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Algebra of Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors1 Mark
MCQ
If $\text{ABCDEF}$ is a regular hexagon, then $\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}+\overrightarrow{\text{FC}}$ equals,
  • A
    $2\overrightarrow{\text{AB}}$
  • B
    $\vec0$
  • C
    $3\overrightarrow{\text{AB}}$
  • ✓
    $4\overrightarrow{\text{AB}}$
✓

Answer

Correct option: D.
$4\overrightarrow{\text{AB}}$
$\overrightarrow{\text{AD}}=2\overrightarrow{\text{BC}}$
$\overrightarrow{\text{EB}}=2\overrightarrow{\text{FA}}$
$\overrightarrow{\text{FC}}=2\overrightarrow{\text{AB}}$
$\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=2\Big(\overrightarrow{\text{BC}}+\overrightarrow{\text{FA}}\Big)$
$=2\Big(\overrightarrow{\text{AO}}+\overrightarrow{\text{FA}}\Big)$ $\Big(\because\ \overrightarrow{\text{BC}}=\overrightarrow{\text{AO}}\Big)$
In triangle $\text{AOF,}$
$\overrightarrow{\text{FA}}+\overrightarrow{\text{AO}}+\overrightarrow{\text{FO}}=0$
$\therefore\ \overrightarrow{\text{FA}}+\overrightarrow{\text{AO}}=-\overrightarrow{\text{FO}}$
$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=-2\overrightarrow{\text{FO}}$
And $\overrightarrow{\text{AB}}=-\overrightarrow{\text{FO}}$
$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=2\overrightarrow{\text{AB}}$
$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}+\overrightarrow{\text{FC}}=2\overrightarrow{\text{AB}}+2\overrightarrow{\text{AB}}$
$=4\overrightarrow{\text{AB}}$

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