Question

$ABCDEF$ is a regular hexagon with centre at the origin such that $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=\lambda \overrightarrow{ED}$. Then, $\lambda$ is equal to?

• A. $2$
• B. $4$
• C. $6$
• D. $3$

Answer 1

The correct option is B

$4$

Explanation for the correct option:

Step-1: Draw a regular hexagon $ABCDEF$:

Now, from the figure, we can see

$\overrightarrow{AD}=2\overrightarrow{OD}$, $\overrightarrow{EB}=2\overrightarrow{EO}$ [$\because O$is the centre]

As, $\overrightarrow{FC}\parallel \overrightarrow{ED}$

[$\because$Opposite sides a regular hexagon are parallel to each other]

Thus, $\overrightarrow{FC}=2\overrightarrow{ED}$

[$\because$The length of a diagonal of a regular hexagon is equal to two times the length of the side]

Step- 2: Find the value of $\lambda$ using vector addition:

$\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=2\overrightarrow{OD}+2\overrightarrow{EO}+2\overrightarrow{ED}$

$=2(\overrightarrow{OD}+\overrightarrow{EO})+2\overrightarrow{ED}$

$=2\overrightarrow{ED}+2\overrightarrow{ED}$

$=4\overrightarrow{ED}$ …..(1)

Given that $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=\lambda \overrightarrow{ED}$ …..(2)

On comparing equation (1) and (2) we get

$\lambda =4$

Hence, option ‘B’ is correct.