Consider the following statements :
1. The cross product of two unit vectors is always a unit vector.
2. The dot product of two unit vectors is always unity.
3. The magnitude of sum of two unit vectors is always greater than the magnitude of their difference.
Which of the above is not correct?

Options

A. `1` and `2` only
B. `2` and `3` only
C. `1` and `3` only
D. `1,2` and `3` only

Answer

To solve the problem, we need to analyze each of the three statements given about unit vectors and determine which one is not correct.

### Step 1: Analyze Statement 1
**Statement 1:** The cross product of two unit vectors is always a unit vector.

Let \( \mathbf{A} \) and \( \mathbf{B} \) be two unit vectors. The magnitude of the cross product \( \mathbf{A} \times \mathbf{B} \) is given by:
\[ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \]
where \( \theta \) is the angle between the two vectors. Since both \( \mathbf{A} \) and \( \mathbf{B} \) are unit vectors, we have:
\[ |\mathbf{A}| = 1 \quad \text{and} \quad |\mathbf{B}| = 1 \]
Thus,
\[ |\mathbf{A} \times \mathbf{B}| = 1 \cdot 1 \cdot \sin \theta = \sin \theta \]
The value of \( \sin \theta \) can range from \( 0 \) to \( 1 \). Therefore, the cross product of two unit vectors is not always a unit vector; it can be zero when \( \theta = 0^\circ \) or \( 180^\circ \).

**Conclusion:** Statement 1 is **not correct**.

### Step 2: Analyze Statement 2
**Statement 2:** The dot product of two unit vectors is always unity.

The dot product \( \mathbf{A} \cdot \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
Again, since both \( \mathbf{A} \) and \( \mathbf{B} \) are unit vectors:
\[ |\mathbf{A}| = 1 \quad \text{and} \quad |\mathbf{B}| = 1 \]
Thus,
\[ \mathbf{A} \cdot \mathbf{B} = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]
The value of \( \cos \theta \) can range from \( -1 \) to \( 1 \). Therefore, the dot product of two unit vectors is not always unity; it can be less than or equal to 1 depending on the angle \( \theta \).

**Conclusion:** Statement 2 is **not correct**.

### Step 3: Analyze Statement 3
**Statement 3:** The magnitude of the sum of two unit vectors is always greater than the magnitude of their difference.

Let’s consider two unit vectors \( \mathbf{A} \) and \( \mathbf{B} \). The magnitude of their sum and difference can be calculated as follows:
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} + \mathbf{B})} = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 + 2(\mathbf{A} \cdot \mathbf{B})} \]
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{(\mathbf{A} - \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B})} = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 - 2(\mathbf{A} \cdot \mathbf{B})} \]
Since \( |\mathbf{A}| = 1 \) and \( |\mathbf{B}| = 1 \):
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{1 + 1 + 2(\mathbf{A} \cdot \mathbf{B})} = \sqrt{2 + 2(\mathbf{A} \cdot \mathbf{B})} \]
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{1 + 1 - 2(\mathbf{A} \cdot \mathbf{B})} = \sqrt{2 - 2(\mathbf{A} \cdot \mathbf{B})} \]
Now, we need to check if:
\[ |\mathbf{A} + \mathbf{B}| > |\mathbf{A} - \mathbf{B}| \]
This inequality holds true because:
\[ \sqrt{2 + 2(\mathbf{A} \cdot \mathbf{B})} > \sqrt{2 - 2(\mathbf{A} \cdot \mathbf{B})} \]
This is true for all angles \( \theta \) between the vectors.

**Conclusion:** Statement 3 is **correct**.

### Final Conclusion
The statements that are not correct are **1 and 2**. Therefore, the answer is that both Statement 1 and Statement 2 are not correct.