Angular bisector vector
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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.
$\blacksquare$
