Dot Product Intuition I'm searching to develop the intuition (rather than memorization) in relating the two forms of a dot product (by an angle theta between the vectors and by the components of the vector ).    
For example, suppose I have vector $\mathbf{a} = (a_1,a_2)$ and vector $\mathbf{b}=(b_1,b_2)$.  What's the physical or geometrical meaning that 
$$a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos(\theta)\;?$$
Why is multiplying $|\mathbf{b}|$ times $|\mathbf{a}|$ in direction of $\mathbf{b}$ the same as multiplying the first and second components of $\mathbf{a}$ and $\mathbf{b}$ and summing ?     
I know this relationship comes out when we use the law of cosines to prove, but even then i cant get a intuition in this relationship.
This image clarifies my doubt:

Thanks

By the scalar product (synonymously, the dot product) of two vectors $\mathbf{A}$ and $\mathbf{B}$, denoted by $\mathbf{A} \cdot\mathbf{B}$, we mean the quantity:
\begin{equation}
\mathbf{A} \cdot\mathbf{B} = |\mathbf{A}| |\mathbf{B}| cos(\mathbf{A},\mathbf{B})
\end{equation}
i.e., the product of the magnitudes of the vectors times the cosine of the angle between them. It follows from the above relationship that the scalar product of $\mathbf{A}$ and $\mathbf{B}$ equals the magnitude of $\mathbf{A}$ times the projection of $\mathbf{B}$ onto the direction of $\mathbf{A}$. As a matter of fact, the projection of $\mathbf{A}$ onto $\mathbf{A}$, denoted by $\mathbf{A_B}$ is length of the segment cut from $\mathbf{B}$ by the planes drawn through the end points of $\mathbf{A}$ perpendicular to $\mathbf{B}$, taken with the plus sign if the direction from the projection (onto $\mathbf{B}$) of the initial point of $\mathbf{A}$ to the projection of the end point of $\mathbf{B}$ coincides with the positive direction of $\mathbf{B}$, and with the minus sign otherwise. We can also conclude that:
\begin{equation}
\mathbf{A} \cdot\mathbf{B} = \mathbf{A_B} \cdot \mathbf{B} =\mathbf{A} \cdot \mathbf{B_A}
\end{equation}
Thus Scalar multiplication of two vectors is commutative. Given a system of rectangular coordinates ${x^1,x^2,x^3}$ let $\mathbf{i_1,i_2,i_3}$ the corresponding basis vectors. Then any vector \mathbf{A} can be represented in the form:
\begin{equation}
\mathbf{A}  = A_1 \mathbf{i_1}+A_2 \mathbf{i_2}+A_3 \mathbf{i_3}
\end{equation}
Since the vectors $\mathbf{i_1,i_2,i_3}$ are orthonormal we have
\begin{equation}
\mathbf{i_j}  \cdot \mathbf{i_k}=\delta_{j,k}
\end{equation}
Then
\begin{equation}
\mathbf{A}  \cdot \mathbf{i_1} = (A_1 \mathbf{i_1}+A_2 \mathbf{i_2}+A_3 \mathbf{i_3}) \cdot \mathbf{i_1} = A_1
\end{equation}
In the similar way we have $\mathbf{A}  \cdot \mathbf{i_2}=A_2 $ and $\mathbf{A}  \cdot \mathbf{i_3}=A_3 $. In other words, $A_1$, $A_2$ and $A_3$ are the projections of the vector $\mathbf{A}$ onto the coordinate axes. Wae can also write:
\begin{equation}
\mathbf{A} = (\mathbf{A}  \cdot \mathbf{i_1})  \cdot \mathbf{i_1} +(\mathbf{A}  \cdot \mathbf{i_2})  \cdot \mathbf{i_2} + (\mathbf{A}  \cdot \mathbf{i_3})  \cdot \mathbf{i_3}
\end{equation}
The scalar product of two vectors A and B can easily be expressed in
terms of their components:
\begin{equation}
\mathbf{A} \cdot \mathbf{B} =  (A_1 \mathbf{i_1}+A_2 \mathbf{i_2}+A_3 \mathbf{i_3}) \cdot (B_1 \mathbf{i_1}+B_2 \mathbf{i_2}+B_3 \mathbf{i_3})
\end{equation}
And when using the orthonormality condition:
\begin{equation}
\mathbf{A} \cdot \mathbf{B} =  A_1 B_1 + A_2 B_2 +A_3 B_3 
\end{equation}
