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$A B C D E F$ is a regular hexagon. The sum of the vectors $\mathbf{B E}, \mathbf{B C}, \mathbf{E F}, \mathbf{B A}, \mathbf{C F}, \mathbf{A F}$
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Verified Answer
The correct answer is:
3BF
Given, $A B C D E F$ is a regular hexagon
$\mathbf{B E}+\mathbf{B C}+\mathbf{E F}+\mathbf{B A}+\mathbf{C F}+\mathbf{A B}$
$\begin{aligned}=O E-O B & +O C-O B+O F-O E \\ & +O A-O B+O F-O C+O F-O A\end{aligned}$
$=3 \mathrm{OF}-3 \mathrm{OB}=3(\mathrm{OF}-\mathrm{OB})=3 \mathrm{BF}$
$\mathbf{B E}+\mathbf{B C}+\mathbf{E F}+\mathbf{B A}+\mathbf{C F}+\mathbf{A B}$
$\begin{aligned}=O E-O B & +O C-O B+O F-O E \\ & +O A-O B+O F-O C+O F-O A\end{aligned}$
$=3 \mathrm{OF}-3 \mathrm{OB}=3(\mathrm{OF}-\mathrm{OB})=3 \mathrm{BF}$
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