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In a regular hexagon `ABCDEF, bar(AB) + bar(AC)+bar(AD)+ bar(AE) + bar(AF)=k bar(AD)` then k is equal to

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In a regular hexagon `ABCDEF, bar(AB) + bar(AC)+bar(AD)+ bar(AE) + bar(AF)=k bar(AD)` then k is equal to
A. 2
B. 3
C. 4
D. 6
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Correct Answer - B
By triangle law, `AB=AD-BD,AC=AD-CD`
image
Therefore, `AB+AC+AD+AE+AF`
`=3AD+(AE-BD)+(AF-CD)=3AD`
hence, `lamda=3" "(becauseAE=BD,AF=CD)`
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