Let `veca` and `vecb` be two non-collinear unit vectors. If `vecu=veca-(veca.vecb)vecb` and `vecv=vecaxxvecb`, then `|vecv|` is

Options

A. `|vecu|`
B. `|vecu|+ |vecu.vecb|`
C. `|vecu| + |vecu .veca|`
D. none of these

Explanation

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

### Step-by-step Solution:

1. **Understanding the Given Information**:
- We have two non-collinear unit vectors \( \vec{a} \) and \( \vec{b} \).
- The vector \( \vec{u} \) is defined as \( \vec{u} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} \).
- We need to find the magnitude of \( \vec{v} = \vec{a} \times \vec{b} \).

2. **Finding the Magnitude of \( \vec{u} \)**:
- We can express the magnitude of \( \vec{u} \):
\[ |\vec{u}| = |\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}| \]
- Using the property of magnitudes, we have:
\[ |\vec{u}| = \sqrt{|\vec{a}|^2 + |(\vec{a} \cdot \vec{b})|^2 |\vec{b}|^2 - 2 |\vec{a}| |\vec{b}| (\vec{a} \cdot \vec{b})} \]
- Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \). Thus:
\[ |\vec{u}| = \sqrt{1 + (\vec{a} \cdot \vec{b})^2 - 2(\vec{a} \cdot \vec{b})} \]

3. **Simplifying the Expression**:
- Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). Then, \( \vec{a} \cdot \vec{b} = \cos \theta \).
- Substitute \( \cos \theta \) into the equation:
\[ |\vec{u}| = \sqrt{1 + \cos^2 \theta - 2\cos \theta} \]
- This simplifies to:
\[ |\vec{u}| = \sqrt{1 - 2\cos \theta + \cos^2 \theta} = \sqrt{(1 - \cos \theta)^2} = |1 - \cos \theta| \]
- Since \( \theta \) is between \( 0 \) and \( \pi \) for non-collinear vectors, \( 1 - \cos \theta \) is always non-negative. Therefore:
\[ |\vec{u}| = 1 - \cos \theta \]

4. **Finding the Magnitude of \( \vec{v} \)**:
- The magnitude of the cross product \( \vec{v} = \vec{a} \times \vec{b} \) is given by:
\[ |\vec{v}| = |\vec{a}| |\vec{b}| \sin \theta \]
- Again, since both \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have:
\[ |\vec{v}| = 1 \cdot 1 \cdot \sin \theta = \sin \theta \]

5. **Relating \( |\vec{u}| \) and \( |\vec{v}| \)**:
- From the previous steps, we have:
\[ |\vec{u}| = 1 - \cos \theta \]
\[ |\vec{v}| = \sin \theta \]
- Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can express \( \sin \theta \) in terms of \( |\vec{u}| \):
\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (1 - |\vec{u}|)^2} \]
- This leads us to conclude that \( |\vec{u}| = |\vec{v}| \).

### Final Result:
Thus, the magnitude \( |\vec{v}| \) is equal to \( |\vec{u}| \).

### Answer:
\[ |\vec{v}| = \sin \theta \]