The figure shows ABCDEF as a regular hexagon. | vedclass

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(D) In a regular hexagon $ABCDEF$ with center $O$,we have the following properties:$\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO}$ (

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$6\overrightarrow {AO}$

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(D) In a regular hexagon $ABCDEF$ with center $O$,we have the following properties:$\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO}$ (by triangle law in $	riangle AOF$ is not correct,rather consider the vectors from center $O$)Alternatively,using the property of the center $O$:$\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} = (\overrightarrow{AO} + \overrightarrow{OB}) + (\overrightarrow{AO} + \overrightarrow{OC}) + (\overrightarrow{AO} + \overrightarrow{OD}) + (\overrightarrow{AO} + \overrightarrow{OE}) + (\overrightarrow{AO} + \overrightarrow{OF})$$= 5\overrightarrow{AO} + (\overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF})$Since $O$ is the center of the regular hexagon,the sum of vectors from the center to the vertices is zero: $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} = 0$.Therefore,$\overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} = -\overrightarrow{OA} = \overrightarrow{AO}$.Substituting this back:$= 5\overrightarrow{AO} + \overrightarrow{AO} = 6\overrightarrow{AO}$.

The figure shows $ABCDEF$ as a regular hexagon. What is the value of $\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF}$ in terms of $\overrightarrow {AO}$?

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