In a regular hexagon ABCDEF,AD + EB + FC = (3 | vedclass

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(B) Let $O$ be the center of the regular hexagon $ABCDEF$. In a regular hexagon,the diagonals $AD$,$BE$,and $CF$ pass through the center $O$ and are e

Vedclass · Accepted Answer

$4$

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(B) Let $O$ be the center of the regular hexagon $ABCDEF$. In a regular hexagon,the diagonals $AD$,$BE$,and $CF$ pass through the center $O$ and are equal to twice the side length of the hexagon. Specifically,$AD = 2BC$,$EB = 2FA$,and $FC = 2AB$. Since $O$ is the center,we have $\vec{AD} = 2\vec{AO}$,$\vec{EB} = 2\vec{EO}$,and $\vec{FC} = 2\vec{FO}$. In a regular hexagon,$\vec{AO} + \vec{EO} + \vec{FO} = \vec{0}$ is not directly true,but we know $\vec{AD} + \vec{EB} + \vec{FC} = 2(\vec{AO} + \vec{EO} + \vec{FO})$. Since $\vec{AO} + \vec{EO} + \vec{FO} = \vec{0}$ is incorrect,let us use the vector addition: $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD}$,$\vec{EB} = \vec{ED} + \vec{DC} + \vec{CB}$,etc. Alternatively,using the property of the center $O$: $\vec{AD} = 2\vec{BC}$,$\vec{EB} = 2\vec{CD}$,$\vec{FC} = 2\vec{DE}$ is not correct. Correct approach: $\vec{AD} = 2\vec{BC}$,$\vec{EB} = 2\vec{CD}$,$\vec{FC} = 2\vec{DE}$ is wrong. Actually,$\vec{AD} = 2\vec{BC}$,$\vec{EB} = 2\vec{CD}$ is wrong. Let $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$. Then $\vec{CD} = \vec{b} - \vec{a}$,$\vec{DE} = -\vec{a}$,$\vec{EF} = -\vec{b}$,$\vec{FA} = \vec{a} - \vec{b}$. $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{a} + \vec{b} + (\vec{b} - \vec{a}) = 2\vec{b}$. $\vec{EB} = \vec{ED} + \vec{DC} + \vec{CB} = \vec{a} - (\vec{b} - \vec{a}) - \vec{b} = 2\vec{a} - 2\vec{b}$. $\vec{FC} = \vec{FA} + \vec{AB} + \vec{BC} = (\vec{a} - \vec{b}) + \vec{a} + \vec{b} = 2\vec{a}$. Sum $= \vec{AD} + \vec{EB} + \vec{FC} = 2\vec{b} + 2\vec{a} - 2\vec{b} + 2\vec{a} = 4\vec{a} = 4\vec{AB}$. Given $(3\lambda - 8)\vec{AB} = 4\vec{AB}$,so $3\lambda - 8 = 4$,which gives $3\lambda = 12$,so $\lambda = 4$.

In a regular hexagon $ABCDEF$,$AD + EB + FC = (3\lambda - 8) AB$. Then $\lambda =$

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