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.
$\blacksquare$