Determine the unknown vertices of a regular hexagon using vectorial methods $OABCDE$ is a regular hexagon, where $\overrightarrow{OA} = {\bf a}$ and $\overrightarrow{OB} = {\bf b}$. What are the position vectors of $C$, $D$, and $E$ in terms of $\bf a$ and $\bf b$?
Is it even possible to answer this question?
 A: More than a hint...Draw an accurate picture of a hexagon to start with. It should be easy for you to show that $\overrightarrow{BC}=-2\underline{a}+\underline{b}$ so that $\overrightarrow{OC}=2\underline{b}-2\underline{a}$. Then $\overrightarrow{OD}=\overrightarrow{OC}-\underline{a}$ etcetera...
A: Pardon the crude sketch—I’m currently on the school bus. I’m going to show you how to think through finding $C$, and then I’ll let you try the rest.

Remember that vectors add tip to tail, and the vector $\overrightarrow{OB}$ to take you from $O$ to $B$ is the vector to take you from $O$ to $A$ plus the vector to take you from $A$ to $B$.
By that logic we can easily find $\overrightarrow{AB}={\bf b}-{\bf a}$.
Then, use the wonderful symmetry of the hexagon to solve the rest. Since $\overrightarrow{BC}$ has the same vertical component as $\overrightarrow{AB}$, adding $\overrightarrow{OB}+\overrightarrow{AB}$ will give us a vector that points to the correct height of $C$ but too far to the right. So, to bring the vector tip back to the left, add the negative $-\bf a$.
That way we have $$\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{AB}-{\bf a}=2{\bf b}-{\bf a}$$
Using that way of thinking, can you try the rest and then come back if you’re still having trouble?
A: Note that
$$AB=b-a$$
$$BC=AB-a=b-2a\implies OC=b+BC=2b-2a$$
$$CD=OC-a=2b-3a$$
$$OE=OC-b=b-2a$$
