MCQ
If $ABCDEF$ is regular hexagon, then $\overrightarrow {AD} \, + \overrightarrow {EB} + \overrightarrow {FC} = $
- A$0$
- B$2\overrightarrow {AB} $
- C$3\overrightarrow {AB} $
- ✓$4\overrightarrow {AB} $
✓
Answer
Correct option: D.
$4\overrightarrow {AB} $
d
(d)A regular hexagon $ABCDEF.$
We know from the hexagon that $\overrightarrow {AD} $ is parallel to $\overrightarrow {BC} $ or $\overrightarrow {AD} = 2\,\overrightarrow {BC} $; $\overrightarrow {EB} $ is parallel to $\overrightarrow {FA} $
(d)A regular hexagon $ABCDEF.$
We know from the hexagon that $\overrightarrow {AD} $ is parallel to $\overrightarrow {BC} $ or $\overrightarrow {AD} = 2\,\overrightarrow {BC} $; $\overrightarrow {EB} $ is parallel to $\overrightarrow {FA} $
or $\overrightarrow {EB} = 2\overrightarrow {FA} $, and $\overrightarrow {FC} $ is parallel to $\overrightarrow {AB} $ or $\overrightarrow {FC} = 2\,\overrightarrow {AB} $.
Thus $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = 2\,\overrightarrow {BC} + 2\,\overrightarrow {FA} + 2\,\overrightarrow {AB} $
$ = 2(\overrightarrow {FA} + \overrightarrow {AB} + \overrightarrow {BC} ) = 2(\overrightarrow {FC} ) = 2(2\overrightarrow {AB} ) = 4\,\overrightarrow {AB} $.
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