Summarized Problem: prove that the two definitions of the dot product $$rs\cos\theta$$ and $$\sum r_{i} s_{i}$$ are equal.

Solution: To do this, we can use a slightly modified version of the law of cosines: $c^2 = {\bf a}^2 + {\bf b}^2 - 2ab\cos \theta$ where $${\bf a}$$ and $${\bf b}$$ are the two-dimensional vectors whose dot-product we are calculating (although this proof could generalize to higher-dimensional vectors as well) and $$c$$ is the magnitude between their heads. Represented like this, we can imagine $${\bf a}$$ and $${\bf b}$$ as two arms of a triangle with angle $$\theta$$ between them, and $$c$$ as the remaining side. To make this easier, let's also assume that the vector $${\bf a}$$ lies on the x-axis, meaning its magnitude is equal to its first value $$a_1$$ and its y-value is 0.

Given this information, we can deduce a few things about $$c$$. First, notice that if we imagine $$c$$ to be the hypotenuse of a right triangle, its height (denoted as $$c_2$$) is equal to $$|b_2|$$, or the magnitude of the height of the vector $${\bf b}$$. Second, we can determine that $$c$$'s base (denoted as $$c_1$$) is equal to to $$|a_1-b_1|$$. This can be demonstrated geometrically relatively easily. To make it easier to conceptualize, imagine adding $${\bf a}$$ and $${\bf -b}$$. The magnitude of the resultant 'x-value' would be equal to the absolute difference between $$a_1$$ and $$b_1$$.

Now that we know all of this, we can prove the equivalence of the two definitions. Let's first rearrange the equation to better match the form of $$rs\cos\theta$$ $\frac{-c^2 + a^2 + b^2}{2} = ab\cos\theta$ Let's then represent $$c^2$$ as the sum of the squares of its height and base. Let's do the same for vectors $${\bf a}$$ and $${\bf b}$$ (the magnitude of the square of a vector is equal to the square of its magntitude). In this equation, we can ignore $$a_2$$ since it's equal to 0 (remember that vector $${\bf a}$$ lies on the x-axis!) $\frac{-[(a_1 - b_1)^2 + b_2^2] + a_1^2 + b_1^2 +b_2^2}{2} = ab\cos\theta$ Expand. $\frac{-[a_1^2 -2a_1b_1 + b_1^2 + b_2^2] + a_1^2 + b_1^2 +b_2^2}{2} = ab\cos\theta$ And simplify. $\frac{-a_1^2 + 2a_1b_1 - b_1^2 - b_2^2 + a_1^2 + b_1^2 +b_2^2}{2} = ab\cos\theta$ $\frac{2a_1b_1}{2} = ab\cos\theta$ $a_1b_1 = ab\cos\theta$ Now if we evaluate $$\sum a_{i} b_{i}$$, we get $$a_1b_1 + a_2b_2 = a_1b_1 + 0b_2 = a_1b_1$$

Therefore, $$rs\cos\theta = \sum r_{i} s_{i}$$