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Question

If $$A_{1}, A_{2}........A_{n}$$ are the vertices of a regular plane polygon with $$n$$ sides and $$O$$ is the center then $$\displaystyle \sum_{i=1}^{n-1}(\vec {OA_{i}}\times\vec {OA_{i+1}})=$$


A
(n+1)(OA1×OA2)
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B
(n+1)(OA2×OA1)
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C
(n1) (OA1×OA2)
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D
(n1) (OA2×OA1)
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Solution

The correct option is B $$(n- 1) $$ $$(\vec {OA}_{1}\times \vec {OA}_{2})$$
$$\sum _{ i=1 }^{ n-1 }{ (\overline { { OA }_{ i } } \times \overline { { OA }_{ i+1 } } ) } $$
$$\overline { { OA }_{ 1 } } \times \overline { { OA }_{ 2 } } =|\overline { { OA }_{ 1 } } ||\overline { { OA }_{ 2 } } |\sin { \theta \hat { n }  } $$
$$\overline { { OA }_{ 1 } } ,\overline { { OA }_{ 2 } } ,......\overline { { OA }_{ n } } $$ will be of same length and subtend same angle at the centre.
$$\therefore \sum _{ i=1 }^{ n-1 }{ (\overline { { OA }_{ i } } \times \overline { { OA }_{ i+1 } } ) } =(n-1)|\overline { { OA }_{ 1 } } ||\overline { { OA }_{ 2 } } |\sin { \theta \hat { n }  } $$
$$\therefore \sum _{ i=1 }^{ n-1 }{ (\overline { { OA }_{ i } } \times \overline { { OA }_{ i+1 } } ) } =(n-1)(\overline { { OA }_{ 1 } } \times \overline { { OA }_{ 2 } }) $$
Hence, option C is correct.

652041_37229_ans_46ab08686f794ee096fc4490e997b02a.png

Mathematics

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