Angle Bisector Vector/Geometric Proof 1

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 $\mathbf v$.

Proof

 * 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:

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

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