The adjacent side vectors `vec(OA) and vec(OB)` of a rectangle OACB are `veca and vecb` respectively, where O is the origin . If `16|veca xx vecb|=3(|veca|+|vecb|)^(2) and theta` be the acute angle between the diagonals OC and AB then the value of `cos(theta/2)` is : (a)`(1)/(sqrt(2))` (b)`(1)/(2)` (c) `(1)/(sqrt(3))` (d)`(1)/(3)`

Options

A. `(1)/(sqrt(2))`
B. `(1)/(2)`
C. `(1)/(sqrt(3))`
D. `(1)/(3)`

Answer

To solve the problem, we need to analyze the given information and apply vector properties. Let's break it down step by step.

### Step 1: Understand the given equation
We are given the equation:
\[ 16 |\vec{a} \times \vec{b}| = 3 (|\vec{a}| + |\vec{b}|)^2 \]

### Step 2: Use the properties of the cross product
Since \(\vec{a}\) and \(\vec{b}\) are adjacent sides of a rectangle, they are perpendicular. Thus, the magnitude of the cross product can be calculated as:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(90^\circ) = |\vec{a}| |\vec{b}| \]
Substituting this into the equation gives:
\[ 16 |\vec{a}| |\vec{b}| = 3 (|\vec{a}| + |\vec{b}|)^2 \]

### Step 3: Expand the right side
Expanding the right side:
\[ 3 (|\vec{a}| + |\vec{b}|)^2 = 3 (|\vec{a}|^2 + 2 |\vec{a}| |\vec{b}| + |\vec{b}|^2) \]
Thus, the equation becomes:
\[ 16 |\vec{a}| |\vec{b}| = 3 |\vec{a}|^2 + 6 |\vec{a}| |\vec{b}| + 3 |\vec{b}|^2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 16 |\vec{a}| |\vec{b}| - 6 |\vec{a}| |\vec{b}| = 3 |\vec{a}|^2 + 3 |\vec{b}|^2 \]
This simplifies to:
\[ 10 |\vec{a}| |\vec{b}| = 3 |\vec{a}|^2 + 3 |\vec{b}|^2 \]

### Step 5: Divide by \( |\vec{b}|^2 \)
Let \( x = \frac{|\vec{a}|}{|\vec{b}|} \). Then we can write:
\[ 10 x = 3 x^2 + 3 \]
Rearranging gives:
\[ 3 x^2 - 10 x + 3 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -10, c = 3 \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} \]
\[ x = \frac{10 \pm 8}{6} \]
This gives us two solutions:
1. \( x = \frac{18}{6} = 3 \)
2. \( x = \frac{2}{6} = \frac{1}{3} \)

### Step 7: Determine \( \cos(\theta/2) \)
Since \( \cos(\theta/2) = \frac{|\vec{a}|}{|\vec{b}|} \), we consider the valid solution:
- \( \frac{|\vec{a}|}{|\vec{b}|} = \frac{1}{3} \) (valid since \( \cos(\theta/2) \) must be ≤ 1)

Thus:
\[ \cos(\theta/2) = \frac{1}{3} \]

### Final Answer
The value of \( \cos(\theta/2) \) is:
\[ \boxed{\frac{1}{3}} \]