The Vector Dot Product in Two Dimensions

Show that the dot product of two unit vectors:

A = (a, b) and C = (c, d)

is $a c + b d$ .

Given both vectors are unit vectors:

a^{2} + b^{2} = c^{2} + d^{2} = 1

The rotation of $C$ by 90° is $R = (- d, c)$

A + m R = l C

Where $l$ is the cosine of the angle between $A$ and $C$ , and $m$ is the sine:

\begin{array}{l} a - m d = l c & \Rightarrow \frac{a}{d} - \frac{l c}{d} = m (1) \\ b + m c = l d & \Rightarrow - \frac{b}{c} + \frac{l d}{c} = m (2) \end{array}

Equate (1) and (2):

\frac{a}{d} + \frac{b}{c} = \frac{l d}{c} + \frac{l c}{d}

Multiply by $c d$ :

c a + b d = l (d^{2} + c^{2})

Since $c^{2} + d^{2} = 1$ :

l = a c + b d

Therefore, the dot product $A \cdot C$ = $a c + b d$ as was to be shown.