Why are vectors of a regular hexagon cos60 of each other?

Question

In a regular hexagon OABCDE $a$ , $b$ denote respectively the position vectors of $A$, $B$ with respect to $O$

I would have assumed that $\overrightarrow {OA} = \overrightarrow {AB} = \overrightarrow {BC} $ since it's a regular hexagon. And, therefore, $\overrightarrow {OC}$ should equal $\overrightarrow {OA} + \overrightarrow {AB} + \overrightarrow {BC} $ and therefore $\overrightarrow {3AB}$.

However, in the literature I was reading, $\overrightarrow{OC}$ is defined as $\overrightarrow {OA}\cos60 + \overrightarrow {AB} + \overrightarrow {BC} \cos60 $, and thus $\frac12 \overrightarrow{AB} + \overrightarrow {AB} + \frac12 \overrightarrow {AB} $ — ultimately $\overrightarrow{2AB}$. Why is this?

An image is included below.

Answer 1

What's true is that the lengths of $OA$ and $AB$ and $BC$ are all equal, but they're not equal as vectors -- they point in three different directions.

Answer 2

The actual vectors are:

$$\overrightarrow{OA}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}$$ $$\overrightarrow{AB}=\begin{pmatrix}1\\0\end{pmatrix}$$ $$\overrightarrow{BC}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}$$

So we have $|\overrightarrow{OA}|=|\overrightarrow {AB}|=|\overrightarrow{BC}|$.

And $\overrightarrow{OA}+\overrightarrow {AB}+\overrightarrow{BC}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}+\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}\frac12\\\frac{-\sqrt{3}}2\end{pmatrix}=\begin{pmatrix}2\\0\end{pmatrix}=2\overrightarrow {AB}$.

The literature uses $OA \cos 60^\circ$ to mean $|\overrightarrow{OA}| \cos 60^\circ$, and not $\overrightarrow{OA} \cos 60^\circ$, to mean the length of $\overrightarrow{OA}$, which is $1$.

Answer 3

Hint:

Since $\overrightarrow{OA}$, $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are not parallel it is not true that: $$ \overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}=3\overrightarrow{OA} $$