Why Does the Cosine Term Disappear in the Lorentz Force Calculation?

[Je m'appelle]

It is known that magnetic fields do no work at a moving particle, all they can do is to change the particle's direction.

So, I've been trying to understand a step on the mathematical explanation, but I'm stuck.

I'm using this source: http://www.tutorvista.com/content/physics/physics-iv/moving-charges-magnetism/lorentz-force.php

What I don't get is the step below

$$m \frac{d}{dt} (v \cdot v) = m(v} \cdot \frac{d v}{dt} + \frac{d v}{dt} \cdot v) = 2m v\frac{d v}{dt}$$

Shouldn't it be

$$2m v \cdot \frac{d v}{dt} = 2m v \frac{d v}{dt} cos \theta$$

What happened to the cosine? The dot product simply disappeared, it's like he considered $$cos \theta = 1$$, but as far as I understood it, the cosine is actually zero and not one.

OBS: 'v' is a vector.

 

[The legend]

your latex image seems to be invalid and I can't view it

 

[Je m'appelle]

your latex image seems to be invalid and I can't view it

I've fixed it already, somehow the vector function in the latex wasn't working.

 

[Jolsa]

It looks like a typo to me. He does it correctly the first time when he says

$$m\frac{d\vec{v}}{dt}\cdot \vec{v}=\frac{m}{2}\frac{d}{dt}(\vec{v} \cdot \vec{v})=0$$

Now $\vec{v} \cdot \vec{v}=v^2$ so it follows that the change in kinetic energy is zero.