Question

$ABCDEF$ is a regular hexagon with centre at the origin such that $\overline{AD}+\overline{EB}+\overline{FC}=\lambda \overline{ED}$. Then $\lambda$ equals

• A. $2$
• B. $4$
• C. $6$
• D. $3$

Answer 1

The correct option is B $4$

$\overline{AD}=2\overline{OD},\overline{EB}=2\overline{EO}$
since $\overline{FC}\parallel \overline{ED}\therefore \overline{FC}=2\overline{ED}$
$\therefore \overline{AD}+\overline{EB}+\overline{FC}=2\overline{OD}+2\overline{EO}+2\overline{ED}$
$2(\overline{OD}+\overline{EO})+2\overline{ED}=2\overline{ED}+2\overline{ED}=4\overline{ED}$
give
$\overline{AD}+\overline{EB}+\overline{FC}=\lambda \overline{ED}\Rightarrow \lambda =4$