Dot Product of Like Vectors
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Theorem
Let $\mathbf a$ and $\mathbf b$ be vector quantities such that $\mathbf a$ and $\mathbf b$ are like.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ and $\mathbf b$.
Then:
- $\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b}$
where $\norm {\, \cdot \,}$ denotes the magnitude of a vector.
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 $\mathbf b$ are like, by definition $\theta = 0$.
The result follows by Cosine of Zero is One, which gives that $\cos 0 \degrees = 1$.
$\blacksquare$
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.3)$