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If $A B C D E F$ is a regular hexagon and $\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{A F}=\lambda \overrightarrow{A D}$, then $\lambda=$
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The correct answer is:
3
By triangle law, $\overrightarrow{A B}=\overrightarrow{A D}-\overrightarrow{B D}, \quad \overrightarrow{A C}=\overrightarrow{A D}-\overrightarrow{C D}$
Therefore, $\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{A F}$
$=3 \overrightarrow{A D}+(\overrightarrow{A E}-\overrightarrow{B D})+(\overrightarrow{A F}-\overrightarrow{C D})=3 \overrightarrow{A D}$
Hence $\lambda=3$, [Since $\overrightarrow{A E}=\overrightarrow{B D}, \overrightarrow{A F}=\overrightarrow{C D}]$
Therefore, $\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{A F}$
$=3 \overrightarrow{A D}+(\overrightarrow{A E}-\overrightarrow{B D})+(\overrightarrow{A F}-\overrightarrow{C D})=3 \overrightarrow{A D}$
Hence $\lambda=3$, [Since $\overrightarrow{A E}=\overrightarrow{B D}, \overrightarrow{A F}=\overrightarrow{C D}]$
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