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Algebra of Vectors — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSAlgebra of Vectors1 Mark
Question
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, $\overrightarrow{\text{AE}}=$
  1. $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
  2. $2\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
  3. $\vec{\text{b}}+\vec{\text{c}}$
  4. $\vec{\text{a}}+2\vec{\text{b}}+2\vec{\text{c}}$
✓

Answer

  1. $\vec{\text{b}}+\vec{\text{c}}$
Solution:
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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