Question

In a regular hexagon ABCDEF, A$\overrightarrow{B}$ = a, B$\overrightarrow{C}$ = $\overrightarrow{b}\mathrm{and}\overrightarrow{CD}=\overrightarrow{c}$. Then, $\overrightarrow{AE}$ =
(a) $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$

(b) $2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$

(c) $\overrightarrow{b}+\overrightarrow{c}$

(d) $\overrightarrow{a}+2\overrightarrow{b}+2\overrightarrow{c}$

Answer 1

Option(c)
Given a regular hexagon $ABCDEF$ such that $\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{BC}=\overrightarrow{b}$ and $\overrightarrow{CD}=\overrightarrow{c}$. Then,
In $△ABC$, we have
$\overrightarrow{AC}=\overrightarrow{a}+\overrightarrow{b}.$
In $△ACD$, we have
$$\begin{gathered} \overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}. \\ \Rightarrow \overrightarrow{AD}=\overrightarrow{AC}+\overrightarrow{c}. \\ \Rightarrow \overrightarrow{AD}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}. \end{gathered}$$
Again, in $△ADE$, we have
$$\begin{gathered} \overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DE}. \\ \Rightarrow \overrightarrow{AE}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}-\overrightarrow{a}. \\ \Rightarrow \overrightarrow{AE}=\overrightarrow{b}+\overrightarrow{c}. \end{gathered}$$
Hence option (c).