Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

From the picture we have

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$.....(1) and $\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}$.....(2) and $\overrightarrow{AD}+\overrightarrow{DE}=-\overrightarrow{EA}$....(3).

And $\overrightarrow{FA}=\overrightarrow{EA}-\overrightarrow{EF}$......(4).

Now, $\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{EA}+\overrightarrow{FA}$

$=\overrightarrow{AB}+(\overrightarrow{AB}+\overrightarrow{BC})+(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD})+2\overrightarrow{EA}-\overrightarrow{EF}$

$=\overrightarrow{AB}+(\overrightarrow{AB}+\overrightarrow{BC})+(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD})-2(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE})-\overrightarrow{EF}$

$=\overrightarrow{AB}-\overrightarrow{CD}-2\overrightarrow{DE}-\overrightarrow{EF}$

$=\overrightarrow{AB}-\overrightarrow{CD}+2\overrightarrow{AB}+\overrightarrow{CD}$ [Since $\overrightarrow{DE}=-\overrightarrow{AB}$ and $\overrightarrow{FA}=-\overrightarrow{CD}$]

$=3\overrightarrow{AB}$.

Comparing with the problem we have $\lambda =3$.