Angle Bisector Vector/Algebraic Proof

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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

Let $\mathbf a = \norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$.

Then:

\(\ds \cos \angle \mathbf u, \mathbf a\) \(=\) \(\ds \frac {\mathbf u \cdot \mathbf a} {\norm {\mathbf u} \norm {\mathbf a} }\) Cosine Formula for Dot Product
\(\ds \) \(=\) \(\ds \frac {\mathbf u \cdot \paren {\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u} } {\norm {\mathbf u} \norm {\mathbf a} }\) $\mathbf a = \norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$
\(\ds \) \(=\) \(\ds \frac {\norm {\mathbf u} \paren {\mathbf u \cdot \mathbf v} + \norm {\mathbf v} \paren {\mathbf u \cdot \mathbf u} } {\norm {\mathbf u} \norm{\mathbf a} }\) Dot Product Associates with Scalar Multiplication
\(\ds \) \(=\) \(\ds \frac {\norm {\mathbf u} \paren {\mathbf u \cdot \mathbf v} + \norm {\mathbf v} \norm {\mathbf u}^2} {\norm {\mathbf u} \norm {\mathbf a} }\) Dot Product of Vector with Itself
\(\ds \) \(=\) \(\ds \frac {\mathbf u \cdot \mathbf v + \norm {\mathbf u} \norm {\mathbf v} } {\norm {\mathbf a} }\)


\(\ds \cos \angle \mathbf a, \mathbf v\) \(=\) \(\ds \frac {\mathbf v \cdot \mathbf a} {\norm {\mathbf v} \norm{\mathbf a} }\) Cosine Formula for Dot Product
\(\ds \) \(=\) \(\ds \frac {\mathbf v \cdot \paren {\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u} } {\norm {\mathbf v} \norm {\mathbf a} }\) $\mathbf a = \norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$
\(\ds \) \(=\) \(\ds \frac {\norm {\mathbf u} \paren {\mathbf v \cdot \mathbf v} + \norm {\mathbf v} \paren {\mathbf u \cdot \mathbf v} } {\norm {\mathbf v} \norm {\mathbf a} }\) Dot Product Associates with Scalar Multiplication
\(\ds \) \(=\) \(\ds \frac {\norm {\mathbf v} \paren {\mathbf u \cdot \mathbf v} + \norm {\mathbf u} \norm {\mathbf v}^2} {\norm {\mathbf v} \norm {\mathbf a} }\) Dot Product of Vector with Itself
\(\ds \) \(=\) \(\ds \frac {\mathbf u \cdot \mathbf v + \norm {\mathbf u} \norm {\mathbf v} } {\norm {\mathbf a} }\)

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):

$\angle \mathbf u, \mathbf a = \angle \mathbf a, \mathbf v$

The result follows.

$\blacksquare$