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Question: If \(ABCDEF\) is a regular hexagon, then \(\overset{\rightarrow}{AD} + \overset{\rightarrow}{EB} + \...

If ABCDEFABCDEF is a regular hexagon, then AD→+EB→+FC→=\overset{\rightarrow}{AD} + \overset{\rightarrow}{EB} + \overset{\rightarrow}{FC} =

A

O→\overset{\rightarrow}{O}

B

2AB→2\overset{\rightarrow}{AB}

C

3AB→3\overset{\rightarrow}{AB}

D

4AB→4\overset{\rightarrow}{AB}

Answer

4AB→4\overset{\rightarrow}{AB}

Explanation

Solution

We have AD→+EB→+FC→\overset{\rightarrow}{AD} + \overset{\rightarrow}{EB} + \overset{\rightarrow}{FC}

=(AB→+BC→+CD→)+(ED→+DC→+CB→)+FC→(\overset{\rightarrow}{AB} + \overset{\rightarrow}{BC} + \overset{\rightarrow}{CD}) + (\overset{\rightarrow}{ED} + \overset{\rightarrow}{DC} + \overset{\rightarrow}{CB}) + \overset{\rightarrow}{FC}

=AB→+(BC→+CB→)+(CD→+DC→)+ED→+FC→\overset{\rightarrow}{AB} + (\overset{\rightarrow}{BC} + \overset{\rightarrow}{CB}) + (\overset{\rightarrow}{CD} + \overset{\rightarrow}{DC}) + \overset{\rightarrow}{ED} + \overset{\rightarrow}{FC}

=AB→+O→+O→+AB→+2AB→\overset{\rightarrow}{AB} + \overset{\rightarrow}{O} + \overset{\rightarrow}{O} + \overset{\rightarrow}{AB} + 2\overset{\rightarrow}{AB}

= 4AB→4\overset{\rightarrow}{AB} [ED→=AB→,FC→=2AB→]\lbrack\overset{\rightarrow}{ED} = \overset{\rightarrow}{AB},\overset{\rightarrow}{FC} = 2\overset{\rightarrow}{AB}\rbrack