# Vector of regular hexagon

#### anemone

##### MHB POTW Director
Staff member
ABCDEF is a regular hexagon with $\vec {BC}$ represents $\underline {b}$ and $\vec {FC}$ represents 2$\underline {a}$. Express, vector
$\vec {AB}$, $\vec {CD}$ and $\vec {EC}$ in terms of $\underline {a}$ and $\underline {b}$.

Before I start, I want to ask if we need to redefined $\underline {b}$ and 2$\underline {a}$? I mean, let $\underline {b}$ as

$$\begin{pmatrix} m \\0 \end{pmatrix}$$.

Thanks.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
You don't have to express $\underline{a}$ and $\underline{b}$ using their coordinates, though it's possible to get the answer that way as well. Suppose $\underline{a}=(k,l)$ and $\underline{b}=(m,0)$ and you express, say, $\vec{AB}$ using k, l and m. The difficulty is that you need express $\vec{AB}$ as a combination of specifically $(k,l)$ and $(m,0)$, not just as some expression of k, l and m.

It is clear that $\vec{EC}=\vec{FB}=\vec{FC}-\vec{BC}=2\underline{a}-\underline{b}$. To get a geometric intuition about the regular hexagon you can also look at this page.

• anemone

#### soroban

##### Well-known member
Hello, anemone!

I don't understand the notation $\underline{b}$.
If $\underline{b}$ represents $\overrightarrow{BC}$, isn't $b$ also a vector?

$ABCDEF\text{ is a regular hexagon with }\vec{b} = \overrightarrow {BC}\text{ and }2\vec{a} = \overrightarrow{FC}$

$\text{Express vectors }\overrightarrow{AB},\;\overrightarrow{CD},\; \overrightarrow{EC}\text{ in terms of }\vec{a}\text{ and }\vec{b}.$

Code:
          A       B
* - - - *
/ \     / \
/   \   /   \ b
/  a  \ /  a  \
F * - - - * - - - * C
\     / \     /
\   /   \   /
\ /     \ /
* - - - *
E       D
$\overrightarrow{AB} \:=\:\vec{a}$

$\overrightarrow{CD} \:=\:\vec{b} - \vec{a}$

$\overrightarrow{EC} \:=\:2\vec{a} - \vec{b}$

• anemone

#### anemone

##### MHB POTW Director
Staff member
You don't have to express $\underline{a}$ and $\underline{b}$ using their coordinates, though it's possible to get the answer that way as well. Suppose $\underline{a}=(k,l)$ and $\underline{b}=(m,0)$ and you express, say, $\vec{AB}$ using k, l and m. The difficulty is that you need express $\vec{AB}$ as a combination of specifically $(k,l)$ and $(m,0)$, not just as some expression of k, l and m.

It is clear that $\vec{EC}=\vec{FB}=\vec{FC}-\vec{BC}=2\underline{a}-\underline{b}$. To get a geometric intuition about the regular hexagon you can also look at this page.
Got it. Thanks, Evgeny.Makarov.
Maybe I'm just trying too hard...and not knowing that I'm actually trying to complicate the simple problem.

---------- Post added at 06:05 AM ---------- Previous post was at 05:59 AM ----------

I don't understand the notation $\underline{b}$.
If $\underline{b}$ represents $\overrightarrow{BC}$, isn't $b$ also a vector?
I had the same reaction as you when I first read the problem!
Anyway, thanks, Soroban.
Now I fully understand with the help of the diagram and it really is as simple as that. 