Area of a parallelogram formed by vectors `(3hati-2hatj+hatk)m` and `(hati+2hatj+3hatk)` m as adjacent sides is

Options

A. `- 8hati-8hatj+8hatk`
B. `-2hati+hatk`
C. `-2hati-hatj+hatk`
D. `2hati-hatj-2hatk`

Explanation

To find the area of the parallelogram formed by the vectors \( \mathbf{A} = 3\hat{i} - 2\hat{j} + \hat{k} \) and \( \mathbf{B} = \hat{i} + 2\hat{j} + 3\hat{k} \), we will use the cross product of the two vectors. The area of the parallelogram can be calculated using the formula:

\[ \text{Area} = |\mathbf{A} \times \mathbf{B}| \]

### Step 1: Write down the vectors
Let:
\[ \mathbf{A} = 3\hat{i} - 2\hat{j} + \hat{k} \]
\[ \mathbf{B} = \hat{i} + 2\hat{j} + 3\hat{k} \]

### Step 2: Set up the determinant for the cross product
The cross product \( \mathbf{A} \times \mathbf{B} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the vectors \( \mathbf{A} \) and \( \mathbf{B} \):

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 1 & 2 & 3 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we can expand it as follows:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -2 & 1 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -2 \\ 1 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} -2 & 1 \\ 2 & 3 \end{vmatrix} = (-2)(3) - (1)(2) = -6 - 2 = -8 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} = (3)(3) - (1)(1) = 9 - 1 = 8 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & -2 \\ 1 & 2 \end{vmatrix} = (3)(2) - (-2)(1) = 6 + 2 = 8 \]

Putting it all together, we have:

\[ \mathbf{A} \times \mathbf{B} = -8\hat{i} - 8\hat{j} + 8\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now, we find the magnitude of the vector \( \mathbf{A} \times \mathbf{B} \):

\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(-8)^2 + (-8)^2 + (8)^2} \]
\[ = \sqrt{64 + 64 + 64} = \sqrt{192} = 8\sqrt{3} \]

### Final Answer
Thus, the area of the parallelogram is:

\[ \text{Area} = 8\sqrt{3} \, \text{m}^2 \]