In this problem there are three particles, with velocities $\vec{v_1}, \vec{v_2}$ and $\vec{u}$.
Relative to the particle moving at $u$, the velocities $v_1$ and $v_2$ are of equal magnitude and are perpendicular. Accordingly, show: $$\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{v_2})\right|^2$$
I couldn't work out whether to put this into components or not. I only got it to work for $v_1\not=v_2$
I have taken $$\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|$$
To mean $$\left(\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right)$$
Then I expand out the LHS that to get:
$$|\vec{u}|^2-\vec{u}\cdot\vec{v_1}-\vec{u}\cdot\vec{v_2}+\frac{1}{4}(\vec{v_1}+\vec{v_2})^2$$
Then I am left with something different to the RHS unless I make some certain assumptions.
The relative velocities are defined as $\vec{v}_1-\vec{u}$ and $\vec{v}_2-\vec{u}$.
If we start from
$$\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{v_2})\right|^2$$
and separate $\vec{u}=\frac{1}{2}\vec{u}+\frac{1}{2}\vec{u}$ on the LHS and add $\frac{1}{2}\vec{u}-\frac{1}{2}\vec{u}=\vec{0}$ on the RHS, we have, grouping the terms opportunely,
$$\left|-\frac{1}{2}(\vec{v_1}-\vec{u})-\frac{1}{2}(\vec{v_2}-\vec{u})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{u})-\frac{1}{2}(\vec{v_2}-\vec{u})\right|^2$$
if we now evaluate the square of the modulus as $|\vec{a}+\vec{b}|^2=\vec{a}^2+\vec{b}^2+2\vec{a}\cdot\vec{b}$, we have
$$\frac{1}{4}(\vec{v_1}-\vec{u})^2+\frac{1}{4}(\vec{v_2}-\vec{u})^2=\frac{1}{4}(\vec{v_1}-\vec{u})^2+\frac{1}{4}(\vec{v_2}-\vec{u})^2$$ where we have taken into account $(\vec{v}_1-\vec{u})\cdot(\vec{v}_2-\vec{u})=0$, as per hypothesis.