The volume of the parallelepiped whose coterminous edges are represented by the vectors `2vecb xx vecc, 3vecc xx veca` and `4veca xx vecb` where
`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`

Answer

To find the values of \( \theta \) in the interval \( (0, \frac{\pi}{2}) \) such that the volume of the parallelepiped formed by the vectors \( 2\vec{b} \times \vec{c}, 3\vec{c} \times \vec{a}, \) and \( 4\vec{a} \times \vec{b} \) is 18 cubic units, we can follow these steps:

### Step 1: Understand the Volume of the Parallelepiped
The volume \( V \) of a parallelepiped defined by three vectors \( \vec{u}, \vec{v}, \vec{w} \) can be calculated using the scalar triple product:
\[ V = |\vec{u} \cdot (\vec{v} \times \vec{w})| \]
In our case, the vectors are \( 2\vec{b} \times \vec{c}, 3\vec{c} \times \vec{a}, \) and \( 4\vec{a} \times \vec{b} \).

### Step 2: Set Up the Volume Equation
The volume can be expressed as:
\[ V = |2\vec{b} \times \vec{c} \cdot (3\vec{c} \times \vec{a}) \times (4\vec{a} \times \vec{b})| \]
This can be simplified to:
\[ V = 24 |\det(\vec{b}, \vec{c}, \vec{a})| \]
Given that the volume is 18 cubic units, we have:
\[ 24 |\det(\vec{b}, \vec{c}, \vec{a})| = 18 \]
Thus,
\[ |\det(\vec{b}, \vec{c}, \vec{a})| = \frac{18}{24} = \frac{3}{4} \]

### Step 3: Express the Vectors
The vectors are given as:
\[ \vec{b} = \sin\left(\theta + \frac{2\pi}{3}\right) \hat{i} + \cos\left(\theta + \frac{2\pi}{3}\right) \hat{j} + \sin\left(2\theta + \frac{4\pi}{3}\right) \hat{k} \]
\[ \vec{c} = \sin\left(\theta - \frac{2\pi}{3}\right) \hat{i} + \cos\left(\theta - \frac{2\pi}{3}\right) \hat{j} + \sin\left(2\theta - \frac{4\pi}{3}\right) \hat{k} \]
\[ \vec{a} = \text{(not provided, assume it is defined)} \]

### Step 4: Calculate the Determinant
To find the determinant \( \det(\vec{b}, \vec{c}, \vec{a}) \), we can set up the matrix:
\[ \begin{vmatrix} \sin\left(\theta + \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) \\ \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta - \frac{4\pi}{3}\right) \\ \text{(components of } \vec{a} \text{)} \end{vmatrix} \]

### Step 5: Solve for \( \theta \)
After calculating the determinant, we set it equal to \( \frac{3}{4} \) and solve for \( \theta \). The determinant will involve trigonometric identities and simplifications.

### Step 6: Find Possible Values of \( \theta \)
From the determinant, we can derive that:
\[ \cos(3\theta) = \pm \frac{1}{2} \]
This leads to:
\[ 3\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \]
Thus,
\[ \theta = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9} \]

### Step 7: Select Values in the Interval \( (0, \frac{\pi}{2}) \)
We only consider values of \( \theta \) in the interval \( (0, \frac{\pi}{2}) \):
- \( \frac{\pi}{9} \)
- \( \frac{2\pi}{9} \)
- \( \frac{4\pi}{9} \)

Thus, the final values of \( \theta \) are:
\[ \theta = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9} \]