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AP EAMCET · Maths · Vector Algebra

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. A \(\mathrm{DE}+\mathrm{FA}\)
  2. B \(\mathrm{CB}+\mathrm{ED}\)
  3. C \(\mathrm{BC}+\mathrm{FA}\)
  4. D \(\mathrm{BC}+\mathrm{DE}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{BC}+\mathrm{DE}\)

Step-by-step Solution

Detailed explanation

In a regular hexagon \(A B C D E F\), We know that, \(\mathbf{A B}+\mathbf{B C}+\mathbf{C D}=\mathbf{A D}=2 \mathbf{B C}\)…