regular hexagon sides as vectors 2

Question

Let ABCDEF be regular hexagon.Then $\overline{AB} +\overline{AC}+\overline{AD}+\overline{EA}+\overline{FA} =?$

my answer is $3\overline{AB}$ but actually it is $4\overline{AB}$

Answer 1

It is easiest to work in a non-rectangular coordinate system where the axes make a 120° angle with each other. For example put the origin at $A$, $(1,0)$ at $B$ and $(0,1)$ at $F$. Then the coordinates of the corners are $$ A(0,0) \quad B(1,0) \quad C(2,1) \quad D(2,2) \quad E(1,2) \quad F(0,1) $$ Your sum is then $$ (1,0) + (2,1) + (2,2) - (1,2) - (0,1) = (4,0) $$

Answer 2

Note that $\def\v#1#2{\overrightarrow{#1#2}}\v AE = \v BD$, as $ABCDEF$ is regular, hence $$ \v AD + \v EA = \v AB + \v BD + \v EA = \v AB $$ On the other hand we have $$ \v AC + \v FA = \v FC $$ As $ABCDEF$ is regular $\v FC = 2\v AB$, so the sum is $$ \v AB + \v AC + \v AD + \v EA + \v FA = \v AB + 2\v AB + \v AB = 4\v AB$$