Dots Products and Duality The dot product of two vectors can be defined either as $$\vec{A}\cdot\vec{B}=\left|\vec{A}\right|\left|\vec{B}\right|\cos \theta$$ or as, $$\vec{A}\cdot\vec{B}=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$$ I want to know how these two definitions are geometrically connected. 
Mr. Sanderson made a video on this, dubbed “Dot products and duality”. However, that is too brief an explanation for me to get a grasp of what was explained. I'm just a high school student, and all of my understanding about linear algebra is merely based on Sal Khan and Grant Sandersons videos. I understand linear transformations (matrix-vector multiplications) and all the mathematical concepts used in the video, but I just don't get how this correlates these two definitions at the end. Thus what I'm asking for, is simply a lengthy explanation.
Also, what courses do I need to take to have a superb geometric understanding of these things? Do tell.
 A: I'm not sure this will answer your question if you are looking for geometric intuition, but with some trickery it is actually straightforward to convince yourself that these definitions are the same:
Start with the definition $a\cdot b=|a||b|\cos\theta$, and notice that it has the property $a\cdot(b+c)=a\cdot b+a\cdot c$ (here $a,b$, and $c$ are vectors). It also has the property $a\cdot(kb)=k(a\cdot b)$, where $a$ and $b$ are vectors and $k$ is a scalar.
Now for arbitrary vectors $a$ and $b$, we can write $a$ and $b$ in terms of some basis vectors. For example, if we are in $\Bbb R^3$, take the usual basis $i=(1,0,0)$, $j=(0,1,0)$, and $k=(0,0,1)$. Then we can write
$$a=a_1i+a_2j+a_3k\qquad\text{and}\qquad b=b_1i+b_2j+b_3k$$
for the appropriate scalars $a_1,a_2,a_3,b_1,b_2$, and $b_3$. Now we can write
\begin{align}
a\cdot b&=(a_1i+a_2j+a_3k)\cdot(b_1i+b_2j+b_3k)\\
&=a_1b_1(i\cdot i)+a_1b_2(i\cdot j)+a_1b_3(i\cdot k)+\cdots+a_3b_3(k\cdot k).
\end{align}
The point is that it suffices to prove the claim for just the basis vectors $i,j$, and $k$. For example, $i\cdot i=|i||i|\cos 0=1$, and this is the same as $1\cdot 1+0\cdot 0+0\cdot 0=1$. Then $i\cdot j=|1||1|\cos(\pi/2)=0$, and this is equal to $1\cdot 0+0\cdot 1+0\cdot 0=0$. All of the other cases are similar.
A: OKey, this is easy if you think why the angle just dissapear from $\vec{A}\cdot\vec{B}=\left|\vec{A}\right|\left|\vec{B}\right|\cos \theta$ to $\vec{A}\cdot\vec{B}=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$.
Let be $\vec{A}$ and $\vec{B}$ two vectors with the following descomposition:
$\vec{A}=A_x\vec{i}+A_y\vec{j}+A_z\vec{k},~\mbox{ and }~\vec{B}=B_x\vec{i}+B_y\vec{j}+B_z\vec{k}$
lets see some natural products before multiply both $\vec{A},\vec{B}$:


*

*$\vec{i}\cdot\vec{i}=\vec{j}\cdot\vec{j}=\vec{k}\cdot\vec{k}=1\cdot1\cdot\cos(0)=1$

*$\vec{i}\cdot\vec{j}=\vec{i}\cdot\vec{k}=1\cdot1\cdot\cos(\pi/2)=0$ 

*$\vec{j}\cdot\vec{i}=\vec{j}\cdot\vec{k}=1\cdot1\cdot\cos(-\pi/2)=0$
finally: 
$$\vec{A}\cdot\vec{B}=(A_x\vec{i}+A_y\vec{j}+A_z\vec{k})(B_x\vec{i}+B_y\vec{j}+B_z\vec{k})$$
$$=A_xB_x(\vec{i}\cdot\vec{i})+A_xB_y(\vec{i}\cdot\vec{j})+\cdots+A_zB_z(\vec{k}\cdot\vec{k})$$
and by distributive law, all products will be equal to zero except those with the same direction:
$$\vec{A}\cdot\vec{B}=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$$
The geometrical connection is because inner product is a multiplication of the parallel side of vector $A$ respect vetor $B$, so if those are written as the sum of their components, then the product will only let "live" those which are parallel.
