# Angle Bisector Vector

(Redirected from Angular 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.

$\blacksquare$