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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors1 Mark
MCQ
In a regular hexagon 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 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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