In a regular hexagon $A B C D E F$, $\mathbf{A D}+\mathbf{E B}+\mathbf{F C}=(\beta \lambda-8) \mathbf{A B}$. Then $\lambda=$

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Aicharya · Accepted Answer

Since, $\mathbf{A D} \| \mathbf{B C}$ and $\mathbf{A D}=2 \mathbf{B C}$ $\begin{gathered} \mathbf{E B} \| \mathbf{F A} \text { and } \mathbf{E B}=2 \mathbf{F A} \ \mathbf{F C} \| \mathbf{A B} \text { and } \mathbf{F C}=2 \mathbf{A B} \end{gathered}$ $\mathbf{A D}+\mathbf{E B}=2(\mathbf{B C}+\mathbf{F A})=2(\mathbf{A O}+\mathbf{F A})$ ...(i) In $\Delta \mathrm{AOF}$, $\begin{aligned} \mathbf{F A}+\mathbf{A O}+\mathbf{O F} & =0 \ \mathbf{F A}+\mathbf{A O} & =-\mathbf{O F} \quad \ldots (ii) \end{aligned}$ Put, Eqs. (ii) in (i) $\begin{aligned} & \mathbf{A D}+\mathbf{E B}=2(-\mathbf{O F})=2 \mathbf{F O} \ & \mathbf{A D}+\mathbf{E B}=2 \mathbf{A B} \end{aligned}$ Now consider, $\begin{array}{rlrl} \mathbf{A D}+\mathbf{E B}+\mathbf{F C} & =2 \mathbf{A B}+2 \mathbf{A B} \quad[\therefore \mathbf{F C}=2 \mathbf{A B}] \ (3 \lambda-8) \mathbf{A B} & =4 \mathbf{A B} \quad {[\therefore \text { given}]} \ 3 \lambda-8 & =4 \ 3 \lambda & =12 & \ \lambda & =4 & \end{array}$ Hence, option (b) is correct.

In a regular hexagon \(A B C D E F\), \(\mathbf{A D}+\mathbf{E B}+\mathbf{F C}=(\beta \lambda-8) \mathbf{A…

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