Let `veca, vecb and vecc` be unit vector such that `veca + vecb - vecc =0`. If the area of triangle formed by vectors `veca and vecb` is A, then what is the value of `4A^(2)` ?

Explanation

To solve the problem, we need to find the value of \(4A^2\) given that \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors satisfying the equation \(\vec{a} + \vec{b} - \vec{c} = 0\) and \(A\) is the area of the triangle formed by vectors \(\vec{a}\) and \(\vec{b}\).

### Step-by-Step Solution:

1. **Rewrite the Equation**:
From the equation \(\vec{a} + \vec{b} - \vec{c} = 0\), we can express \(\vec{c}\) as:
\[ \vec{c} = \vec{a} + \vec{b} \]

2. **Use the Property of Unit Vectors**:
Since \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors, we know:
\[ |\vec{a}| = 1, \quad |\vec{b}| = 1, \quad |\vec{c}| = 1 \]
Therefore, we can write:
\[ |\vec{c}|^2 = |\vec{a} + \vec{b}|^2 = 1 \]

3. **Expand the Magnitude Squared**:
Using the property of dot products, we have:
\[ |\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 1 \]
Substituting the magnitudes:
\[ 1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1 \]
Simplifying gives:
\[ 2 + 2(\vec{a} \cdot \vec{b}) = 1 \]
\[ 2(\vec{a} \cdot \vec{b}) = -1 \]
\[ \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

4. **Calculate the Area of the Triangle**:
The area \(A\) of the triangle formed by vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ A = \frac{1}{2} |\vec{a} \times \vec{b}| \]
The magnitude of the cross product can be expressed as:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
Since \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\), we have:
\[ |\vec{a} \times \vec{b}| = \sin \theta \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). We can find \(\sin \theta\) using the cosine we found earlier:
\[ \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3} \text{ (or } 120^\circ\text{)} \]
Thus:
\[ \sin \theta = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Therefore, the area \(A\) becomes:
\[ A = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \]

5. **Calculate \(4A^2\)**:
Now we need to find \(4A^2\):
\[ 4A^2 = 4\left(\frac{\sqrt{3}}{4}\right)^2 = 4 \cdot \frac{3}{16} = \frac{3}{4} \]

### Final Answer:
\[ 4A^2 = \frac{3}{4} \]