Identify Type of Quadrilateral Formed by Given Vectors

To determine the type of quadrilateral formed by the given points $P, Q, R,$ and $S$, we first write their position vectors:
$\vec{r}_P = -2\hat{i} - \hat{j}$
$\vec{r}_Q = 4\hat{i}$
$\vec{r}_R = 3\hat{i} + 3\hat{j}$
$\vec{r}_S = -3\hat{i} + 2\hat{j}$

Next, we calculate the vectors representing the side lengths of the quadrilateral $PQRS$:

1. Vector $\vec{PQ}$:
   $\vec{PQ} = \vec{r}_Q - \vec{r}_P = (4\hat{i}) - (-2\hat{i} - \hat{j}) = 6\hat{i} + \hat{j}$

2. Vector $\vec{SR}$:
   $\vec{SR} = \vec{r}_R - \vec{r}_S = (3\hat{i} + 3\hat{j}) - (-3\hat{i} + 2\hat{j}) = 6\hat{i} + \hat{j}$

Since $\vec{PQ} = \vec{SR}$, one pair of opposite sides is equal and parallel. Therefore, the quadrilateral $PQRS$ is a parallelogram.

Now, let's determine if $PQRS$ is a rhombus or a rectangle by examining adjacent sides:
3. Vector $\vec{PS}$:
$\vec{PS} = \vec{r}_S - \vec{r}_P = (-3\hat{i} + 2\hat{j}) - (-2\hat{i} - \hat{j}) = -\hat{i} + 3\hat{j}$

• Check for Rhombus (equality of adjacent side lengths):
  The lengths of adjacent sides $PQ$ and $PS$ are:
  $|\vec{PQ}| = \sqrt{6^2 + 1^2} = \sqrt{37}$
  $|\vec{PS}| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$
  Since $|\vec{PQ}| \neq |\vec{PS}|$, the adjacent sides are not equal in length. Thus, $PQRS$ is not a rhombus (and hence not a square).

• Check for Rectangle (orthogonality of adjacent sides):
  We compute the dot product of $\vec{PQ}$ and $\vec{PS}$:
  $\vec{PQ} \cdot \vec{PS} = (6\hat{i} + \hat{j}) \cdot (-\hat{i} + 3\hat{j}) = 6(-1) + 1(3) = -6 + 3 = -3 \neq 0$
  Since the dot product is non-zero, adjacent sides are not perpendicular. Thus, $PQRS$ is not a rectangle.

Therefore, the quadrilateral $PQRS$ is a parallelogram, which is neither a rhombus nor a rectangle.

Correct Answer: A