# Geometric proof of dot product and projection

Given two vectors $u = (x_1,y_1)$ and $v=(x_2,y_2)$ there is the dot product $u\cdot v = x_1 x_2 + y_1 y_2$. This has the geometric interpretation $u \cdot v = |u||v|\cos\theta$. The standard proof of that geometric interpretation goes through the law of cosines which itself is proved with the pythagorean theorem or the techniques used in the proof of the pythagorean theorem. For example, the following image illustrates a proof of the law of cosines using the orthocenter. But how can one prove $x_1 x_2 + y_1 y_2 = |u||v|\cos\theta$ directly geometrically? There should be some way to prove it using my horribly drawn diagram below. The top left shaded area is $|u||v|\cos\theta$. And the other two shaded areas are $x_1 x_2$ and $y_1 y_2$. How does one prove that using geometry and trigonometry like in the first diagram for the law of cosines. • How do you define the cosine? My guess is if you get that clearly in your head, the geometric proof will follow. This might give you some yucks: arxiv.org/abs/1001.0201 Dec 21, 2016 at 20:45
• So I don't know what proof you intended. But I think I see how to prove it in a way analogous to Euclid's proof of the pythagorean theorem. It requires showing that $|u|(|v|cosθ) = |v|(|u|cosθ)$ in the sense of two rectangles have the same area which requires indroducing trigonometry. The top left area is divided into two parts each corresponding to one of the other areas. A difference is that in Euclid's proof of the pythagorean theorem he in effect showed that in a special scenario without ever introducing trigonometric functions. Dec 21, 2016 at 22:49
• Have you seen the geometric derivation for the angle difference formula $\cos(x-y) = \cos x \cos y - \sin x \sin y$? The dot product is exactly this expression. Hence if you would accept a geometric proof on the angle difference formula, then that is what you want here. Oct 8, 2018 at 7:31