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Q. Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{ i }-\hat{ j }, 4 \hat{ i }, 3 \hat{ i }+3 \hat{ j }$ and $-3 \hat{ i }+2 \hat{ j }$, respectively. The quadrilateral $P Q R S$ must be a

Vector Algebra

Solution:

$P Q=6 \hat{i}+\hat{j}$
$Q R =-\hat{i}+3 \hat{j} $
$R S =-6 \hat{i}-\hat{j} $
$S P =\hat{i}-3 \hat{j} $
$|P Q| =\sqrt{37}=|R S| $
$|Q R| =\sqrt{10}=|S P|$
$P Q \cdot Q R =-6+3=-3 \neq 0$
$P Q$ is parallel to $R S$ and their magnitudes are equal.
$\Rightarrow$ Quadrilateral $P Q R S$ must be a parallelogram, which is neither a rhombus nor a rectangle.