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Question
Five forces \[\overrightarrow{AB,} \overrightarrow { AC,} \overrightarrow{ AD,}\overrightarrow{AE}\] and \[\overrightarrow{AF}\] act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is 6 \[\overrightarrow{AO,}\] where O is the centre of hexagon.
Solution
\[\overrightarrow{AB} + \overrightarrow {AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF}\]
Consider ∆ADE,
\[\begin{array}{l}\overrightarrow{AD} + \overrightarrow{DE} + \overrightarrow{EA} = 0 \\ \overrightarrow{AD} + \overrightarrow{DE} = \overrightarrow{AE} \\ 2 \overrightarrow{AO} + \left( - \overrightarrow{AB} \right) =\overrightarrow{AE} \left[ \overrightarrow{AD} = 2 \overrightarrow{AO} \hspace{0.167em} \hspace{0.167em}\text{ and }ED ||\hspace{0.167em}AB \hspace{0.167em} \hspace{0.167em} \overrightarrow{DE} = - \overrightarrow{AB} \right] \\ \therefore \hspace{0.167em} \hspace{0.167em} \overrightarrow{AE} + \overrightarrow{AB} = 2 \overrightarrow{AO} . . . . . (1)\end{array}\]
Now, consider ∆ADC
\[\overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DA} = 0\]
\[ \overrightarrow{AC} + \overrightarrow{CD} = \overrightarrow{AD} \left[ \because \overrightarrow{CD} = \overrightarrow{AF} \right]\]
\[ \overrightarrow{AC} + \overrightarrow{AF} = 2 \overrightarrow{AO} . . . . . \left( 2 \right)\]
Using (1) and (2),
\[\overrightarrow{AB} +\overrightarrow{AE} + \overrightarrow{AC} +\overrightarrow{AF}+\overrightarrow{AD} \]
\[ 2\overrightarrow{AO} + 2\overrightarrow{AO} + 2\overrightarrow{AO}\]
= \[6\overrightarrow{AO}\]
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