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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
$A B C D$ is a rhombus whose diagonals intersects at $E$. Then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals to
  • A
    $\overrightarrow{0}$
  • B
    $\overrightarrow{ AD }$
  • C
    $2 \overrightarrow{B D}$
  • D
    $2 \overrightarrow{A D}$
✓

Answer

Given, $A B C D$ is a rhombus whose diagonals bisect each other. $|\overrightarrow{E A}|=|\overrightarrow{E C}|$ and $|\overrightarrow{E B}|=|\overrightarrow{E D}|$ but since they are opposite to each other so they are of opposite signs.
Image
$\Rightarrow \overrightarrow{E A}=-\overrightarrow{E C}$ and $\overrightarrow{E B}=-\overrightarrow{E D}$.     
$\Rightarrow \quad \overrightarrow{E A}+\overrightarrow{E C}=\vec{O}$   ..........(i)
and $\overrightarrow{E B}+\overrightarrow{E D}=\vec{O}$    .......(ii)
Adding (i) and (ii), we get $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}=\vec{O}$.
$=2(1+\cos \theta)=4 \cos ^2 \frac{\theta}{2} \quad\left[\because 1+\cos \theta=2 \cos ^2 \frac{\theta}{2}\right]$
$\therefore|\vec{a}+\vec{b}|=2 \cos \frac{\theta}{2}$

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