Let `bara and barb` be two non coplanar unit vectors IF `baru=bara-(bara.barb)barb and barv=bara times barb` then `|barv|` is

Options

A. `|baru|`
B. `|baru|=|baru.bara|`
C. `|baru|=|baru.barb|`
D. `|baru|+baru.(bara+barb)`

Answer

To solve the problem, we need to find the modulus of the vector \( \mathbf{v} \) given the definitions of vectors \( \mathbf{u} \) and \( \mathbf{v} \).

### Step-by-Step Solution:

1. **Understand the Given Vectors**:
We are given two non-coplanar unit vectors \( \mathbf{a} \) and \( \mathbf{b} \). This means:
\[ |\mathbf{a}| = 1 \quad \text{and} \quad |\mathbf{b}| = 1 \]

2. **Define the Vectors**:
The vector \( \mathbf{u} \) is defined as:
\[ \mathbf{u} = \mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{b} \]
The vector \( \mathbf{v} \) is defined as:
\[ \mathbf{v} = \mathbf{a} \times \mathbf{b} \]

3. **Find the Modulus of \( \mathbf{u} \)**:
First, we need to calculate \( |\mathbf{u}|^2 \):
\[ |\mathbf{u}|^2 = |\mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{b}|^2 \]
Expanding this using the formula \( |\mathbf{x} - \mathbf{y}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y} \):
\[ |\mathbf{u}|^2 = |\mathbf{a}|^2 + |(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}|^2 - 2 \mathbf{a} \cdot ((\mathbf{a} \cdot \mathbf{b}) \mathbf{b}) \]
Since \( |\mathbf{a}|^2 = 1 \) and \( |(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}|^2 = (\mathbf{a} \cdot \mathbf{b})^2 \):
\[ |\mathbf{u}|^2 = 1 + (\mathbf{a} \cdot \mathbf{b})^2 - 2(\mathbf{a} \cdot \mathbf{b})^2 \]
Simplifying gives:
\[ |\mathbf{u}|^2 = 1 - (\mathbf{a} \cdot \mathbf{b})^2 \]

4. **Find the Modulus of \( \mathbf{v} \)**:
Now we calculate \( |\mathbf{v}|^2 \):
\[ |\mathbf{v}|^2 = |\mathbf{a} \times \mathbf{b}|^2 \]
The magnitude of the cross product of two vectors is given by:
\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta \]
Since both \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors:
\[ |\mathbf{v}|^2 = 1 \cdot 1 \cdot \sin^2\theta = \sin^2\theta \]

5. **Relate \( |\mathbf{u}| \) and \( |\mathbf{v}| \)**:
From the previous steps, we have:
\[ |\mathbf{u}|^2 = 1 - (\mathbf{a} \cdot \mathbf{b})^2 \]
and
\[ |\mathbf{v}|^2 = \sin^2\theta \]
We know that \( \sin^2\theta = 1 - \cos^2\theta \) and \( \cos\theta = \mathbf{a} \cdot \mathbf{b} \). Therefore:
\[ |\mathbf{u}|^2 = |\mathbf{v}|^2 \]

6. **Conclusion**:
Since \( |\mathbf{u}|^2 = |\mathbf{v}|^2 \), we conclude:
\[ |\mathbf{u}| = |\mathbf{v}| \]
Thus, the modulus of \( \mathbf{v} \) is equal to the modulus of \( \mathbf{u} \).

### Final Answer:
\[ |\mathbf{v}| = |\mathbf{u}| \]