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In a Regular Hexagon Abcdef, a → B = A, B → C = → B and → C D = → C . Then, → a E = - Mathematics

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प्रश्न

In a regular hexagon ABCDEF, A \[\vec{B}\] = a, B \[\vec{C}\] = \[\overrightarrow{b}\text{ and }\overrightarrow{CD} = \vec{c}\].
Then, \[\overrightarrow{AE}\] =

विकल्प

  • \[\vec{a} + \vec{b} + \vec{c}\]
  • \[2 \vec{a} + \vec{b} + \vec{c}\]

  • \[\vec{b} + \vec{c}\]

     

  • \[\vec{a} + 2 \vec{b} + 2 \vec{c}\]

MCQ
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उत्तर

\[\vec{b} + \vec{c}\]
Given a regular hexagon ABCDEF such that \[\overrightarrow{AB} = \vec{a} , \overrightarrow{BC} = \vec{b}\] and \[\overrightarrow{CD} = \vec{c}\].
Then,
In \[\bigtriangleup ABC\], we have \[\overrightarrow{AC} = \vec{a} + \vec{b} .\]
\[\bigtriangleup ACD\], we have
\[\overrightarrow{AC} + \overrightarrow{CD} = \vec{AD} . \]
\[ \Rightarrow \overrightarrow{AD} = \overrightarrow{AC} + \vec{c} . \]
\[ \Rightarrow \overrightarrow{AD} = \vec{a} + \vec{b} + \vec{c} .\]
Again, in \[\bigtriangleup ADE\], we have
\[\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE} . \]
\[ \Rightarrow \overrightarrow{AE} = \vec{a} + \vec{b} + \vec{c} - \vec{a} . \]
\[ \Rightarrow \overrightarrow{AE} = \vec{b} + \vec{c} .\]
Hence option (c).

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अध्याय 23: Algebra of Vectors - MCQ [पृष्ठ ७८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 23 Algebra of Vectors
MCQ | Q 8 | पृष्ठ ७८

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