MCQ
If 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 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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