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(D) In a regular hexagon $ABCDEF$,the center $O$ is the origin. Let the position vectors of the vertices be $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{

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$\vec{BC} + \vec{DE}$

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(D) In a regular hexagon $ABCDEF$,the center $O$ is the origin. Let the position vectors of the vertices be $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{e}, \vec{f}$.Since it is a regular hexagon,$\vec{AB} = \vec{FO} = \vec{OC} = \vec{b} - \vec{a}$.Alternatively,using vector addition in the hexagon:$\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = \vec{0}$.We know that in a regular hexagon,$\vec{AB} = \vec{ED}$ and $\vec{AF} = \vec{CD}$ is not generally true,but $\vec{AB} + \vec{AF} = \vec{AO}$.Let's use the property: $\vec{AB} = \vec{FO}$ and $\vec{EF} = \vec{OC}$.Actually,a simpler approach: $\vec{AB} + \vec{AF} = \vec{AO}$.Also,$\vec{CD} + \vec{EF} = \vec{CD} + \vec{CB} = \vec{CO}$ (since $\vec{EF} = \vec{CB}$ is false,rather $\vec{EF} = \vec{BC}$ is false).Correct approach: $\vec{AB} = \vec{ED}$ is false. In a regular hexagon,$\vec{AB} + \vec{AF} = \vec{AO}$.$\vec{CD} + \vec{EF} = \vec{CO}$.Thus,$\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF} = \vec{AO} + \vec{CO} = \vec{BC} + \vec{DE}$.

Let $ABCDEF$ be a regular hexagon with the vertices $A, B, C, D, E, F$ in counterclockwise order. Then the vector $\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF}$ is equal to

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