Representing a point in cartesian space as a position vector in spherical coordinates I had a quiz in one of my physics classes the other day, and one of the questions is still bugging me.
Say we have a point in 3-dimensional cartesian space with coordinates $(0, 2m, 0)$. How you would represents a position vector pointing from the origin to this point in spherical coordinates: $(r, \theta, \phi )$.
Before writing down any equations, it's obvious to see that the magnitude of the position vector would be 2m, and it would be pointing in the $\hat y$ direction. The position vector in cartesian coordinates would be: $$\vec r=2m \hat y$$
Well if we define $\phi = 0$ as the positive x-axis and $\theta =0$ as the positive z-axis, then each angle should by $90^\circ$ or $\frac \pi2$ radians. Then I think the vector would be: $$\vec r=2m\hat r+\frac \pi2\hat \theta+\frac \pi2\hat \phi$$
Which shows the exact position of that point relative to the origin. I got this question wrong. I talked to my professor, and she said the real right answer should be $$\vec r=2m\hat r$$
Which doesn't specify the position. That's just the distance from the origin. That could be used to talk about an infinite amount of points. Her argument was that, since you can place the coordinate axes anywhere you want, the angles don't really matter. By that logic, since we're talking about a defined point in space with a specific location, we could just move the origin to that point. In that case, it would just be $\vec r=0$. It seems extremely wishy-washy and I feel kind of cheated. I trust my professor, but I didn't receive a very satisfying explanation for how my answer is incorrect. Can anyone give me a more reasonable explanation to clear up why writing the vector like this is okay? If the $\vec r=2m\hat r$ is correct, does that mean that $\vec r=2m\hat r+\frac \pi2\hat \theta+\frac \pi2\hat \phi$ is wrong?
 A: Look at the following graphics

the unit vector of the spherical coordinates are the unit vector tangent to the lines where two coordinate are constants and the third change.
For example, if you only change $\theta$, with $r=2m$ and $\phi=\pi/2$ fixed, you obtain the quarter of circle in the graphics, so $e_\theta$ is tangent to this circle.
If you only change $\phi$, with $r=2m$ and $\theta=\pi/2$ fixed, you obtain the circle that appears as an ellipse in the graphics, so $e_\phi$ is tangent to this circle.
If you only change $r$, with $\phi=\pi/2$ and $\theta=\pi/2$ fixed, you obtain the positive $y$ axis, so $e_r$ is tangent to this line.
You can see that the vector $r=2me_y$ has component only along $e_r$, so the result of the problem.
In general, the unit vectors in a generic position are given by
\begin{align}
e_r      &= \sin\theta(e_x\cos\phi+e_y\sin\phi)+e_z\cos\theta, \\
e_\theta &= \cos\theta(e_x\cos\phi+e_y\sin\phi)-e_z\sin\theta, \\
e_\phi   &= -e_x\sin\phi+e_y\cos\phi,
\end{align}
and conversely
\begin{align}
e_x &= \cos\phi(e_r\sin\theta+e_\theta\cos\theta)-e_\phi\sin\phi, \\
e_y &= \sin\phi(e_r\sin\theta+e_\theta\cos\theta)+e_\phi\cos\phi, \\
e_z &= e_r\cos\theta-e_\theta\sin\theta.
\end{align}
If you put $\theta=\phi=\pi/2$ in these equations, you get the $e_y=e_r$ in that point.
