Let $\|\cdot\|$ be the euclidean norm on $\mathbb{R^n}$.
Let $a,b \in \mathbb{R^n}$,
My question is do we always have for all $t\in [0,1]$ , the following identity ? $$\|ta+(1-t)b\|^2=t\|a\|^2+(1-t)\|b\|^2-t(1-t)\|a-b\|^2$$
I tried to test it for $t=1/2$ and it gives me the parallelogram identity.
The identity can be proved by noting that $||x||^2 = x\cdot x$, hence we have
$$\begin{align} ||ta+(1-t)b||^2 &= (ta+(1-t)b)\cdot (ta+(1-t)b) \\ &= t^2 a\cdot a + 2t(1-t)a\cdot b + (1-t)^2 b \cdot b \\ &= ta\cdot a + (1-t)b\cdot b-t(1-t)a\cdot a - t(1-t)b\cdot b) + 2t(1-t)a\cdot b\\ &= ta\cdot a + (1-t)b\cdot b-t(1-t)(a\cdot a - 2a\cdot b + b\cdot b) \\ &= ta\cdot a + (1-t)b\cdot b-t(1-t)(a-b)\cdot (a-b) \\ &=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2 \end{align}$$