If ABCDEF is a regular hexagon,then overright | vedclass

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(D) In a regular hexagon $ABCDEF$,let $O$ be the center of the hexagon. We can express the vectors in terms of the sides of the hexagon.We know that $

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$4\overrightarrow {AB}$

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(D) In a regular hexagon $ABCDEF$,let $O$ be the center of the hexagon. We can express the vectors in terms of the sides of the hexagon.We know that $\overrightarrow {AD} = 2\overrightarrow {AO}$,$\overrightarrow {EB} = 2\overrightarrow {EO}$,and $\overrightarrow {FC} = 2\overrightarrow {FO}$.Alternatively,using the properties of a regular hexagon:$\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {AF} = 2\overrightarrow {BC} + 2\overrightarrow {AB}$ (since $\overrightarrow {CD} = \overrightarrow {AF} + \overrightarrow {AB}$ is not direct,let's use the center $O$).Actually,$\overrightarrow {AD} = 2\overrightarrow {BC}$,$\overrightarrow {EB} = 2\overrightarrow {FA}$,and $\overrightarrow {FC} = 2\overrightarrow {AB}$.Summing these: $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = 2\overrightarrow {BC} + 2\overrightarrow {FA} + 2\overrightarrow {AB}$.$= 2(\overrightarrow {FA} + \overrightarrow {AB} + \overrightarrow {BC}) = 2(\overrightarrow {FC}) = 2(2\overrightarrow {AB}) = 4\overrightarrow {AB}$.

If $ABCDEF$ is a regular hexagon,then $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = $

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