Why is $ \frac{\vec{r}}{r^3} = \frac{1}{r^2} $? I know it's surely a beginner's question but I don't see why you can write 
\begin{align}
\frac{\vec{r}}{r^3} = \frac{1}{r^2}\cdot \frac{|\vec{r}|}{r}
\end{align}
Could someone explain it please? It would help understand quite a few things ... 
 A: What you wrote is not true. The $r$ without an arrow is only a scalar - the length of a vector. So the right-hand side is a scalar. The left-hand side is a vector $\vec{r}$ divided by its length cubed. So the result is still a vector. However, the length of the resulting vector in the left is $1/r^2$. This is because the $\vec{r}$ carries a length of $r$ with it. So if you take the norm, you get 
$$\frac{|\vec{r}|}{r^3} = \frac{r}{r^3} = \frac{1}{r^2}$$
A: $$\frac{\vec{r}}{r^3} = \frac{1}{r^2}\frac{\vec{r}}{r}=\frac{1}{r^2} \vec{u_r}$$
where $\vec{u_r}$ is an unitary vector with the direction of $\vec{r}$
A: Another possibility is: being $\overline r= r\mathbf{\widehat r}$, then
$$\frac{\overline r}{r^3}=\frac{r\mathbf{\widehat r}}{r^3}=\frac{\not r\mathbf{\widehat r}}{r^{\not 3\,2}}=\frac{\mathbf{\widehat r}}{r^{2}}.$$
A: Answer: Equality: $\frac{\vec r}{r^3}= \frac{1}{r^2}$ is not true
Explanation: A vector quantity (having both magnitude & direction) & a scalar quantity (having only magnitude) can never be equated.   
Now, $\frac{\vec r}{r^3}$ is a vector quantity while $\frac{1}{r^2}$ is a scalar quantity hence 
$$\therefore \frac{\vec r}{r^3}\ne \frac{1}{r^2}$$
Similarly $|\vec r|=r$ is the magnitude of vector quantity $\vec r$ so $\frac{1}{r^2}\frac{|\vec r|}{r}$ is a scalar quantity 
$$\therefore \frac{\vec r}{r^3}\ne \frac{1}{r^2}\frac{|\vec r|}{r}$$
However your equality seems to probably be similar to 
$$\frac{|\vec r|}{r^3}= \frac{1}{r^2}$$
