Dot Product of Perpendicular Vectors
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Theorem
Let $\mathbf a$ and $\mathbf b$ be vector quantities such that $\mathbf a \ne \bszero$ and $\mathbf b \ne \bszero$.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ and $\mathbf b$.
Then:
- $\mathbf a \cdot \mathbf b = 0$
- $\mathbf a$ and $\mathbf b$ are perpendicular.
Proof
By definition of dot product:
- $\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.
When $\mathbf a$ and $\mathbf b$ be perpendicular, by definition $\theta = 90 \degrees$.
The result follows by Cosine of Right Angle, which gives that $\cos 90 \degrees = 0$.
$\blacksquare$
Sources
- 1927: C.E. Weatherburn: Differential Geometry of Three Dimensions: Volume $\text { I }$ ... (previous) ... (next): Introduction: Vector Notation and Formulae: Products of Vectors
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.2)$
- 1970: George Arfken: Mathematical Methods for Physicists (2nd ed.) ... (previous) ... (next): Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product
- 1992: Frederick W. Byron, Jr. and Robert W. Fuller: Mathematics of Classical and Quantum Physics ... (previous) ... (next): Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product