In a regular hexagon ABCDEF,overrightarrow{AB | vedclass

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Question

(A) In a regular hexagon $ABCDEF$,the center $O$ is such that $\overrightarrow{OA}=\overrightarrow{BC}=\vec{b}$ and $\overrightarrow{AB}=\overrightarr

Vedclass · Accepted Answer

$\vec{a}-\vec{b}$

Vedclass · Answer

(A) In a regular hexagon $ABCDEF$,the center $O$ is such that $\overrightarrow{OA}=\overrightarrow{BC}=\vec{b}$ and $\overrightarrow{AB}=\overrightarrow{OC}=\vec{a}$.Also,$\overrightarrow{CD}=\overrightarrow{AF}$ and $\overrightarrow{DE}=\overrightarrow{BA}=-\vec{a}$.By the polygon law of vector addition,the sum of vectors along the sides of a closed polygon is zero:$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FA} = 0$Since $\overrightarrow{CD} = \overrightarrow{AF} = -\overrightarrow{FA}$,we have:$\vec{a} + \vec{b} + \overrightarrow{CD} - \vec{a} + \overrightarrow{EF} - \overrightarrow{CD} = 0$Alternatively,using the property of a regular hexagon,$\overrightarrow{FA} = \overrightarrow{CD} - \overrightarrow{BC} = \vec{a} - \vec{b}$.

In a regular hexagon $ABCDEF$,$\overrightarrow{AB}=\vec{a}$ and $\overrightarrow{BC}=\vec{b}$,then $\overrightarrow{FA}=$

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