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, \(rs\cos\theta = \sum r_{i} s_{i}\)