About an Interesting Proof of a Basic Dot Product Property

It is a basic fact that, given two vectors $u,v\in\mathbb R^3$, the angle $\theta$ between them and their dot product is related by the expression

$$ u\cdot v=\sum u_iv_i = |u||v|\cos\theta. $$

This is sometimes also considered to be the definition of angles, especially in higher dimensional Euclidean space. The typical proof, assuming a definition of angles in $\mathbb R^3$ independent of this property (of which there are many), is expressing everything explicitly in terms of the coordinates $u_i,v_i$ and using the Law of Cosines. This is of course a perfectly good proof, even the simplest one, but perhaps unenlightening, as it seems like the result pops out of nowhere but some magical algebraic manipulation!

The following is an alternative that I have come up with recently. It trades off the need to perform complicated manipulations with the need for either some linear algebra knowledge, or just taking an property of angles on faith. We first show the two dimensional case, then show that the three dimensional case reduces thus.

Lemma. The angle $\theta$ between two vectors $u,v\in\mathbb R^2$ satisfies $\sum u_iv_i = |u||v|\cos\theta$.

Proof. The claimed expression is homogeneous, so it suffices to prove the claim for unit vectors $u,v$ followed by a necessary scaling to get a general case. Therefore, let $u=(\cos\phi_1,\sin\phi_1)$ and $v=(\cos\phi_2,\sin\phi_2)$. Then $\sum u_iv_i = \cos\phi_1\cos\phi_2+\sin\phi_1\sin\phi_2=\cos(\phi_1-\phi_2)$, as desired.

Theorem. The angle $\theta$ between two vectors $u,v\in\mathbb R^3$ satisfies $\sum u_iv_i = |u||v|\cos\theta$.

Proof. An orthogonal change of coordinates places the span of $u,v$ on the $xy$-plane so that their third coordinate is zero. This reduces the problem to the preceding Lemma.

The proof of the Theorem reducing to the two dimensional case is actually somewhat nontrivial to justify, in that one needs to show an orthogonal transformation $M\colon\mathbb R^3\to\mathbb R^3$ (satisfying $M^TM=I$ by definition and hence preserving inner products) really exists with the desired properties. But the language in which the proof is phrased makes it quite intuitively obvious that the claim is true, at least.

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