Angle Bisector Vector/Geometric Proof 1

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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

Angular Bisector Vector Diagram.png

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.

$\blacksquare$