Step 1: Understanding the Concept:
In a rectangle with adjacent sides \( \vec{a} \) and \( \vec{b} \), the vectors are perpendicular (\( \vec{a} \cdot \vec{b} = 0 \)). The diagonals are given by \( \vec{OC} = \vec{a} + \vec{b} \) and \( \vec{AB} = \vec{b} - \vec{a} \). We are given a relationship between the side lengths and need to find the angle between diagonals.
Step 2: Key Formula or Approach:
1. Let \( |\vec{a}| = l \) and \( |\vec{b}| = w \). Since it's a rectangle, \( |\vec{a} \times \vec{b}| = lw \).
2. Use the given equation: \( 16lw = 3(l + w)^2 \).
3. Solve for the ratio \( l/w \).
4. Use the diagonal angle formula: \( \cos \theta = \frac{|\vec{OC} \cdot \vec{AB}|}{|\vec{OC}||\vec{AB}|} \).
Step 3: Detailed Explanation:
The given equation is \( 16lw = 3(l^2 + w^2 + 2lw) \).
\( 16lw = 3l^2 + 3w^2 + 6lw \implies 3l^2 - 10lw + 3w^2 = 0 \).
This is a quadratic in \( l/w \). Let \( r = l/w \):
\( 3r^2 - 10r + 3 = 0 \).
Factoring: \( (3r - 1)(r - 3) = 0 \).
So, \( r = 3 \) or \( r = 1/3 \). This means the length of one side is 3 times the other. Let \( l = 3w \).
Now find the angle \( \theta \) between diagonals \( \vec{OC} = \vec{a} + \vec{b} \) and \( \vec{AB} = \vec{b} - \vec{a} \).
\[ \vec{OC} \cdot \vec{AB} = (\vec{a} + \vec{b}) \cdot (\vec{b} - \vec{a}) = |\vec{b}|^2 - |\vec{a}|^2 = w^2 - (3w)^2 = -8w^2 \]
The magnitudes are:
\[ |\vec{OC}| = \sqrt{l^2 + w^2} = \sqrt{9w^2 + w^2} = \sqrt{10}w \]
\[ |\vec{AB}| = \sqrt{w^2 + l^2} = \sqrt{10}w \]
The cosine of the angle is:
\[ \cos \theta = \frac{|-8w^2|}{(\sqrt{10}w)(\sqrt{10}w)} = \frac{8w^2}{10w^2} = \frac{4}{5} = 0.8 \]
We need \( \tan(\theta/2) \). Using half-angle identity:
\[ \tan^2(\theta/2) = \frac{1 - \cos \theta}{1 + \cos \theta} = \frac{1 - 0.8}{1 + 0.8} = \frac{0.2}{1.8} = \frac{1}{9} \]
Taking the square root:
\[ \tan(\theta/2) = \frac{1}{3} \]
Step 4: Final Answer:
The value of \( \tan(\theta/2) \) is 1/3.