$u,v,w\in R^n, \text{ } \|u\| = \|w\| = 7, \text{ } \|v\| = 1, \text{ } \|u - v + w\| = \|u + v + w\| \text{ }$
The angle between $u$, $v$ is $\frac{\pi}{3}$.
What is the angle between $v$, $w$ ?
This is what i have:
Let $s = v + w$
$\|u - s\| = \|u + s\|$
Using the Polarization identity
$u \cdot s = \frac{1}{4} (\|u + s \|^2 - \|u - s\|^2)$
$u \cdot s = 0$
$u \cdot (v + w) = 0$
$u \cdot v + u \cdot w = 0$
$u \cdot v = -(u \cdot w)$
$\cos(\frac{\pi}{3}) \|u\| \|v\| = -\cos(\alpha) \|u\| \|w\|$
$\alpha = \arccos(-\frac{\cos(\frac{\pi}{3}) \|v\|}{\|w\|})$
$\alpha = \arccos(-\frac{1}{14})$
$\alpha\text{ }$ is the angle between $u$, $w$
How do i find the angle between $v$, $w$ ?
Squaring both sides of $\left \| u-v+w \right \|=\left \| u+v+w \right \|$ yields after canceling $\left \| \cdot\right \|^2$s $$-2u\cdot v-2w\cdot v+2u\cdot w=2u\cdot v+2w\cdot v+2u\cdot w$$ And so $w\cdot v = -u\cdot w$. From here use the definition of the dot product like you did.
Edit: There's an algebric mistake. are you sure there isn't a degree of freedom for the angle?