Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Here,we will use triangle vector property Consider $△ACD:$

We get $\overline{AC}+\overline{CD}=\overline{AD}-\left(i\right)$

In $△ADE:$

We get $\overline{AE}+\overline{ED}=\overline{AD}-\left(ii\right).$

Now adding $\left(i\right)\&\left(ii\right)$

$⟹\overline{AC}+\overline{CD}+\overline{AE}+\overline{ED}=2\overline{AD}-\left(iii\right)\because$

It is a regular hexagon,$

$\therefore \overline{AF}=\overline{CD}(\therefore$Equal and parallel opposite sides)$

Similarily,$\overline{ED}=\overline{AB}$

Putting in (iii):

$⟹\overline{AC}+\overline{AF}+\overline{AE}+\overline{AB}=2\overline{AD}$

Adding $\overline{AD}$ on both sides

$⟹\overline{AC}+\overline{AB}+\overline{AD}+\overline{AE}+\overline{AF}=3\overline{AD}⟹\lambda =3$