Proof of $(u.v)=(u_{x}.v_{x}+u_{y}.v_{y}+u_{z}.v_{z})$, assuming $(u.v)=|u||v|cos\theta$. I would really love any sort of proof of this. 
I have a very elementary geometric proof for $R^{2}$. That's mainly because I can easily represent $cos\theta$ in the form of $\frac{u_{x}v_{x}+u_{y}v_{y}}{\sqrt{(u_{x}^{2}+u_{y}^{2})(v_{x}^{2}+v_{y}^{2})}}$. 
I don't know a way of representing $cos\theta$ in $R^{3}$ without using dot product.

Since the OP still seems to have some doubt, I've put together a community wiki answer based on the suggestions made in comments.
Make the definition
$$u\cdot v:=\Vert u\Vert\Vert v\Vert\cos\theta,$$
where $\theta$ is the angle between $u$ and $v.$  Then $u\cdot u=\Vert u\Vert\Vert u\Vert\cos0=\Vert u\Vert^2=u_x^2+u_y^2+u_z^2,$
where the last equality follows from the Pythagorean Theorem.
Therefore
$$\begin{aligned}
(u-v)\cdot(u-v)&=(u_x-v_x)^2+(u_y-v_y)^2+(u_z-v_z)^2\\
&=(u_x^2+u_y^2+u_z^2)+(v_x^2+v_y^2+v_z^2)-2(u_xv_x+u_yv_y+u_zv_z)\\
&=u\cdot u+v\cdot v-2(u_xv_x+u_yv_y+u_zv_z).
\end{aligned}
$$
On the other hand
$$
\begin{aligned}
(u-v)\cdot(u-v)&=\Vert u-v\Vert^2\\
&=\Vert u\Vert^2+\Vert v\Vert^2-2\Vert u\Vert\Vert v\Vert\cos\theta\\
&=u\cdot u+v\cdot v-2u\cdot v,
\end{aligned}
$$
where the law of cosines was used to obtain line $2.$  Comparing these expressions gives the result.
