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Image of the proof

I was reading this proof in my textbook. They say that $$\vec{r} \cdot \dot{\vec{r}} = |\vec{r}||\dot{\vec{r}}|.$$ Doesn't that mean $\vec{r}$ is parallel to $\dot{\vec{r}}$, and if so, then the line before $3.80$ is $0$ ($\vec{r}$ crossed with $\dot{\vec{r}}$ is $0$). How can $\vec{r}$ possibly not be parallel to $\dot{\vec{r}}$ as this is the only way the proof makes sense?

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You don't have the notation quite right. The end result is that $$\vec{r}\cdot\dot{\vec{r}} = |\vec{r}|\frac{d|\vec{r}|}{dt}.$$ In other words, $\dot{r}$ is the rate of change of the distance from the origin, not the speed of the object. If the object is undergoing circular motion about the origin at a constant speed, $\dot{r} = 0$.

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    $\begingroup$ The deduction of the expression for $\vec{r} \cdot \dot{\vec{r}}$ is critical to the relevant vector analysis. Denoting $r:=|\vec{r}|$ and $\hat{r} := \frac{\vec{r}}{|\vec{r}|}$, we can see that $\dot{|\hat{r}|} := \frac{d |\hat{r}|}{dt} = \dot{1} = 0 = \vec{r} \cdot \dot{\vec{r}}$, which implies that $\dot{\hat{r}} \perp \hat{r}$. Indeed, in general, the time derivative of a vector of constant magnitude is perpendicular to the vector. Therefore, $\vec{r} \cdot \dot{\vec{r}} = r \hat{r} \cdot (\dot{r} \hat{r} + r \dot{\hat{r}})$, so that $\vec{r} \cdot \dot{\vec{r}} = r \dot{r}$. $\endgroup$
    – kbakshi314
    Commented Mar 23, 2020 at 3:24
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$\dot r$ means $$\frac{d|\mathbf r|}{dt},$$ not $$\left|\frac{d\mathbf r}{dt}\right|.$$ Your notation $|\dot{\vec{r}}|$ means the latter, not the former. The latter is the speed. The former is the radial speed.

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They say that $\vec{r} \cdot \dot{\vec{r}} = |\vec{r}|\frac{d}{dt}|\vec{r}|,$ not that $\vec{r} \cdot \dot{\vec{r}} = |\vec{r}||\frac{d}{dt}\vec{r}|.$

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