If `vecu=veca-vecb`, `vecv=veca+vecb` and `|veca|=|vecb|=2`, then `|vecuxxvecv|` is equal to

Options

A. `2sqrt(16-(veca.vecb)^(2))`
B. `sqrt([16-(veca.vecb)^(2)])`
C. `2sqrt([4-(veca.vecb)^(2)])`
D. `[4-(veca.vecb)^(2)]`

Answer

To solve the problem, we need to find the magnitude of the cross product of two vectors defined as follows:

1. **Given:**
- \(\vec{u} = \vec{a} - \vec{b}\)
- \(\vec{v} = \vec{a} + \vec{b}\)
- \(|\vec{a}| = |\vec{b}| = 2\)

2. **Find:**
- \(|\vec{u} \times \vec{v}|\)

### Step-by-Step Solution:

**Step 1: Write the expression for the cross product.**
\[ \vec{u} \times \vec{v} = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \]

**Step 2: Use the distributive property of the cross product.**
\[ \vec{u} \times \vec{v} = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \]

**Step 3: Simplify using properties of the cross product.**
- Recall that \(\vec{a} \times \vec{a} = \vec{0}\) and \(\vec{b} \times \vec{b} = \vec{0}\).
- Also, \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\).

Thus, we have:
\[ \vec{u} \times \vec{v} = 0 + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} + 0 = \vec{a} \times \vec{b} + \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{b}) \]

**Step 4: Find the magnitude of the cross product.**
\[ |\vec{u} \times \vec{v}| = |2(\vec{a} \times \vec{b})| = 2 |\vec{a} \times \vec{b}| \]

**Step 5: Calculate \(|\vec{a} \times \vec{b}|\).**
Using the formula for the magnitude of the cross product:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

Given \(|\vec{a}| = |\vec{b}| = 2\):
\[ |\vec{a} \times \vec{b}| = 2 \cdot 2 \cdot \sin \theta = 4 \sin \theta \]

**Step 6: Substitute back into the magnitude expression.**
\[ |\vec{u} \times \vec{v}| = 2 |\vec{a} \times \vec{b}| = 2 \cdot 4 \sin \theta = 8 \sin \theta \]

**Step 7: Relate to the dot product.**
Using the relation between dot product and angle:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 2 \cdot 2 \cdot \cos \theta = 4 \cos \theta \]
Thus,
\[ \cos^2 \theta = \left(\frac{\vec{a} \cdot \vec{b}}{4}\right)^2 \]

**Step 8: Substitute into the expression for \(|\vec{u} \times \vec{v}|\).**
Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[ |\vec{u} \times \vec{v}| = 8 \sqrt{1 - \left(\frac{\vec{a} \cdot \vec{b}}{4}\right)^2} \]

**Final Result:**
Thus, we can summarize the result:
\[ |\vec{u} \times \vec{v}| = 2 \sqrt{16 - (\vec{a} \cdot \vec{b})^2} \]

### Conclusion:
The magnitude of \(|\vec{u} \times \vec{v}|\) is equal to \(2 \sqrt{16 - (\vec{a} \cdot \vec{b})^2}\).