If the figure formed by the four points `hati+hatj-hatk,2hati+3hatj,3hati+5hatj-2hatk and hatk-hatj` is

Options

A. rectangle
B. parallelogram
C. trapezium
D. none of these

Explanation

To determine the figure formed by the four points given in the question, we will analyze the vectors step by step.

### Step 1: Define the Points
We have the following points represented as vectors:
- Point A: \( \mathbf{A} = \hat{i} + \hat{j} - \hat{k} \)
- Point B: \( \mathbf{B} = 2\hat{i} + 3\hat{j} \)
- Point C: \( \mathbf{C} = 3\hat{i} + 5\hat{j} - 2\hat{k} \)
- Point D: \( \mathbf{D} = \hat{k} - \hat{j} \)

### Step 2: Calculate the Vectors AB, BC, CD, and DA
To find the sides of the figure, we will calculate the vectors between these points.

- **Vector AB**:
\[ \mathbf{AB} = \mathbf{B} - \mathbf{A} = (2\hat{i} + 3\hat{j}) - (\hat{i} + \hat{j} - \hat{k}) = (2 - 1)\hat{i} + (3 - 1)\hat{j} + (0 + 1)\hat{k} = \hat{i} + 2\hat{j} + \hat{k} \]

- **Vector BC**:
\[ \mathbf{BC} = \mathbf{C} - \mathbf{B} = (3\hat{i} + 5\hat{j} - 2\hat{k}) - (2\hat{i} + 3\hat{j}) = (3 - 2)\hat{i} + (5 - 3)\hat{j} + (-2 - 0)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k} \]

- **Vector CD**:
\[ \mathbf{CD} = \mathbf{D} - \mathbf{C} = (\hat{k} - \hat{j}) - (3\hat{i} + 5\hat{j} - 2\hat{k}) = (-3)\hat{i} + (-5 - (-1))\hat{j} + (1 + 2)\hat{k} = -3\hat{i} - 4\hat{j} + 3\hat{k} \]

- **Vector DA**:
\[ \mathbf{DA} = \mathbf{A} - \mathbf{D} = (\hat{i} + \hat{j} - \hat{k}) - (\hat{k} - \hat{j}) = \hat{i} + (1 + 1)\hat{j} + (-1 - 1)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k} \]

### Step 3: Analyze the Vectors
Now we will analyze the vectors to determine the nature of the figure.

1. **Check if AB is parallel to CD**:
- \( \mathbf{AB} = \hat{i} + 2\hat{j} + \hat{k} \)
- \( \mathbf{CD} = -3\hat{i} - 4\hat{j} + 3\hat{k} \)
- They are not scalar multiples of each other, hence not parallel.

2. **Check if BC is parallel to DA**:
- \( \mathbf{BC} = \hat{i} + 2\hat{j} - 2\hat{k} \)
- \( \mathbf{DA} = \hat{i} + 2\hat{j} - 2\hat{k} \)
- They are equal, hence parallel.

### Step 4: Conclusion
Since \( \mathbf{AB} \) is not parallel to \( \mathbf{CD} \) but \( \mathbf{BC} \) is parallel to \( \mathbf{DA} \), the figure formed by the points A, B, C, and D is a trapezium.

### Final Answer
The figure formed by the four points is a trapezium.

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