# Angular bisector vector

## Theorem

Given two vectors $\mathbf u, \mathbf v$ of non-zero length, the direction of the vector $\| \mathbf u \| \mathbf v + \| \mathbf v \| \mathbf u$ is their angular bisector, where $\|\mathbf u\|$ and $\|\mathbf v\|$ are the lengths of $\mathbf u$ and $\mathbf v$ respectively.

## Geometrical Proof 1

Let the angle between $\mathbf u$ and $\mathbf v$ be $\gamma$, the angle between $\| \mathbf u \| \mathbf v$ and $\| \mathbf v \| \mathbf u$ be $\alpha$ and the angle between $\mathbf u$ and $\| \mathbf u \| \mathbf v + \| \mathbf v \| \mathbf u$ be $\beta$ as shown above.

Note that $\left\Vert{ \mathbf u }\right\Vert \mathbf v$ is $\mathbf v$ multiplied by the length of $\mathbf u$.

The vectors $\| \mathbf u \| \mathbf v$ and $\| \mathbf v \| \mathbf u$ have equal length by Vector Times Magnitude Same Length As Magnitude Times Vector, so $\| \mathbf u \| \mathbf v$, $\| \mathbf v \| \mathbf u$ and $\| \mathbf u \| \mathbf v + \| \mathbf v \| \mathbf u$ make an isosceles triangle. Therefore,

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle 2\beta + \alpha$$ $$=$$ $$\displaystyle$$ $$\displaystyle 180^\circ$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \beta$$ $$=$$ $$\displaystyle$$ $$\displaystyle 90^\circ - \frac 1 2 \alpha$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle 2\beta$$ $$=$$ $$\displaystyle$$ $$\displaystyle 180^\circ - \alpha$$ $$\displaystyle$$ $$\displaystyle$$

But since $\mathbf v$ and $\left\vert{ \mathbf u }\right\vert \mathbf v$ are parallel, we also have:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \delta$$ $$=$$ $$\displaystyle$$ $$\displaystyle \alpha$$ $$\displaystyle$$ $$\displaystyle$$ from Euclid I.29 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \delta + \gamma$$ $$=$$ $$\displaystyle$$ $$\displaystyle 180^\circ$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \alpha + \gamma$$ $$=$$ $$\displaystyle$$ $$\displaystyle 180^\circ$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \gamma$$ $$=$$ $$\displaystyle$$ $$\displaystyle 180^\circ - \alpha$$ $$\displaystyle$$ $$\displaystyle$$

Thus it is obvious that $\gamma = 2\beta$, and the result follows.

$\blacksquare$

### Notes

In order to get the second angular bisector vector, one can multiply either $\mathbf u$ or $\mathbf v$ by $-1$.

## Geometrical Proof 2

The vectors $\| \mathbf u \| \mathbf v$ and $\| \mathbf v \| \mathbf u$ have equal length from Vector Times Magnitude Same Length As Magnitude Times Vector.

Thus $\| \mathbf u \| \mathbf v + \| \mathbf v \| \mathbf u$ is the diagonal of a rhombus.

In a rhombus, diagonals are the angular bisectors of the sides.

The result follows.

$\blacksquare$

## Algebraic Proof

Let $\mathbf a = \| \mathbf u \| \mathbf v + \| \mathbf v \| \mathbf u$.

Then:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \cos \angle \mathbf u, \mathbf a$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\mathbf u \cdot \mathbf a} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ from Cosine Formula for Dot Product $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\mathbf u \cdot \left({ \left\Vert{ \mathbf u }\right\Vert \mathbf v + \left\Vert{ \mathbf v }\right\Vert \mathbf u }\right)} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a} \right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf u }\right) } {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ from Properties of Dot Product $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf u }\right\Vert^2} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ from Dot Product of Vector with Itself $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\mathbf u \cdot \mathbf v + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert} {\left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \cos \angle \mathbf a, \mathbf v$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\mathbf v \cdot \mathbf a} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ from Cosine Formula for Dot Product $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\mathbf v \cdot \left({ \left\Vert{ \mathbf u }\right\Vert \mathbf v + \left\Vert{ \mathbf v }\right\Vert \mathbf u }\right)} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf v \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right)} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ from Properties of Dot Product $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert^2} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$ Dot Product of Vector with Itself $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\mathbf u \cdot \mathbf v + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert} {\left\Vert{ \mathbf a }\right\Vert}$$ $$\displaystyle$$ $$\displaystyle$$

Comparing the two expressions gives us:

$\cos \angle \mathbf u, \mathbf a = \cos \angle \mathbf a, \mathbf v$

Since the angle used in the dot product is always taken to be between $0$ and $\pi$ and cosine is injective on this interval (from Shape of Cosine Function), we have:

$\angle \mathbf u, \mathbf a = \angle \mathbf a, \mathbf v$

The result follows.

$\blacksquare$