Question

In a regular hexagon $ABCDEF,\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=(3\lambda -8)\overrightarrow{AB}$ Then $\lambda =$

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (B)

Since, $AD\Vert BC$ and $AD=2BC$
$EB\Vert FA$ and $EB=2FA$
$FC\Vert AB$ and $FC=2AB$
$AD+EB=2(BC+FA)=2(AO+FA)\dots$ (i)

In $△AOF$
$FA+AO+OF=0$
$FA+AO=-OF\dots$ (ii)
Put, Eqs. (ii) in (i)
$AD+EB=2(-OF)=2FO$
$AD+EB=2AB$
Now consider,
$AD+EB+FC=2AB+2AB[\therefore FC=2AB]$
$(3\lambda -8)AB=4AB[\therefore$ given $]$
$3\lambda -8=4$
$3\lambda =12$
$\lambda =4$