The volume of the parallelepiped whose coterminous edges are represented by the vectors `2vecb xx vecc, 3vecc xx veca` and `4veca xx vecb` where `veca=(1+sintheta)hati+costhetahatj+sin2thetahatk` , `vecb=sin(theta+(2pi)/(3))hati+cos(theta+(2pi)/(3))hatj+sin(2theta+(4pi)/(3))hatk`, `vecc=sin(theta-(2pi)/(3))hati+cos(theta-(2pi)/(3))hatj + sin(2theta-(4pi)/(3))hatk` is 18 cubic units, then the values of `theta`, in the interval `(0,pi/2)`, is/are

Options

A. `pi/9`
B. `(2pi)/(9)`
C. `pi/3`
D. `(4pi)/9`

Explanation

To solve the problem, we need to find the volume of the parallelepiped formed by the vectors given, and then determine the values of \(\theta\) that satisfy the condition of the volume being 18 cubic units.

### Step 1: Write the volume formula
The volume \(V\) of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) can be expressed using the scalar triple product:
\[ V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]
In this case, we have the vectors defined as:
- \(\vec{a} = (1 + \sin \theta) \hat{i} + \cos \theta \hat{j} + \sin 2\theta \hat{k}\)
- \(\vec{b} = \sin\left(\theta + \frac{2\pi}{3}\right) \hat{i} + \cos\left(\theta + \frac{2\pi}{3}\right) \hat{j} + \sin\left(2\theta + \frac{4\pi}{3}\right) \hat{k}\)
- \(\vec{c} = \sin\left(\theta - \frac{2\pi}{3}\right) \hat{i} + \cos\left(\theta - \frac{2\pi}{3}\right) \hat{j} + \sin\left(2\theta - \frac{4\pi}{3}\right) \hat{k}\)

### Step 2: Set up the volume equation
Given that the volume is \(18\) cubic units, we can express this as:
\[ |2(\vec{b} \times \vec{c}) + 3(\vec{c} \times \vec{a}) + 4(\vec{a} \times \vec{b})| = 18 \]

### Step 3: Simplify the volume expression
We can factor out the coefficients from the volume expression:
\[ |2(\vec{b} \times \vec{c}) + 3(\vec{c} \times \vec{a}) + 4(\vec{a} \times \vec{b})| = 18 \]
This can be rewritten as:
\[ \frac{1}{6} |2(\vec{b} \times \vec{c}) + 3(\vec{c} \times \vec{a}) + 4(\vec{a} \times \vec{b})| = 18 \]
Thus,
\[ |2(\vec{b} \times \vec{c}) + 3(\vec{c} \times \vec{a}) + 4(\vec{a} \times \vec{b})| = 108 \]

### Step 4: Calculate the scalar triple product
To find the scalar triple product, we can use the determinant of the matrix formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). However, given the complexity of the vectors, we can use the relationship:
\[ abc = \sqrt{3} \cos(3\theta) \]
We know from the volume condition:
\[ \sqrt{3} \cos(3\theta) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \cos(3\theta) = \frac{1}{2} \]

### Step 5: Solve for \(3\theta\)
The values of \(3\theta\) that satisfy \(\cos(3\theta) = \frac{1}{2}\) are:
\[ 3\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \text{ or } 4\pi/3 \]

### Step 6: Solve for \(\theta\)
Dividing these values by \(3\) gives us:
\[ \theta = \frac{\pi}{9}, \frac{5\pi}{9}, \text{ or } \frac{4\pi}{9} \]

### Step 7: Check the interval
Since we are looking for values in the interval \((0, \frac{\pi}{2})\), we only consider:
\[ \theta = \frac{\pi}{9} \]

### Final Answer
Thus, the value of \(\theta\) in the interval \((0, \frac{\pi}{2})\) is:
\[ \theta = \frac{\pi}{9} \]