Perpendicularity of the acceleration and velocity vectors given a derivative of a scalar product of two time-dependant vectors

Question

So, Consider a pair of vectors L and M that are dependent on time. Then the scalar product L⋅M is also time-dependent. Now, coming onto the derivative with respect to time of the scalar product. Show that

$$ \frac{\mathrm{d}}{\mathrm{d} t} (L \cdot M) = \frac{\mathrm{d} L}{\mathrm{d} t} \cdot M + L \cdot \frac{\mathrm{d} M}{\mathrm{d} t} $$

Note: I did this by dividing it into components and then using the product rule and regrouping the values.

But i cant figure out this part: I cant for one figure out if the velocity and acceleration vectors are either L or M. Secondly, i cant form a proof for this.

Q. Using the derived relation(previous part) express/indicate that if the speed (magnitude of vector velocity) is fixed, then either the acceleration is zero, or acceleration and velocity vectors are in perpendicularity.

Answer 1

Try $L = M = v$ the velocity. Then $L \cdot M = v^2$ is the speed (squared). Now what does your formula tell you?