In a regular hexagon ABCDEF,overrightarrow{AE | vedclass

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(B) In a regular hexagon $ABCDEF$,we can express the vector $\overrightarrow{AE}$ using the triangle law of vector addition.Consider the path $A 	o C

Vedclass · Accepted Answer

$\overrightarrow{AC} + \overrightarrow{AF} - \overrightarrow{AB}$

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(B) In a regular hexagon $ABCDEF$,we can express the vector $\overrightarrow{AE}$ using the triangle law of vector addition.Consider the path $A 	o C 	o D 	o E$. Thus,$\overrightarrow{AE} = \overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DE}$.In a regular hexagon,opposite sides are parallel and equal in magnitude. Therefore,$\overrightarrow{CD} = \overrightarrow{AF}$ and $\overrightarrow{DE} = -\overrightarrow{AB}$ (since $\overrightarrow{AB} = -\overrightarrow{ED} = \overrightarrow{DE}$ is incorrect,rather $\overrightarrow{ED} = \overrightarrow{AB}$,so $\overrightarrow{DE} = -\overrightarrow{AB}$).Substituting these into the expression,we get $\overrightarrow{AE} = \overrightarrow{AC} + \overrightarrow{AF} - \overrightarrow{AB}$.

In a regular hexagon $ABCDEF$,$\overrightarrow{AE} = $

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