by:
$$
\hat n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\quad
\hat t=(\sin\phi,-\cos\phi,0),\quad
\hat s=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta).
$$
(Notice that here $\hat t={1\over\sin\theta}\hat n\times\hat z$, to obtain a
unit vector).
Suppose then you have a vector $v_{nts}=(v_n,v_t,v_s)$ expressed in $nts$ coordinates. From the above formulas it follows that its cartesian components are given by 
$v_{xyz}=Mv_{nts}$, where:
$$
M=\pmatrix
{\sin\theta\cos\phi & \sin\phi & \cos\theta\cos\phi \\
\sin\theta\sin\phi & -\cos\phi & \cos\theta\sin\phi \\
\cos\theta & 0 & -\sin\theta}.
$$
And of course: $v_{nts}=M^{-1}v_{xyz}$.
