Dot product of two vectors How does one show that the dot product of two vectors is $ A \cdot B = |A| |B| \cos (\Theta) $?
 A: Think about a triangle with sidelengths $|\textbf{a}|,|\textbf{b}|,|\textbf{c}|$. Then we can use the law of cosines. 
$$
\begin{align}
|\textbf{c}|^2&=|\textbf{a}|^2+|\textbf{b}|^2-2|\textbf{a}||\textbf{b}| \cos \theta \\
\implies 2|\textbf{a}||\textbf{b}| \cos \theta &= |\textbf{a}|^2+|\textbf{b}|^2-|\textbf{c}| = \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - \textbf{c}\cdot \textbf{c} 
\end{align}
$$
By the properties of the dot product and from the fact that $\textbf{c}=\textbf{b}-\textbf{a}$ we find that 
$$
\begin{align}
  \textbf{c}\cdot \textbf{c} &=(\textbf{b}-\textbf{a}) \cdot (\textbf{b}-\textbf{a}) \\
  &=(\textbf{b}-\textbf{a})\cdot \textbf{b} - (\textbf{b}-\textbf{a}) \cdot \textbf{a} \\
  &= \textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}.
\end{align}
$$
By substituting $\textbf{c}\cdot \textbf{c}$, we get
$$
\begin{align}
  2|\textbf{a}||\textbf{b}| \cos \theta &= \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - (\textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}) \\
  &=2 \textbf{a}\cdot \textbf{b}.
\end{align}
$$
A: One way of showing this requires taking a Geometric look at stuff:
Think of $\vec{a},\vec{b}$ as the vertices of a triangle with one corner in the origin
and sides of length $|\vec a|,|\vec b|,|\vec{a-b}|$. Now, use the Law of cosines, and inner product properties (over $\mathbb R$) to calculate $|\vec{a-b}|$:
The law of cosines gives you:
$$|\vec{a-b}|^2=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\cos\theta(a,b)$$
Inner product gives:
$$|\vec{a-b}|^2=\langle\vec{a-b},\vec{a-b}\rangle=|\vec{a}|^2+|\vec{b}|^2-2\langle\vec{a},\vec{b}\rangle$$
from there on it's an easy proof (I left you the technical details).
Hope that helps
