Prove dot product identity $u\cdot v = \tfrac14(|u \cdot v|^2 − |u − v|^2).$

Question

$$u\cdot v = \dfrac14\left(|u \cdot v|^2 − |u − v|^2\right).$$

So far I've only gotten the RHS to $\tfrac14((u \cdot v)(u \cdot v) − |u|^2 + 2(u \cdot v) - |v|^2)$

Only way I see this working is if $(u \cdot v)(u \cdot v)$ is equal to $|u + v|^2$ but that doesn't seem right either.

Answer 1

Directly, and assuming the dot product is over a real vector space:

$$\frac14\left(||u+v||^2-||u-v||^2\right)=\frac14\left(\color{red}{||u||^2}+\color{green}{||v||^2}+2u\cdot v-\color{red}{||u||^2}-\color{green}{||v||^2}+2u\cdot v\right)$$

and we're done

Answer 2

See parallelogram law and polarization identity for your (corrected) formula's point, geometric interpretation, usage and generalizations.