[Math] Proving a vector bisect two other vectors

Question

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!

Answer 1

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.