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Abcde is a Pentagon, Prove that → a B + → a E + → B C + → D C + → E D + → a C = 3 → a C - Mathematics

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प्रश्न

ABCDE is a pentagon, prove that 
\[\overrightarrow{AB} + \overrightarrow{AE} + \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{ED} + \overrightarrow{AC} = 3\overrightarrow{AC}\]

बेरीज

उत्तर

Given: ABCDE is a pentagon.
To Prove: \[\overrightarrow{AB} + \overrightarrow{AE} + \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{ED} + \overrightarrow{AC} = 3 \overrightarrow{AC} .\]
Proof: We have,
\[LHS = \overrightarrow{AB} + \overrightarrow{AE} + \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{ED} + \overrightarrow{AC} \] 
\[= \left( \overrightarrow{AB} + \overrightarrow{BC} \right) + \left( \overrightarrow{AE} + \overrightarrow{ED} \right) + \overrightarrow{DC} + \overrightarrow{AC}\]
\[= \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{DC} + \overrightarrow{AC}\]      [∵​ \[\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\]  and  \[\overrightarrow{AE} + \overrightarrow{ED} = \overrightarrow{AD}\]]

\[\overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AC} - \overrightarrow{AD} + \overrightarrow{AC}\]                       [∵ \[\overrightarrow{AD} + \overrightarrow{DC} = \overrightarrow{AC} \Rightarrow \overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD} \]]
\[= 3 \overrightarrow{AC} = RHS\]
Hence proved.

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पाठ 23: Algebra of Vectors - Exercise 23.2 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 23 Algebra of Vectors
Exercise 23.2 | Q 7.2 | पृष्ठ १७

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