If ABCDEF is a regular hexagon,the length of | vedclass

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(D) In a regular hexagon $ABCDEF$ with side length $a$,the interior angle at each vertex is $120^\circ$.Consider the vectors $\overrightarrow{AB}$ and

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$0$

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(D) In a regular hexagon $ABCDEF$ with side length $a$,the interior angle at each vertex is $120^\circ$.Consider the vectors $\overrightarrow{AB}$ and $\overrightarrow{AF}$. The angle between these two vectors is $120^\circ$.The dot product is given by $\overrightarrow{AB} \cdot \overrightarrow{AF} = |\overrightarrow{AB}| |\overrightarrow{AF}| \cos(120^\circ) = (a)(a) \left(-\frac{1}{2}\right) = -\frac{1}{2}a^2$.The term $\overrightarrow{BC}^2$ represents the square of the magnitude of vector $\overrightarrow{BC}$,which is $|\overrightarrow{BC}|^2 = a^2$.Therefore,$\overrightarrow{AB} \cdot \overrightarrow{AF} + \frac{1}{2} \overrightarrow{BC}^2 = -\frac{1}{2}a^2 + \frac{1}{2}(a^2) = 0$.The correct option is $D$.

If $ABCDEF$ is a regular hexagon,the length of whose side is $a$,then $\overrightarrow{AB} \cdot \overrightarrow{AF} + \frac{1}{2} \overrightarrow{BC}^2 = $

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