Question on interpretation of Dot Product Everyone says that the dot product is interpreted as the projection of A onto B (if you are dot producting A and B), but isn't that length just equal to |A|$\cos \left( \theta  \right)$?  Why does the dot product have an extra |B|?
 A: If the length of $B$ is $1$ then $\langle A,B\rangle$ is the coordinate of $A$ in direction $B$, not the projection of $A$ onto $B$; that would be $\langle A,B\rangle B$.
There is a nice interpretation of the scalar product where $B$ has arbitrary length.  Let $B=(b_1,b_2)$, then define $J(B):=(-b_2,b_1)$; you'll get $J(B)$ by rotating $B$ counterclockwise by $\pi/2$.  Observe that
$$\langle A, B\rangle=\det\bigl(A,J(B)\bigr),$$
that is: the dot product is the (orientated) area of the parallelogram spanned by $A$ and $J(B)$.
A: If we didn't account for the length of $B$ then the order of the operands to the dot product would matter.  $A$ dot $B$ would not be equal to $B$ dot $A$, and that would be irritating.
I suppose you could think of taking the dot product of $A$ and $B$ as going in three steps.  First scaling $B$ to be unit length, then measuring the projection of $A$ onto (unit length) $B$, then undoing that first scaling.  I'm not sure if that's the most intuitive way to go.  The main thing to remember is that you've got to do something to account for the length of $B$, or the dot product couldn't be commutative.
A: I have also heard people say "the dot product is as the projection of A onto B .."  But, of course, that is not entirely correct.  Yes -- the dot product is fundamentally a projection; but the actual computation of the dot product is given by the equation
$$a \cdot b = \| a \| \| b \| \cos \theta = \sum_i a_i b_i$$ 
So it is the projection of [$a$ onto $b$] multiplied by [the magnitude of $b$].
If you are interested, there is a wonderful, intuitive explanation of the dot product at this site: https://betterexplained.com/articles/vector-calculus-understanding-the-dot-product/
