Angle Between Vectors in Terms of Dot Product
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Theorem
The angle between two non-zero vector quantities can be calculated by:
- $\theta = \arccos \frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }$
where:
- $\mathbf v \cdot \mathbf w$ represents the dot product of $\mathbf v$ and $\mathbf w$
- $\norm {\, \cdot \,}$ represents vector length
- $\arccos$ represents arccosine.
Proof
| \(\ds \norm {\mathbf v} \norm {\mathbf w} \cos \theta\) | \(=\) | \(\ds \mathbf v \cdot \mathbf w\) | Cosine Formula for Dot Product | |||||||||||
| \(\ds \cos \theta\) | \(=\) | \(\ds \frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }\) | because $\mathbf v, \mathbf w \ne \mathbf 0 \implies \norm {\mathbf v}, \norm {\mathbf w} \ne 0$ | |||||||||||
| \(\ds \map \arccos {\cos \theta}\) | \(=\) | \(\ds \map \arccos {\frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} } }\) | because $0 \le \theta \le \pi$ $\implies -1 \le \cos \theta \le 1$ | |||||||||||
| \(\ds \theta\) | \(=\) | \(\ds \map \arccos {\frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} } }\) | Composite of Bijection with Inverse is Identity Mapping |
$\blacksquare$
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