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If $A B C D E F$ is a regular hexagon with $\mathbf{A B}=\mathbf{a}$ and $\mathbf{B C}=\mathbf{b}$, then $\mathbf{C E}$ equals to
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The correct answer is:
$\mathbf{b}-2 \mathbf{a}$
Given, $A B C D E F$ is a regular hexagon
$\mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}$
$\begin{aligned} & \mathbf{A C}=\mathbf{a}+\mathbf{b} \\ & \mathbf{A B}=2 \mathbf{b} \\ & \mathbf{C D}=\mathbf{A D}-\mathbf{A C}=2 \mathbf{b}-\mathbf{a}-\mathbf{b}=\mathbf{b}-\mathbf{a} \\ & \mathbf{D E}=\mathbf{- a}\end{aligned}$
$\therefore \quad \mathbf{C E}=\mathbf{C D}+\mathrm{DE}=\mathbf{b}-\mathbf{a}-\mathbf{c}=\mathbf{b}-2 \mathbf{a}$
$\mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}$
$\begin{aligned} & \mathbf{A C}=\mathbf{a}+\mathbf{b} \\ & \mathbf{A B}=2 \mathbf{b} \\ & \mathbf{C D}=\mathbf{A D}-\mathbf{A C}=2 \mathbf{b}-\mathbf{a}-\mathbf{b}=\mathbf{b}-\mathbf{a} \\ & \mathbf{D E}=\mathbf{- a}\end{aligned}$
$\therefore \quad \mathbf{C E}=\mathbf{C D}+\mathrm{DE}=\mathbf{b}-\mathbf{a}-\mathbf{c}=\mathbf{b}-2 \mathbf{a}$
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