Inner Products Define Angles Theorem

If $u,v$ are non-zero Vectors in ${\mathbb{R}}^n$ Then, the angle $\theta$ between $u,v$ satisfies: $\cos (\theta )=\frac{⟨u,v⟩}{\mid \mid u\mid \mid \mid \mid v\mid \mid }$

Intuition

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By applying the law of cosines: $\mid \mid u-v\mid {\mid }^2=\mid \mid u\mid {\mid }^2+\mid \mid v\mid {\mid }^2-2\mid \mid u\mid \mid \mid \mid v\mid \mid \cos (\theta )$ in a plane containing u and v

We have: $\mid \mid u-v\mid {\mid }^2-⟨u-v,u-v⟩-⟨u-v,u⟩-⟨u=v,v⟩$ $=⟨u,v⟩-⟨v,u⟩-⟨u,v⟩+⟨v,v⟩$ as they are linear $=\mid \mid u\mid {\mid }^2-2⟨u,v⟩+\mid \mid v\mid {\mid }^2$ by defn of || || $=\mid \mid u\mid {\mid }^2+\mid \mid v\mid {\mid }^2-2\mid \mid u\mid \mid \mid \mid v\mid \mid \cos (\theta )$ by Law Of Cosines

$⟹-2⟨u,v⟩=-2\mid \mid v\mid \mid \mid \mid v\mid \mid \cos (\theta )⟹\cos (\theta )=\frac{⟨u,v⟩}{\mid \mid u\mid \mid \mid \mid v\mid \mid }$ This also requires: $\overrightarrow{u},\overrightarrow{v}\ne 0$