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(A) In a regular hexagon $ABCDEF$,let the center be $O$. Given $\overrightarrow{AB} = a$ and $\overrightarrow{BC} = b$. In $	riangle ABC$,by triangle

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$2\,b - a$

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(A) In a regular hexagon $ABCDEF$,let the center be $O$. Given $\overrightarrow{AB} = a$ and $\overrightarrow{BC} = b$. In $	riangle ABC$,by triangle law of vector addition,$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = a + b$. Since $O$ is the center,$\overrightarrow{AO} = \overrightarrow{OC} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}(a + b)$. Also,$\overrightarrow{AF} = \overrightarrow{OC} = \frac{1}{2}(a + b)$ and $\overrightarrow{FE} = \overrightarrow{AB} = a$. Now,in $	riangle AFE$,by triangle law of vector addition,$\overrightarrow{AE} = \overrightarrow{AF} + \overrightarrow{FE}$. Substituting the values,$\overrightarrow{AE} = \frac{1}{2}(a + b) + a = \frac{1}{2}a + \frac{1}{2}b + a = \frac{3}{2}a + \frac{1}{2}b$. Wait,let us re-evaluate using the property of the hexagon: $\overrightarrow{AE} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE}$. Alternatively,$\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE}$. Since $\overrightarrow{AD} = 2\overrightarrow{BC} = 2b$ and $\overrightarrow{DE} = -\overrightarrow{AB} = -a$,Therefore,$\overrightarrow{AE} = 2b - a$.

If the vectors represented by the sides $AB$ and $BC$ of the regular hexagon $ABCDEF$ are $a$ and $b$ respectively,then the vector represented by $\overrightarrow{AE}$ will be

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