Let P, Q, R and S be the points on the plane with position vectors `-2hati-hatj,4hati,3hati+3hatj and -3hati+2hatj` respectively. The quadrilateral PQRS must be a

Options

A. Parallelogram, which is neither a rhombus nor a rectangle
B. square
C. rectangle, but not a square
D. rhombus, but not a square.

Explanation

To determine the type of quadrilateral formed by the points P, Q, R, and S with the given position vectors, we will follow these steps:

### Step 1: Identify the position vectors
The position vectors of the points are given as follows:
- \( \vec{P} = -2\hat{i} - \hat{j} \)
- \( \vec{Q} = 4\hat{i} \)
- \( \vec{R} = 3\hat{i} + 3\hat{j} \)
- \( \vec{S} = -3\hat{i} + 2\hat{j} \)

### Step 2: Calculate the vectors PQ and RS
To find the vectors \( \vec{PQ} \) and \( \vec{RS} \):
- \( \vec{PQ} = \vec{Q} - \vec{P} = (4\hat{i}) - (-2\hat{i} - \hat{j}) = 4\hat{i} + 2\hat{i} + \hat{j} = 6\hat{i} + \hat{j} \)
- \( \vec{RS} = \vec{S} - \vec{R} = (-3\hat{i} + 2\hat{j}) - (3\hat{i} + 3\hat{j}) = -3\hat{i} + 2\hat{j} - 3\hat{i} - 3\hat{j} = -6\hat{i} - \hat{j} \)

### Step 3: Check if PQ is parallel to RS
To check if \( \vec{PQ} \) is parallel to \( \vec{RS} \):
- The vector \( \vec{PQ} = 6\hat{i} + \hat{j} \)
- The vector \( \vec{RS} = -6\hat{i} - \hat{j} \)
- Since \( \vec{RS} = -1 \cdot \vec{PQ} \), we conclude that \( \vec{PQ} \parallel \vec{RS} \).

### Step 4: Calculate the vectors PS and QR
Next, we calculate the vectors \( \vec{PS} \) and \( \vec{QR} \):
- \( \vec{PS} = \vec{S} - \vec{P} = (-3\hat{i} + 2\hat{j}) - (-2\hat{i} - \hat{j}) = -3\hat{i} + 2\hat{j} + 2\hat{i} + \hat{j} = -\hat{i} + 3\hat{j} \)
- \( \vec{QR} = \vec{R} - \vec{Q} = (3\hat{i} + 3\hat{j}) - (4\hat{i}) = 3\hat{i} + 3\hat{j} - 4\hat{i} = -\hat{i} + 3\hat{j} \)

### Step 5: Check if PS is parallel to QR
To check if \( \vec{PS} \) is parallel to \( \vec{QR} \):
- The vector \( \vec{PS} = -\hat{i} + 3\hat{j} \)
- The vector \( \vec{QR} = -\hat{i} + 3\hat{j} \)
- Since \( \vec{QR} = 1 \cdot \vec{PS} \), we conclude that \( \vec{PS} \parallel \vec{QR} \).

### Step 6: Conclusion about the quadrilateral
Since both pairs of opposite sides \( \vec{PQ} \parallel \vec{RS} \) and \( \vec{PS} \parallel \vec{QR} \), the quadrilateral PQRS is a parallelogram.

### Step 7: Check for rhombus or rectangle
To check if it is a rhombus or rectangle, we calculate the magnitudes:
- Magnitude of \( \vec{PQ} = \sqrt{(6^2 + 1^2)} = \sqrt{36 + 1} = \sqrt{37} \)
- Magnitude of \( \vec{PS} = \sqrt{((-1)^2 + 3^2)} = \sqrt{1 + 9} = \sqrt{10} \)

Since the magnitudes of \( \vec{PQ} \) and \( \vec{PS} \) are not equal, PQRS is not a rhombus.

Also, since the dot product \( \vec{PQ} \cdot \vec{PS} \) is not zero, it is not a rectangle either.

### Final Answer:
Thus, the quadrilateral PQRS must be a **parallelogram** which is neither a rhombus nor a rectangle.

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