Angle Bisector Vector
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 $\mathbf v$.
Geometric Proof 1
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.
$\blacksquare$
Geometric Proof 2
The vectors $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$ have equal length from Vector Times Magnitude Same Length As Magnitude Times Vector.
Thus $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the diagonal of a rhombus.
The result follows from Diagonals of Rhombus Bisect Angles.
$\blacksquare$
Algebraic 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.
$\blacksquare$