Question:

Step 1: From the given condition, \(\vec{a} + \vec{b} = \vec{c}\). Step 2: Since all vectors are unit vectors: \(|\vec{a}| = |\vec{b}| = |\vec{c}| = 1\). Step 3: Taking magnitude: \(|\vec{a} + \vec{b}|^2 = |\vec{c}|^2 = 1\). \[1 + 2\vec{a} \cdot \vec{b} + 1 = 1\] \[\vec{a} \cdot \vec{b} = -\frac{1}{2}\] Step 4: The area of the triangle formed by \(\vec{a}\) and \(\vec{b}\) is: \[A = \frac{1}{2}|\vec{a} \times \vec{b}| = \frac{1}{2}\sqrt{1 - (\vec{a} \cdot \vec{b})^2} = \frac{1}{2}\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{4}\] Step 5: Therefore, \(A^2 = \frac{3}{16}\) and \(16A^2 = 3\).