Question

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

Options

• \[\vec{a} + \vec{b} + \vec{c}\]

• \[2 \vec{a} + \vec{b} + \vec{c}\]

• \[\vec{b} + \vec{c}\]

   

• \[\vec{a} + 2 \vec{b} + 2 \vec{c}\]

Answer 1

\[\vec{b} + \vec{c}\]
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).