Question

If ABCDEF is a regular hexagon and if $\overline{AB}+\overline{AC}+\overline{AD}+\overline{AE}+\overline{AF}=\lambda \overline{AD}$ then $\lambda$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

3

$ABCDEF$ is a regular hexagon.
$\overline{AB}+\overline{AC}+\overline{AD}+\overline{AE}+\overline{AF}$
$=\overline{AB}+\overline{AB}+\overline{BC}+\overline{AD}+\overline{AE}+\overline{AE}+\overline{EF}$
$=\overline{AB}+\overline{AB}+\overline{AD}+\overline{AE}+\overline{AE}$ ($\because \overline{BC}+\overline{EF}=0$)
$=3\overline{AD}+2(\overline{AB}+\overline{DE})$
$=3\overline{AD}$ ($\because \overline{AB}+\overline{DE}=0$)
$\therefore \lambda =3$
Hence, option D.