If ABCDEF is a regular hexagon with −−→AB=→a and −−→BC=→b, (where |→a|=|→b|) then −−→CE
A
→v−→b
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B
−→b
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C
→b−2→a
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D
None of these
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Solution
The correct option is C→b−2→a In ΔABC,−−→AB+−−→BC=−−→AC ⇒−−→AC=→a+→b.....(i) AD is parallel to BC and AD=2BC, therefore −−→AD=2→b. In ΔACD,−−→AC+−−→CD=−−→AD ⇒−−→CD=2→b−(→a+→b)=→b−→a ∴ In ΔCDE,−−→CE=−−→CD+−−→DE=→b−→a+(−→a)=→b−2→a