Proof for Dot Product

Let AB, be a vector from A to B. Based on the cos rule. Cos Rule

\vec{A B} = \vec{O B} - \vec{O A}

| A B |^{2} = | O A |^{2} + | O B |^{2} - 2 | O A | | O B | \cos θ

Let $O A = A$ and $O B = B$ .

\begin{aligned} | A | | B | \cos θ & = \frac{1}{2} (| A |^{2} + | B |^{2} - | B - A |^{2}) \\ = \frac{1}{2} (\sum_{n = 1} (a_{n}^{2} + b_{n}^{2}) - \sum_{n = 1} (b_{n} - a_{n})^{2}) \\ = \frac{1}{2} (\sum_{n = 1} (a_{n}^{2} + b_{n}^{2}) - \sum_{n = 1} (b_{n}^{2} - 2 a_{n} b_{n} + a_{n}^{2})) \\ = \sum_{n = 1} a_{n} b_{n} \end{aligned}

Thus

A \cdot B = \sum_{n = 1} a_{n} b_{n}