Let \(A B C D E F\) be a regular hexagon with the vertices \(A, B, C, D, E, F\) counterclock-wise. Then the…

Let \(A B C D E F\) be a regular hexagon with the vertices \(A, B, C, D, E, F\) counterclock-wise. Then the vector \(\mathbf{A B}+\mathbf{A F}+\mathbf{C D}+\mathbf{E F}\) is equal to
  1. \(\mathrm{DE}+\mathrm{FA}\)
  2. \(\mathrm{CB}+\mathrm{ED}\)
  3. \(\mathrm{BC}+\mathrm{FA}\)
  4. \(\mathrm{BC}+\mathrm{DE}\)

Solution

In a regular hexagon \(A B C D E F\), We know that, \(\mathbf{A B}+\mathbf{B C}+\mathbf{C D}=\mathbf{A D}=2 \mathbf{B C}\)
\(\begin{array}{rrr} \therefore & \mathbf{A B}+\mathbf{C D} & =\mathbf{B C} \quad \ldots (i) \\ \text {Similarly, as } & \mathbf{A F} & =\mathbf{C D} \quad \ldots (ii) \end{array}\) and as we know that \(\begin{array}{l} \mathrm{CD}+\mathrm{DE}+\mathrm{EF}=\mathrm{CF}=2 \mathrm{DF} \\ \Rightarrow \quad \mathrm{CD}+\mathrm{EF}=\mathrm{DE} \quad \ldots (iii) \end{array}\) from Eqs. (i) and (iii), we get \(\begin{aligned} & \mathbf{A B}+\mathbf{C D}+\mathbf{C D}+\mathbf{E F}=\mathbf{B C}+\mathbf{D E} \\ & \Rightarrow \quad \mathbf{A B}+\mathbf{A F}+\mathbf{C D}+\mathbf{E F}=\mathbf{B C}+\mathbf{D E} \\ \end{aligned}\) {from Eq. (ii)}

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

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