If the sides of a regular hexagon ABCDEF are | vedclass

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Question

(A) In a regular hexagon $ABCDEF$,the center $O$ is such that $\vec{OA} = \vec{BC} = \bar{b}$ and $\vec{AB} = \bar{a}$.Since it is a regular hexagon,t

Vedclass · Accepted Answer

$\bar{b} - \bar{a}$

Vedclass · Answer

(A) In a regular hexagon $ABCDEF$,the center $O$ is such that $\vec{OA} = \vec{BC} = \bar{b}$ and $\vec{AB} = \bar{a}$.Since it is a regular hexagon,the vector $\vec{FA}$ is parallel and equal in magnitude to $\vec{CD}$.Using the properties of vectors in a regular hexagon,we know that $\vec{CD} = \vec{BC} - \vec{AB} = \bar{b} - \bar{a}$.However,looking at the geometry,$\vec{FA} = \vec{AB} - \vec{BC}$ is incorrect based on standard vector addition.Let us re-evaluate: $\vec{AB} = \bar{a}$ and $\vec{BC} = \bar{b}$.In a regular hexagon,$\vec{CD} = \bar{b} - \bar{a}$.Since $\vec{FA} = \vec{CD}$,we have $\vec{FA} = \bar{b} - \bar{a}$.Wait,let's check the direction: $\vec{FA} = \vec{OA} - \vec{OF}$.Actually,$\vec{FA} = \vec{AB} - \vec{BC}$ is not correct. The correct relation is $\vec{FA} = \bar{b} - \bar{a}$.

If the sides of a regular hexagon $ABCDEF$ are given by $\vec{AB} = \bar{a}$ and $\vec{BC} = \bar{b}$,then $\vec{FA} = .....$

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