Q:

Proving a vector bisect two other vectors

How can I prove the vector:
$$
\vec{w}=|\vec{u}|\vec{v} + |\vec{v}| \vec{u}
$$
bisects the angle between the vectors $\vec{u}$ and $\vec{v}$ ?
I have trying using the scalar product, but it does not help me. Thanks! 

A:

We want to compute $\theta_{u,w}$, and we know that $\cos\theta_{u,w} = \frac{u\cdot w}{|u||w|}$. Let's compute $u\cdot w$ in terms of $u$ and $v$:
\begin{align*}
    u\cdot w &= |u|u\cdot v + |v|u\cdot u\\
             &= |u|^2|v|\cos\theta_{u,v} + |u|^2|v|\\
             &= |u|^2|v|(1 + \cos\theta_{u,v})
\end{align*}
Then, we can also compute
    $$ |w| = \sqrt{w\cdot w} = \sqrt{2}|u||v|\sqrt{1+\cos\theta_{u,v}}.$$
From this we then have
    \begin{align*}
     \cos\theta_{u,w} &= \frac{u\cdot w}{|u||w|}\\
                    & = \frac{|u|^2|v|(1 + \cos\theta_{u,v})}{\sqrt{2}|u|^2|v|\sqrt{1+\cos\theta_{u,v}}}\\
               &= \sqrt{\frac{1+\cos\theta_{u,v}}{2}}
\end{align*}
Thus $\theta_{u,w} = \theta_{u,v}/2$ by the half angle formula.

