Resultant in Terms of Dot Product
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Theorem
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Let their resultant be $\mathbf v$:
- $\mathbf v = \mathbf a + \mathbf b$
Then:
- $\mathbf v^2 = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2$
where:
- $\mathbf v^2$ denotes the square of $\mathbf v$
- $\mathbf a \cdot \mathbf b$ denotes the dot product of $\mathbf a$ and $\mathbf b$
Proof
\(\ds \mathbf v^2\) | \(=\) | \(\ds \paren {\mathbf a + \mathbf b}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\mathbf a + \mathbf b} \paren {\mathbf a + \mathbf b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf a \cdot \mathbf a + \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf a + \mathbf b \cdot \mathbf b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2\) |
$\blacksquare$
Also presented as
This result can also be presented as:
- $v^2 = a^2 + 2 a b \cos \theta + b^2$
where:
- $v$, $a$ and $b$ are the magnitudes of $\mathbf v$, $\mathbf a$ and $\mathbf b$ respectively
- $\theta$ is the angle between the directions of $\mathbf a$ and $\mathbf b$.
Also see
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product