The Dot Product I'm sure this is pretty basic and has been answered before but I couldn't find any satisfactory answer. Purely for the purpose of understanding a geometric representation of the dot product could someone show me why $A \cdot B  > A \cos \theta$. Essentially why is $A \cos \theta$ not the projection? A diagram with the two of them would be nice.
 A: From law of cosines we know that $||A - B||^2 = ||A||^2+||B||^2 -2||A||||B||\cos (\theta)$
From this it follows that $||A||||B||\cos (\theta) =\frac{||A||^2+||B||^2-||A-B||^2}{2} = \frac{A\cdot A + B\cdot B-(A-B)\cdot(A-B)}{2} = \frac{A\cdot A+B\cdot B-A\cdot A+A\cdot B + B\cdot A - B \cdot B}{2} = \frac{2A \cdot B}{2} =  A \cdot B$
So overall, $||A|| ||B|| \cos (\theta) = A \cdot B$
Does this answer your question? The geometric intuition comes from law of cosines.
A: I think the reason for taking $A\cdot B$ to be $|A||B|\cos\theta$ and not just $|A|\cos\theta$ is not geometric, but algebraic: although the projection is more intuitive in a visual sense, the actual definition produces a symmetric product, which does not gives preference to one of the vectors over the other, and so is more suitable when making computations.
A: By definition, the dot product of two vectors $A = [a, b, c]$ and $B = [d, e, f]$ is $A\cdot B = ad + be + cf$. Or, the sum of each component sequentially multiplied together.
Note that $A\cdot B = |A||B|\cos(\theta)$ follows from the definition (and vice versa), so this is also a definition. This ends up being the area of the parallelogram that the vectors create, so this is yet another way to define the dot product. 
You have noticed that one vector's projection onto a line is given by $|A|\cos(\theta)$. Another way to write that using the dot product is $\frac{A\cdot B}{|B|}$ by rearranging the definition of the dot product. 
