Since A B C D E F ABCDEF A BC D EF is a regular hexagon, we have: A B ⃗ = p ⃗ \vec{AB} = \vec{p} A B = p B C ⃗ = q ⃗ \vec{BC} = \vec{q} BC = q C D ⃗ = − p ⃗ \vec{CD} = -\vec{p} C D = − p D E ⃗ = − q ⃗ \vec{DE} = -\vec{q} D E = − q E F ⃗ = − A B ⃗ = − p ⃗ \vec{EF} = -\vec{AB} = -\vec{p} EF = − A B = − p F A ⃗ = − B C ⃗ = − q ⃗ \vec{FA} = -\vec{BC} = -\vec{q} F A = − BC = − q A C ⃗ = A B ⃗ + B C ⃗ = p ⃗ + q ⃗ \vec{AC} = \vec{AB} + \vec{BC} = \vec{p} + \vec{q} A C = A B + BC = p + q A D ⃗ = A B ⃗ + B C ⃗ + C D ⃗ = p ⃗ + q ⃗ − p ⃗ = 2 q ⃗ \vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} - \vec{p} = 2\vec{q} A D = A B + BC + C D = p + q − p = 2 q (Since A D ⃗ \vec{AD} A D is parallel to B C ⃗ \vec{BC} BC , and twice the length of B C ⃗ \vec{BC} BC .) Alternatively, A D ⃗ = 2 B C ⃗ \vec{AD} = 2\vec{BC} A D = 2 BC . E A ⃗ = B C ⃗ − A B ⃗ = q ⃗ − p ⃗ \vec{EA} = \vec{BC} - \vec{AB} = \vec{q} - \vec{p} E A = BC − A B = q − p
In a regular hexagon, we have:
C D ⃗ = − A B ⃗ = − p ⃗ \vec{CD} = -\vec{AB} = -\vec{p} C D = − A B = − p D E ⃗ = − B C ⃗ = − q ⃗ \vec{DE} = -\vec{BC} = -\vec{q} D E = − BC = − q E F ⃗ = − A B ⃗ = − p ⃗ \vec{EF} = -\vec{AB} = -\vec{p} EF = − A B = − p F A ⃗ = − B C ⃗ = − q ⃗ \vec{FA} = -\vec{BC} = -\vec{q} F A = − BC = − q A D ⃗ = 2 B C ⃗ = 2 q ⃗ \vec{AD} = 2 \vec{BC} = 2\vec{q} A D = 2 BC = 2 q E A ⃗ = q ⃗ − p ⃗ \vec{EA} = \vec{q} - \vec{p} E A = q − p A C ⃗ = A B ⃗ + B C ⃗ = p ⃗ + q ⃗ \vec{AC} = \vec{AB} + \vec{BC} = \vec{p} + \vec{q} A C = A B + BC = p + q 