Calculate the magnitude of the vector field $\vec F= ({-GM/r^3})\vec r$

Question

Apparently, if $\vec F= ({-GM/r^3})\vec r$ then $||\vec F || = GM/r^2$

Why is that? The magnitude of a vector (and thus a vector field, I think) is the root of the sum of the squares of the components of the vector. But the definition of this vector field gave some general vector $\vec r$, so I'm not too sure how to apply that root of sum of squares formula here.

I can see that $GM/r^2$ is the antiderivative of $-GM/r^3$, but again there's the $\vec r$ to consider so I don't even know if that's relevant here.

Any help is appreciated, I'm rather confused.

Just for reference, here is the exact question from my text:

I also don't understand why the textbook says that differentiating the rate of change of f towards the origin gives you the rate of change of f away from the origin. Maybe I'm misinterpreting what they're saying?

Answer 1

We have that

$$\left|\vec F\right|=\left| {\frac{-GM}{r^3}}\vec r\right|=GM\frac{|\vec r|}{r^3}=GM\frac{r}{r^3}$$