In a regular hexagon ABCDEF, AB = a , BC = b and CD = c . Then, A… - Vidyadip

MCQ

In a regular hexagon $\text{ABCDEF,} \overrightarrow{\text{AB}}=\vec{\text{a}},\ \overrightarrow{\text{BC}}=\vec{\text{b}}$ and $\overrightarrow{\text{CD}}=\vec{\text{c}}$. Then, $\overrightarrow{\text{AE}}=$

A
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
B
$2\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
✓
$\vec{\text{b}}+\vec{\text{c}}$
D
$\vec{\text{a}}+2\vec{\text{b}}+2\vec{\text{c}}$

✓

Answer

Correct option: C.

$\vec{\text{b}}+\vec{\text{c}}$

Given a regular hexagon $\text{ABCDEF,} \overrightarrow{\text{AB}}=\vec{\text{a}},\ \overrightarrow{\text{BC}}=\vec{\text{b}}$ and $\overrightarrow{\text{CD}}=\vec{\text{c}}$. Then,
In $\triangle{\text{ABC}}$, we have
$\overrightarrow{\text{AC}}=\vec{\text{a}}+\vec{\text{b}}$
In $\triangle{\text{ACD}}$, we have
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}$
$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{AC}}+\vec{\text{c}}$
$\Rightarrow\overrightarrow{\text{AD}}=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
Again, in $\triangle{\text{ADE}}$, we have
$\overrightarrow{\text{AE}}=\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}$
$\Rightarrow\overrightarrow{\text{AE}}=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AE}}=\vec{\text{b}}+\vec{\text{c}}$
Hence option $(c).$

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