differential geometry - Do $vec r$ and $d vec r$ have the same direction?

One question is bugging me for a long time but I couldn't make out anything nor could my friends. Here it goes:

We know, $\vec r$ is regarded as the position vector. So we can say,

$$\vec r \cdot\vec r = r^2$$

Differentiating both sides with respect to time $t$, we get

$$\vec r \cdot\frac{d \vec r}{dt} + \frac{d \vec r}{dt}\cdot\vec r = 2r \frac{dr}{dt}$$ or, $$2 \vec r \cdot\frac{d \vec r}{dt} = 2r \frac{dr}{dt}$$ or, $$\vec r \cdot\frac{d \vec r}{dt} = r \,\frac{dr}{dt}$$ or, $$\vec r \cdot\ d \vec r = r \,\ dr$$ or, $$r \,\ dr \cos \theta = r \,\ dr$$ where $\theta$ is angle between $\vec r$ and $d \vec r$

or,

$$\cos \theta = 1$$ or, $$\theta = 0^\circ$$

Question no.1:So can I conclude that $\vec r$ and $d \vec r$ have the same direction? The above calculation suggests so but the diagram below does not. Why?

Also if $\vec r$ and $d \vec r$ have the same direction, then

$$\vec r \times \frac{d \vec r}{dt} = 0$$ Now, $$\vec r = r \hat r$$ where $$\hat r = \hat i \cos \theta + \hat j \sin \theta$$ So, $$\frac{d \vec r}{dt} = \frac{dr}{dt}\hat r + r \hat \theta \frac{d \theta}{dt}$$ and $$\vec r \times \frac{d \vec r}{dt} = r^2 \frac{d \theta}{dt} (\hat r \times \hat \theta) \not = 0 \,\ \text{(in general)}$$

Why does this contradiction arise?

Answer

Despite what some of the other answers are mentioning, the following equation you have is correct

$$\vec{r} \cdot d \vec{r} = r dr$$ You can check this by noting $$\vec{r} = x {\hat i} + y {\hat j} + z {\hat k} \implies d \vec{r} = d x {\hat i} + d y {\hat j} + d z {\hat k}$$ Then $$\vec{r} \cdot d \vec{r} = x dx + y dy + z dz$$ Further note $$r = \sqrt{ x^2 + y^2 + z^2 } \implies dr = \frac{1}{r} \left( x dx + y dy + z dz \right)$$ implying that it is true that $\vec{r} \cdot d \vec{r} = r dr$.

Where you go wrong is the next step. You say

$$\vec{r} \cdot d \vec{r} = r dr \cos\theta$$ In doing this you are assuming $$| d \vec{r} | = d r$$ Is this really true? Let us check. We have already computed $d r$. We have $$| d \vec{r} | = | d x {\hat i} + d y {\hat j} + d z {\hat k} | = \sqrt{ dx^2 + dy^2 + dz^2 }$$ This is clearly not equal to $dr$, thereby making any conclusions based on this incorrect.