Let $u\neq0 $ and $v\neq 0$ be two vectors in a vector space with inner product such as $\left \| u \right \|=\left \| v \right \|=\left \| u-v \right \|$. Determine the angle between $u$ and $v$.
Progress:
$\cos\theta=\frac{\left \langle u,v \right \rangle}{\left \| u \right \|\left \| v \right \|}=\frac{\left \langle u,v \right \rangle}{\left \| u-v \right \|^2}=\frac{\left \| u-v \right \|^2-2\left \| u \right \|^2}{2\left \| u \right \|^2}=\frac{-\left \| u \right \|^2}{2 \left \| u \right \|^2}=-\frac{1}{2}$
$\cos\theta=-\frac{1}{2}\rightarrow \theta=\frac{2\pi }{3}$. Correct answer is $\theta=\frac{\pi }{3}$
Best Answer
$$\|u-v\|^2=\|u\|^2+\|v\|^2-2(u\cdot v)$$ implies that
$$ \|v\|^2=2(u\cdot v) $$
and then simplify and reorganize this
$$ \frac{\|v\|^2}{\|u\|\|v\|}=\frac{2(u\cdot v)}{\|u\|\|v\|} $$ to get the correct value of $1/2$.
The problem in your computation seems to be after the second equality:
$$ \|u-v\|^2-2\|v\|^2=-2(u\cdot v) $$
but not $2(u\cdot v)$ as you wanted.
You would be on the right track to use
$$2\|v\|^2-\|u-v\|^2$$
and then the sign error is gone.