Angle Bisector Vector

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Theorem

Let $\mathbf u$ and $\mathbf v$ be vectors of non-zero length.

Let $\left\Vert{\mathbf u}\right\Vert$ and $\left\Vert{\mathbf v}\right\Vert$ be their respective lengths.


Then $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ is the angle bisector of $\mathbf u$ and $\mathbf v$.


Geometric Proof 1

Angular Bisector Vector Diagram.png

As shown above:

Let the angle between $\mathbf u$ and $\mathbf v$ be $\gamma$.
Let the angle between $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ be $\alpha$.
Let the angle between $\mathbf u$ and $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ be $\beta$.

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

By Vector Times Magnitude Same Length As Magnitude Times Vector the vectors $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ have equal length.

So $\left\Vert{\mathbf u}\right\Vert \mathbf v$, $\left\Vert{\mathbf v}\right\Vert \mathbf u$ and $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ make an isosceles triangle.

Therefore:

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

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

\(\displaystyle \delta\) \(=\) \(\displaystyle \alpha\) Parallelism implies Equal Corresponding Angles
\(\displaystyle \delta + \gamma\) \(=\) \(\displaystyle 180^\circ\)
\(\displaystyle \alpha + \gamma\) \(=\) \(\displaystyle 180^\circ\)
\(\displaystyle \gamma\) \(=\) \(\displaystyle 180^\circ - \alpha\)

Thus $\gamma = 2 \beta$, and the result follows.

$\blacksquare$


Geometric Proof 2

The vectors $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ have equal length from Vector Times Magnitude Same Length As Magnitude Times Vector.

Thus $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ is the diagonal of a rhombus.


The result follows from Diagonals of Rhombus Bisect Angles.

$\blacksquare$


Algebraic Proof

Let $\mathbf a = \left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert\mathbf u$.

Then:

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


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

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.

$\blacksquare$