In middle-school algebra (at least in the US) students are taught the FOIL method (first, outer, inner, last) to multiply two binomals:
$(a+b)(c+d) = ac+ad+bc+bd$.
It works like this:
$(a+b)(c+d) = a(c+d) + b(c+d)$ (by the distributive law)
$a(c+d) + b(c+d) = (ac + ad) + (bc + bd)$ (by the distributive law)
$(ac + ad) + (bc + bd) = ac + ad + bc + bd$ (by the associative property)
Dot products and vector addition have similar properties:
$(\mathbf{a} + \mathbf{b}) + (\mathbf{c+d}) = \mathbf{a+b+c+d}$ (vector addition is associative)
$\mathbf{a} \cdot (\mathbf{b + c}) = \mathbf{a\cdot b + a \cdot c}$ (the dot product is distributive)
And for what it matters, $\mathbf{a \cdot b = b \cdot a}$. This follows from the algebraic definition.
So does the dot product have an analogue of FOIL like this:
$\mathbf{(a+b)\cdot(c+d) = a\cdot c + a \cdot d + b \cdot c + b \cdot d}$
Edit: I think it should given these properties, am I correct?
Best Answer
The dot product for $x, y\in R^n$(where $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$) is defined to be $x\cdot y=x_1y_1+\dots +x_ny_n$.
For your particular question, take four vectors $a, b, x, y\in R^n$. Then use the definition of the dot product to compute $(a+b)\cdot (x + y)=(a_1+b_1)(x_1+y_1)+\dots +(a_n+b_n)(x_n+y_n)$
The above representation could be written as $$a_1x_1+a_1y_1+b_1x_1+b_1y_1+\dots+a_nx_n+a_ny_n+b_nx_n+b_ny_n$$
$$=(a_1x_1+\dots +a_nx_n)+(a_1y_1+\dots+a_ny_n)+(b_1x_1+\dots+b_nx_n)+(b_1y_1+\dots+b_ny_n)$$
$$=ax+ay+bx+by$$
And you get your desired property!