Square of Sum of Vectors

Theorem

Let $\mathbf a$ and $\mathbf b$ be vector quantities.

Then:

$\paren {\mathbf a + \mathbf b}^2 = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2$

where:

$\mathbf a \cdot \mathbf b$ denotes dot product

$\mathbf a^2$ denotes the square of $\mathbf a$, that is: $\mathbf a \cdot \mathbf a$.

$\ds \paren {\mathbf a + \mathbf b}^2$

$=$

$\ds \paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a + \mathbf b}$

$\ds $

$=$

$\ds \mathbf a \cdot \paren {\mathbf a + \mathbf b} + \mathbf b \cdot \paren {\mathbf a + \mathbf b}$

$\ds $

$=$

$\ds \mathbf a \cdot \mathbf a + \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf a + \mathbf b \cdot \mathbf b$

$\ds $

$=$

$\ds \mathbf a \cdot \mathbf a + 2 \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf b$

$\ds $

$=$

$\ds \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2$

$\blacksquare$