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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ and $\vec{\text{d}}$ are the position vector of points $\text{A, B, C, D}$ such that no three of them are collinear and $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$, then $\text{ABCD}$ is a,
  • A
    Rhombus.
  • B
    Rectangle.
  • C
    Square.
  • ✓
    Parallelogram.
✓

Answer

Correct option: D.
Parallelogram.
Given :
$\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\vec{\text{c}}-\vec{\text{d}}=\vec{\text{b}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
And $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\vec{\text{c}}-\vec{\text{b}}=\vec{\text{d}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}$
Also, since $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\frac{1}2\big(\vec{\text{a}}+\vec{\text{c}}\big)=\frac{1}2\big(\vec{\text{b}}+\vec{\text{d}}\big)$
So, position vector of mid $-$ point of $\text{BD} =$ position vector of mid $-$ point of $\text{AC}.$
Hence diagonals bisect each other.
The given $\text{ABCD}$ is a parallelogram.

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