How do I convert a vector like $\vec{A}=6 \hat{i} + \hat{j}$ to spherical coordinates? Given a vector,
$$\vec{A}=6 \hat{i} + \hat{j}$$
My question is how do I convert it into spherical form and is my conversion correct?
My approach:
I calculated $r$, $\theta$ and $\phi$ and it came out to be $\sqrt{37}$, $\frac{\pi}{2}$ and $\arctan(\frac{1}{6})$,
According to me the converted vector is, 
$$\vec{A}=\sqrt{37} \hat{r} + \arctan\left(\frac{1}{6}\right)\hat{\phi} + \frac{\pi}{2}\hat{\theta}$$
Do I have to convert i, j, k to spherical form?
 A: In spherical coordinates, the unit vectors themselves depend on the position $(\theta, \phi)$. So, the result depends on which point the vector is associated with. For example, consider the unit vector $\mathbf{\hat i}$. If it is located at $(1,0,0)$ in Cartesian coordinates, it would be represented as $\mathbf{\hat r}$ in spherical coordinates. If it were instead located at $(0,-1,0)$, it would be represented as $\mathbf{\hat ϕ}$ instead. So the answer depends on where the vector is. 
Identify the point associated with the vector. Find the coordinates of this point in $(r, \theta, \phi)$ first, and then convert the unit vectors. The standard conversions apply. 
\begin{align}
r & = \sqrt{x^2 + y^2 + z^2}\\
\theta & = \arccos{\frac{z}{\sqrt{x^2 + y^2 + z^2}}}\\
\phi & = \arctan{\frac{y}{x}}
\end{align}
\begin{align}
\mathbf{\hat r} & = \sin\theta \cos\phi \, \mathbf{\hat x} + \sin\theta \sin\phi \, \mathbf{\hat y} + \cos\theta \, \mathbf{\hat z}
\\
\mathbf{\hat θ} & = \cos\theta \cos\phi \, \mathbf{\hat x} + \cos\theta \sin\phi \, \mathbf{\hat y} - \sin\theta \, \mathbf{\hat z}
\\
\mathbf{\hat ϕ} & = -\sin\phi \, \mathbf{\hat x} + \cos\phi \, \mathbf{\hat y}
\end{align}
Note: The correct quadrant needs to be accounted for when finding $\phi$. See $\mathrm {atan2}$.
