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प्रश्न
Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.
उत्तर
Given: A regular octagon of eight sides with centre O.
To show: \[\overrightarrow{OA} +\overrightarrow{OB} + \overrightarrow{OC} +\overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH} = \overrightarrow{0} .\]
Proof: We know centre of the regular octagon bisects all the diagonals passing through it.
\[\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} +\overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH} =\overrightarrow{0} .\] and \[\overrightarrow{OC} = -\overrightarrow{OG} .\]
\[\overrightarrow{OA} + \overrightarrow{OE} = \overrightarrow{0} , \overrightarrow{OB} + \overrightarrow{OF} = \overrightarrow{0} , \overrightarrow{OD} + \overrightarrow{OH} = \vec{0}\]
\[\overrightarrow{OA} = - \overrightarrow{OE} , \overrightarrow{OB} = - \overrightarrow{OF} , \overrightarrow{OD} = - \overrightarrow{OH}\] and \[\overrightarrow{OC} +\overrightarrow{OG} = \overrightarrow{0} .\] ............(i)
Now,
\[\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} +\overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH} \]
\[ = \left(\overrightarrow{OA} +\overrightarrow{OE} \right) + \left( \overrightarrow{OB} + \overrightarrow{OF} \right) + \left( \overrightarrow{OC} + \overrightarrow{OG} \right) + \left( \overrightarrow{OD} + \overrightarrow{OH} \right)\]
\[ = \overrightarrow{0} +\overrightarrow{0} +\overrightarrow{0} + \overrightarrow{0} \]
\[ = \overrightarrow{0}\]
Hence proved.\]
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