Question

In a regular hexagon $A B C D E F, \overrightarrow{A D}+\overrightarrow{E B}+\overrightarrow{F C}=(3 \lambda-8) \overrightarrow{A B}$ Then $\lambda=$

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (B)

Since, $A D \| B C$ and $A D=2 B C$
$E B \| F A$ and $E B=2 F A$
$F C \| A B$ and $F C=2 A B$
$A D+E B=2(B C+F A)=2(A O+F A) \ldots$ (i)

In $\triangle A O F$
$F A+A O+O F=0$
$F A+A O=-O F \ldots$ (ii)
Put, Eqs. (ii) in (i)
$A D+E B=2(-O F)=2 F O$
$A D+E B=2 A B$
Now consider,
$A D+E B+F C=2 A B+2 A B[\therefore F C=2 A B]$
$(3 \lambda-8) A B=4 A B[\therefore $ given $]$
$3 \lambda-8=4$
$3 \lambda=12$
$\lambda=4$