A regular hexagon OABCDE has side OA = 3 . Given…

A regular hexagon $OABCDE$ has side $OA = 3$. Given $\overrightarrow{OA}=\vec{m}$, $\overrightarrow{AB}=\vec{n}$ and $\overrightarrow{BC}=\vec{p}$, find the relations between vectors $\vec{m}$, $\vec{n}$ and $\vec{p}$, and express vectors $\overrightarrow{OD}$, $\overrightarrow{DA}$, $\overrightarrow{EB}$ and $\overrightarrow{OC}$ in terms of them.

Step 1: Setup and coordinates

Place vertex $O$ at the origin with $OA$ along the $x$-axis. Walking around a regular hexagon, the direction turns by $60^\circ$ at each vertex, so

\[ O = (0,0), \qquad A = (3,0), \qquad B = \left(\tfrac{9}{2},\tfrac{3\sqrt{3}}{2}\right), \]

\[ C = \left(3,3\sqrt{3}\right), \qquad D = \left(0,3\sqrt{3}\right), \qquad E = \left(-\tfrac{3}{2},\tfrac{3\sqrt{3}}{2}\right). \]

Step 2: Base vectors

\[ \vec{m} = \overrightarrow{OA} = (3,\,0), \qquad \vec{n} = \overrightarrow{AB} = \left(\tfrac{3}{2},\,\tfrac{3\sqrt{3}}{2}\right), \qquad \vec{p} = \overrightarrow{BC} = \left(-\tfrac{3}{2},\,\tfrac{3\sqrt{3}}{2}\right) \]

All three have the same magnitude: \[|\vec{m}| = 3, \qquad |\vec{n}| = |\vec{p}| = \sqrt{\tfrac{9}{4}+\tfrac{27}{4}} = 3.\]

Step 3: Relations between $\vec{m}$, $\vec{n}$ and $\vec{p}$

Adding $\vec m$ and $\vec p$: \[\vec{m}+\vec{p} = (3,0)+\left(-\tfrac{3}{2},\tfrac{3\sqrt{3}}{2}\right) = \left(\tfrac{3}{2},\tfrac{3\sqrt{3}}{2}\right) = \vec{n}\]

\[\vec{n} = \vec{m} + \vec{p} \qquad\Longleftrightarrow\qquad \vec{m} = \vec{n} - \vec{p} \qquad\Longleftrightarrow\qquad \vec{p} = \vec{n} - \vec{m}\]

Angles between the vectors

The interior angle of a regular hexagon is $120^\circ$, so consecutive side vectors (placed tail to tail) make an angle of $180^\circ - 120^\circ = 60^\circ$: \[\vec{m}\cdot\vec{n} = 3\cdot3\cdot\cos60^\circ = \tfrac{9}{2}, \qquad \text{check: } (3)(\tfrac{3}{2})+(0)(\tfrac{3\sqrt{3}}{2}) = \tfrac{9}{2} \;\checkmark\]

\[ \vec{n}\cdot\vec{p} = \left(\tfrac{3}{2}\right)\left(-\tfrac{3}{2}\right)+\left(\tfrac{3\sqrt{3}}{2}\right)^2 = -\tfrac{9}{4}+\tfrac{27}{4} = \tfrac{9}{2} = 3\cdot3\cdot\cos60^\circ \;\checkmark \]

\[ \angle(\vec{m},\vec{n}) = 60^\circ, \qquad \angle(\vec{n},\vec{p}) = 60^\circ, \qquad \angle(\vec{m},\vec{p}) = 120^\circ \]

Step 4: Express the required vectors

$\overrightarrow{OC}$

\[ \overrightarrow{OC} = \overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC} = \vec{m}+\vec{n}+\vec{p} = \vec m+(\vec m+\vec p)+\vec p = 2(\vec m+\vec p) = 2\vec n \]

Check: $(3,0)+\left(\tfrac{3}{2},\tfrac{3\sqrt{3}}{2}\right)+\left(-\tfrac{3}{2},\tfrac{3\sqrt{3}}{2}\right) = (3,3\sqrt{3}) = C$ $\checkmark$

\[\overrightarrow{OC} = \vec{m}+\vec{n}+\vec{p} = 2\vec{n}\]

$\overrightarrow{OD}$

Side $CD$ is parallel to $OA$ and points the opposite way, so $\overrightarrow{CD} = -\vec{m}$. Along the path $O \to A \to B \to C \to D$: \[\overrightarrow{OD} = \vec{m}+\vec{n}+\vec{p}-\vec{m} = \vec{n}+\vec{p} = \vec m+2\vec p\]

Check: $\vec{n}+\vec{p} = \left(0,\,3\sqrt{3}\right) = D$ $\checkmark$

\[\overrightarrow{OD} = \vec{n}+\vec{p} = \vec{m}+2\vec{p}\]

$\overrightarrow{DA}$

\[ \overrightarrow{DA} = \overrightarrow{DO}+\overrightarrow{OA} = -(\vec{n}+\vec{p})+\vec{m} = \vec{m}-\vec{n}-\vec{p} = \vec m-(\vec m+\vec p)-\vec p = -2\vec{p} \]

Check: $-2\vec{p} = -2\left(-\tfrac{3}{2},\tfrac{3\sqrt{3}}{2}\right) = (3,-3\sqrt{3}) = A - D$ $\checkmark$

\[\overrightarrow{DA} = \vec{m}-\vec{n}-\vec{p} = -2\vec{p}\]

$\overrightarrow{EB}$

Side $EO$ is parallel to $BC$ and points the opposite way, so $\overrightarrow{EO} = -\vec{p}$. Check: $O - E = \left(\tfrac{3}{2},-\tfrac{3\sqrt{3}}{2}\right) = -\vec p$ $\checkmark$. Along the path $E \to O \to A \to B$:

\[ \overrightarrow{EB} = -\vec{p}+\vec{m}+\vec{n} = \vec{m}+(\vec{n}-\vec{p}) = \vec m+\vec m = 2\vec{m} \]

Check: $B - E = (6,\,0) = 2\vec{m}$ $\checkmark$

\[\overrightarrow{EB} = \vec{m}+\vec{n}-\vec{p} = 2\vec{m}\]