Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is B $3$

Given that,

$ABCDEF$ is a regular hexagon.

Now,

$\begin{array}{c}\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=\lambda \overrightarrow{AD}\\ \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\\ =\overrightarrow{AB}+\left(\overrightarrow{AC}+\overrightarrow{\infty }\right)+\overrightarrow{AD}+\overrightarrow{AE}\left[AF=\infty \right]\\ =\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AD}+\overrightarrow{AE}\\ =2\overrightarrow{AD}+\left(\overrightarrow{AB}+\overrightarrow{AE}\right)\\ =2\overrightarrow{AD}+\left(\overrightarrow{E\infty }+\overrightarrow{AE}\right)\left[AB=E\infty \right]\\ =2\overrightarrow{AD}+\overrightarrow{AD}\\ =3\overrightarrow{AD}\\ \therefore \lambda =3\end{array}$