Angular bisector vector
Contents |
Theorem
Given two vectors $\vec u,\vec v$ of non-zero length, the direction of the vector $\| \vec u \| \vec v + \| \vec v \| \vec u$ is their angular bisector, where $\|\vec u\|$ and $\|\vec v\|$ are the lengths of $\vec u$ and $\vec v$ respectively.
Geometrical Proof 1
Let the angle between $\vec u$ and $\vec v$ be $\gamma$, the angle between $\| \vec u \| \vec v$ and $\| \vec v \| \vec u$ be $\alpha$ and the angle between $\vec u$ and $\| \vec u \| \vec v + \| \vec v \| \vec u$ be $\beta$ as shown above.
Note that $\left\Vert{ \vec u }\right\Vert \vec v$ is $\vec v$ multiplied by the length of $\vec u$.
The vectors $\| \vec u \| \vec v$ and $\| \vec v \| \vec u$ have equal length by Vector Times Magnitude Same Length As Magnitude Times Vector, so $\| \vec u \| \vec v$, $\| \vec v \| \vec u$ and $\| \vec u \| \vec v + \| \vec v \| \vec u$ make an isosceles triangle. Therefore,
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle 2\beta + \alpha\) | \(=\) | \(\displaystyle 180^\circ\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \beta\) | \(=\) | \(\displaystyle 90^\circ - \frac 1 2 \alpha\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle 2\beta\) | \(=\) | \(\displaystyle 180^\circ - \alpha\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
But since $\vec v$ and $\left\vert{ \vec u }\right\vert \vec v$ are parallel, we also have:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \delta\) | \(=\) | \(\displaystyle \alpha\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Euclid I.29 | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \delta + \gamma\) | \(=\) | \(\displaystyle 180^\circ\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \alpha + \gamma\) | \(=\) | \(\displaystyle 180^\circ\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \gamma\) | \(=\) | \(\displaystyle 180^\circ - \alpha\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
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 $\vec u$ or $\vec v$ by $-1$.
Geometrical Proof 2
The vectors $\| \vec u \| \vec v$ and $\| \vec v \| \vec u$ have equal length from Vector Times Magnitude Same Length As Magnitude Times Vector.
Thus $\| \vec u \| \vec v + \| \vec v \| \vec u$ is the diagonal of a rhombus.
In a rhombus, diagonals are the angular bisectors of the sides.
The result follows.
$\blacksquare$
Algebraic Proof
Let $\vec a = \| \vec u \| \vec v + \| \vec v \| \vec u$.
Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \cos \angle \vec u, \vec a\) | \(=\) | \(\displaystyle \frac {\vec u \cdot \vec a} {\left\Vert{ \vec u }\right\Vert \left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Cosine Formula for Dot Product | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\vec u \cdot \left({ \left\Vert{ \vec u }\right\Vert \vec v + \left\Vert{ \vec v }\right\Vert \vec u }\right)} {\left\Vert{ \vec u }\right\Vert \left\Vert{ \vec a} \right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left\Vert{ \vec u }\right\Vert \left({ \vec u \cdot \vec v }\right) + \left\Vert{ \vec v }\right\Vert \left({ \vec u \cdot \vec u }\right) } {\left\Vert{ \vec u }\right\Vert \left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Properties of Dot Product | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left\Vert{ \vec u }\right\Vert \left({ \vec u \cdot \vec v }\right) + \left\Vert{ \vec v }\right\Vert \left\Vert{ \vec u }\right\Vert^2} {\left\Vert{ \vec u }\right\Vert \left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Dot Product of a Vector with Itself | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\vec u \cdot \vec v + \left\Vert{ \vec u }\right\Vert \left\Vert{ \vec v }\right\Vert} {\left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \cos \angle \vec a, \vec v\) | \(=\) | \(\displaystyle \frac {\vec v \cdot \vec a} {\left\Vert{ \vec v }\right\Vert \left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Cosine Formula for Dot Product | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\vec v \cdot \left({ \left\Vert{ \vec u }\right\Vert \vec v + \left\Vert{ \vec v }\right\Vert \vec u }\right)} {\left\Vert{ \vec v }\right\Vert \left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left\Vert{ \vec u }\right\Vert \left({ \vec v \cdot \vec v }\right) + \left\Vert{ \vec v }\right\Vert \left({ \vec u \cdot \vec v }\right)} {\left\Vert{ \vec v }\right\Vert \left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Properties of Dot Product | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left\Vert{ \vec v }\right\Vert \left({ \vec u \cdot \vec v }\right) + \left\Vert{ \vec u }\right\Vert \left\Vert{ \vec v }\right\Vert^2} {\left\Vert{ \vec v }\right\Vert \left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Dot Product of a Vector with Itself | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\vec u \cdot \vec v + \left\Vert{ \vec u }\right\Vert \left\Vert{ \vec v }\right\Vert} {\left\Vert{ \vec a }\right\Vert}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Comparing the two expressions gives us:
- $\cos \angle \vec u, \vec a = \cos \angle \vec a, \vec v$
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:
- $\angle \vec u, \vec a = \angle \vec a, \vec v$
The result follows.
$\blacksquare$
