Let `hataandhatb` be two non -collinear unit vectors .If `u=hata-(hata*hatb)hatbandv=hataxxhatb`, then `|v|` is equal to

Options

A. `|u|`
B. `|u|+|v.hata|`
C. `2|v|`
D. `|v|+u.(hata+hatb)`

Answer

To solve the problem, we need to find the magnitude of the vector \( v \), where \( v = \hat{a} \times \hat{b} \) and \( \hat{a} \) and \( \hat{b} \) are two non-collinear unit vectors.

### Step-by-Step Solution:

1. **Understanding the Cross Product**:
The cross product of two vectors \( \hat{a} \) and \( \hat{b} \) is given by:
\[ \hat{a} \times \hat{b} = |\hat{a}| |\hat{b}| \sin \theta \, \hat{n} \]
where \( \theta \) is the angle between the two vectors and \( \hat{n} \) is the unit vector perpendicular to the plane formed by \( \hat{a} \) and \( \hat{b} \).

2. **Magnitude of Unit Vectors**:
Since \( \hat{a} \) and \( \hat{b} \) are unit vectors, we have:
\[ |\hat{a}| = 1 \quad \text{and} \quad |\hat{b}| = 1 \]

3. **Substituting Values**:
Substituting the magnitudes into the cross product formula:
\[ v = \hat{a} \times \hat{b} = 1 \cdot 1 \cdot \sin \theta \, \hat{n} = \sin \theta \, \hat{n} \]

4. **Finding the Magnitude of \( v \)**:
The magnitude of \( v \) is:
\[ |v| = |\sin \theta \, \hat{n}| = |\sin \theta| \cdot |\hat{n}| \]
Since \( \hat{n} \) is a unit vector, \( |\hat{n}| = 1 \):
\[ |v| = |\sin \theta| \]

5. **Conclusion**:
Therefore, the magnitude of \( v \) is:
\[ |v| = \sin \theta \]

### Final Answer:
\[ |v| = \sin \theta \]