If $A B C D E F$ is a regular hexagon with $\mathbf{A B}=\mathbf{a}$ and $\mathbf{B C}=\mathbf{b}$, then…
- $\mathbf{b}-\mathbf{a}$
- $-\mathbf{b}$
- $\mathbf{b}-2 \mathbf{a}$
- $\mathbf{a}-2 \mathbf{b}$
Solution

$\begin{aligned} & \mathbf{A C}=\mathbf{a}+\mathbf{b} \\ & \mathbf{A B}=2 \mathbf{b} \\ & \mathbf{C D}=\mathbf{A D}-\mathbf{A C}=2 \mathbf{b}-\mathbf{a}-\mathbf{b}=\mathbf{b}-\mathbf{a} \\ & \mathbf{D E}=\mathbf{- a}\end{aligned}$ $\therefore \quad \mathbf{C E}=\mathbf{C D}+\mathrm{DE}=\mathbf{b}-\mathbf{a}-\mathbf{c}=\mathbf{b}-2 \mathbf{a}$
Asked in: AP EAMCET 2021 (24 Aug Shift 2)