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Question
If ABCDEF is a regular hexagon, then \[\overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC}\] equals
Options
\[2 \overrightarrow{AB}\]
- \[\vec{0}\]
\[3 \overrightarrow{AB}\]
- \[4 \overrightarrow{AB}\]
Solution
\[\begin{array}{l}\overrightarrow{AD} = 2 \overrightarrow{BC} \\ \overrightarrow{EB} = 2 \overrightarrow{FA} \\ \overrightarrow{FC} = 2 \overrightarrow{AB}\end{array}\]
\[\begin{array}{cl}\overrightarrow{AD} + \overrightarrow{EB} & = 2\left( \overrightarrow{BC} + \overrightarrow{FA} \right) \\ = & 2\left( \overrightarrow{AO} + \overrightarrow{FA} \right) \left( \because \hspace{0.167em} \overrightarrow{BC} = \overrightarrow{AO} \right)\end{array}\]
In triangle AOF,
\[\begin{array}{l}\overrightarrow{FA} + \overrightarrow{AO} + \overrightarrow{FO} = 0 \\ \therefore \hspace{0.167em} \hspace{0.167em} \overrightarrow{FA} + \overrightarrow{AO} = - \overrightarrow{FO} \\ \therefore \hspace{0.167em} \hspace{0.167em} \overrightarrow{AD} + \overrightarrow{EB} = - 2 \overrightarrow{FO}\end{array}\]
And \[\overrightarrow{AB} = - \overrightarrow{FO}\]
\[\begin{array}{l}\therefore \hspace{0.167em} \hspace{0.167em} \overrightarrow{AD} + \overrightarrow{EB} = 2 \overrightarrow{AB} \\ \therefore \hspace{0.167em} \hspace{0.167em} \overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC} = 2 \overrightarrow{AB} + 2 \overrightarrow{AB} = 4 \overrightarrow{AB}\end{array}\]
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