# W= |u| * v +|v| * u Where u and v and w are all non-zero vectors Show that w bisects the angle between u and v ?

May 25, 2018

See below:

#### Explanation:

The vector $\vec{w}$ is in the plane defined by $\vec{u}$ and $\vec{v}$.

Unless $\vec{u} = - \vec{v}$, we have $\vec{w} \ne 0$.

Let ${\theta}_{u}$ and ${\theta}_{v}$ be the angles the vector $\vec{w}$ makes with $\vec{u}$ and $\vec{v}$, respectively. Then we have

$\vec{w} \cdot \vec{u} = | \vec{w} | | \vec{u} | \cos {\theta}_{u}$
$q \quad q \quad = \left(| \vec{u} | \vec{v} + | \vec{v} | \vec{u}\right) \cdot \vec{u}$
$q \quad q \quad = | \vec{u} | \left(\vec{v} \cdot \vec{u}\right) + | \vec{v} | \left(\vec{u} \cdot \vec{u}\right)$
$q \quad q \quad = | \vec{u} | \left(\vec{v} \cdot \vec{u} + | \vec{v} | | \vec{u} |\right) \implies$

$| \vec{w} | \cos {\theta}_{u} = \vec{v} \cdot \vec{u} + | \vec{v} | | \vec{u} |$

By interchanging $\vec{u}$ and $\vec{v}$ we find

$| \vec{w} | \cos {\theta}_{v} = \vec{u} \cdot \vec{v} + | \vec{u} | | \vec{v} |$

and so

$| \vec{w} | \cos {\theta}_{u} = | \vec{w} | \cos {\theta}_{v} \implies$

$\textcolor{red}{{\theta}_{u} = {\theta}_{v}}$

Thus $\vec{w}$ bisects the angle between $\vec{u}$ and $\vec{v}$ (except in the special case $\vec{u} = - \vec{v}$ )