Question

If ABCDEF is a regular hexagon, then \[\overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC}\] equals

 

विकल्प

• \[2 \overrightarrow{AB}\]

• \[\vec{0}\]

• \[3 \overrightarrow{AB}\]

• \[4 \overrightarrow{AB}\]

Answer 1

\[4 \overrightarrow{AB}\]

\[\begin{array}{l}\overrightarrow{AD} = 2 \overrightarrow{BC} \\ \overrightarrow{EB} = 2 \overrightarrow{FA} \\ \overrightarrow{FC} = 2 \overrightarrow{AB}\end{array}\]
\[\begin{array}{cl}\overrightarrow{AD} + \overrightarrow{EB} & =  2\left( \overrightarrow{BC} + \overrightarrow{FA} \right) \\ = & 2\left( \overrightarrow{AO} + \overrightarrow{FA} \right) \left( \because \hspace{0.167em} \overrightarrow{BC} = \overrightarrow{AO} \right)\end{array}\]
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 \[\overrightarrow{AB} = - \overrightarrow{FO}\]
\[\begin{array}{l}\therefore \hspace{0.167em} \hspace{0.167em} \overrightarrow{AD} + \overrightarrow{EB} = 2 \overrightarrow{AB} \\ \therefore \hspace{0.167em} \hspace{0.167em} \overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC} = 2 \overrightarrow{AB} + 2 \overrightarrow{AB} = 4 \overrightarrow{AB}\end{array}\]