There are two cases, the first where the two vectors are not scalar multiples of each other, and the second where they are. === Case 1 === :500px Let the two vectors $\mathbf v$ and $\mathbf w$ not be scalar multiples of each other.  Then by the definition of angle between vectors, we have $\theta$ defined as in the triangle as shown above. (Note that from Angle Between Non-Zero Vectors Always Defined, such a triangle is guaranteed to exist). By the Law of Cosines: :$\norm {\mathbf v - \mathbf w}^2 = \norm {\mathbf w}^2 + \norm {\mathbf v}^2 - 2 \norm {\mathbf v} \norm {\mathbf w} \cos \theta$ Now, observe that: {{begin-eqn}} {{eqn | l = \norm {\mathbf v - \mathbf w}^2       | r = \paren {\mathbf v - \mathbf w} \cdot \paren {\mathbf v - \mathbf w}       | c = Dot Product of Vector with Itself }} {{eqn | r = \mathbf v \cdot \mathbf v - 2 \paren {\mathbf v \cdot \mathbf w} + \mathbf w \cdot \mathbf w       | c = Properties of Dot Product }} {{eqn | r = \norm {\mathbf v}^2 + \norm {\mathbf w}^2 - 2 \paren {\mathbf v \cdot \mathbf w}       | c = Dot Product of Vector with Itself }} {{end-eqn}} Equating these two expressions for $\norm {\mathbf v - \mathbf w}^2$ gives: {{begin-eqn}} {{eqn | l = \norm {\mathbf w}^2 + \norm {\mathbf v}^2 - 2 \norm {\mathbf v} \norm {\mathbf w} \cos \theta       | r = \norm {\mathbf v}^2 + \norm {\mathbf w}^2 - 2 \paren {\mathbf v \cdot \mathbf w}       | c = }} {{eqn | l = -2 \paren {\norm {\mathbf v} \norm {\mathbf w} \cos \theta}       | r = -2 \paren {\mathbf v \cdot \mathbf w}       | c = }} {{eqn | l = \norm {\mathbf v} \norm {\mathbf w} \cos \theta       | r = \mathbf v \cdot \mathbf w       | c = }} {{end-eqn}} which is exactly the desired result. {{qed|lemma}} === Case 2 === Let $\mathbf v = \paren {v_1, v_2, \ldots, v_n}$ and $\mathbf w = \paren {w_1, w_2, \ldots, w_n}$. {{WLOG}}, let $\mathbf v = c \mathbf w$, where $c$ is some scalar.  If $c > 0$, then by the definition of angle between vectors: :$\theta = 0 \implies \cos \theta = 1$ If $c < 0$, then by the definition of angle between vectors: :$\theta = \pi \implies \cos \theta = -1$ (Note that $c$ cannot be $0$ because we have stipulated $\mathbf v$ and $\mathbf w$ to be non-zero). Then: {{begin-eqn}} {{eqn | l = \mathbf v \cdot \mathbf w       | r = \paren {c \mathbf w} \cdot \mathbf w       | c = }} {{eqn | r = c \paren {\mathbf w \cdot \mathbf w}       | c = Properties of Dot Product }} {{eqn | r = c \paren {\norm {\mathbf w}^2}       | c = Dot Product of Vector with Itself }} {{eqn | r = \paren {c \norm {\mathbf w} } \norm {\mathbf w}       | c = }} {{eqn | r = c \sqrt {\sum_{i \mathop = 1}^n w_i^2} \norm {\mathbf w}       | c ={{Defof|Vector Length}} in $\R^n$ }} {{eqn | r = \sgn \paren c \sqrt{c^2 \sum_{i \mathop = 1}^n w_i^2} \norm {\mathbf w}       | c = where $\sgn \paren c$ is the signum function, that is the algebraic sign of $c$ }} {{eqn | r = \sgn \paren c \sqrt {\sum_{i \mathop = 1}^n \paren {c w_i}^2} \norm {\mathbf w}       | c = }} {{eqn | r = \sgn \paren c \sqrt {\sum_{i \mathop = 1}^n v_i^2} \norm {\mathbf w}       | c = by hypothesis }} {{eqn | r = \sgn \paren c \norm {\mathbf v} \norm {\mathbf w}       | c = Definition of $\norm {\mathbf u}$ }} {{eqn | r = \norm {\mathbf v} \norm {\mathbf w} \cos \theta       | c = $\sgn \paren c = \cos \theta$ }} {{end-eqn}} {{qed}}