Resultant in Terms of Dot Product

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

• Law of Cosines

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product