What is the dot product of a vector and its derivative equal to the product of the magnitude of the vector and the magnitude of its derivative?

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On page 15 of "Fundamentals of Astrodynamics" by Bate et al., the energy constant of motion is derived. In the second step of the derivation, Bate asserts the following:

Since in general $\mathbf{a}\dot{}\mathbf{\dot{a}} = a\dot{a}$, $\mathbf{v}=\mathbf{\dot{r}}$ and $\mathbf{\dot{v}}=\mathbf{\ddot{r}}$, then...

I am familiar with the latter two identities, which express kinematic relationships, but unfamiliar with the former:

$\mathbf{a}\dot{}\mathbf{\dot{a}} = a\dot{a}$

In the preceding text, Bate does not precise a physical meaning of $\mathbf{a}$, so it would appear that he intends a general vector.

(A similar, well-known identity is of course $\mathbf{a}\dot{}\mathbf{a} = a^2$, but this doesn't seem to suggest that the above identity should be valid.)

Prove or refute Bate's identity.

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Using the product rule for scalair multiplication $$2aa'=(a^2)' = (\mathbf{a}\cdot \mathbf{a})' = \mathbf{a}\cdot \mathbf{a}' + \mathbf{a}'\cdot \mathbf{a} = 2(\mathbf{a}\cdot \mathbf{a}')$$