Angle Bisector Vector/Geometric Proof 1
Theorem
Let $\mathbf u$ and $\mathbf v$ be vectors of non-zero length.
Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective lengths.
Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the angle bisector of $\mathbf u$ and $\mathbf v$.
Proof
As shown above:
- Let $\gamma$ be the angle between $\mathbf u$ and $\mathbf v$.
- Let $\alpha$ be the angle between $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$.
- Let $\beta$ be the angle between $\mathbf u$ and $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$.
Note that $\norm {\mathbf u} \mathbf v$ is $\mathbf v$ multiplied by the length of $\mathbf u$.
By Vector Times Magnitude Same Length As Magnitude Times Vector the vectors $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$ have equal length.
So $\norm {\mathbf u} \mathbf v$, $\norm {\mathbf v} \mathbf u$ and $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ make an isosceles triangle.
Therefore:
\(\ds 2 \beta + \alpha\) | \(=\) | \(\ds 180 \degrees\) | ||||||||||||
\(\ds \beta\) | \(=\) | \(\ds 90 \degrees - \frac 1 2 \alpha\) | ||||||||||||
\(\ds 2 \beta\) | \(=\) | \(\ds 180 \degrees - \alpha\) |
But since $\mathbf v$ and $\norm {\mathbf u} \mathbf v$ are parallel, we also have:
\(\ds \delta\) | \(=\) | \(\ds \alpha\) | Parallelism implies Equal Corresponding Angles | |||||||||||
\(\ds \delta + \gamma\) | \(=\) | \(\ds 180 \degrees\) | ||||||||||||
\(\ds \alpha + \gamma\) | \(=\) | \(\ds 180 \degrees\) | ||||||||||||
\(\ds \gamma\) | \(=\) | \(\ds 180 \degrees - \alpha\) |
Thus $\gamma = 2 \beta$, and the result follows.
$\blacksquare$