Find the area oif the parallelogram determined `A=2hati+hatj-3hatk` and `B=12hatj-2hatk`

Options

A. 43
B. 56
C. 38
D. 74

Explanation

To find the area of the parallelogram determined by the vectors \( \mathbf{A} = 2\hat{i} + \hat{j} - 3\hat{k} \) and \( \mathbf{B} = 12\hat{j} - 2\hat{k} \), we can use the formula for the area of a parallelogram formed by two vectors, which is given by the magnitude of their cross product:

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

### Step 1: Write the vectors in component form
The vectors are:
\[ \mathbf{A} = 2\hat{i} + 1\hat{j} - 3\hat{k} \quad \text{(which can be represented as (2, 1, -3))} \]
\[ \mathbf{B} = 0\hat{i} + 12\hat{j} - 2\hat{k} \quad \text{(which can be represented as (0, 12, -2))} \]

### 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 vectors \( \mathbf{A} \) and \( \mathbf{B} \):

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 0 & 12 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we expand it as follows:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 1 & -3 \\ 12 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 0 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 0 & 12 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & -3 \\ 12 & -2 \end{vmatrix} = (1)(-2) - (-3)(12) = -2 + 36 = 34 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & -3 \\ 0 & -2 \end{vmatrix} = (2)(-2) - (-3)(0) = -4 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 2 & 1 \\ 0 & 12 \end{vmatrix} = (2)(12) - (1)(0) = 24 \]

Putting it all together, we have:
\[ \mathbf{A} \times \mathbf{B} = 34\hat{i} + 4\hat{j} + 24\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now we need to find the magnitude of \( \mathbf{A} \times \mathbf{B} \):
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(34)^2 + (4)^2 + (24)^2} \]

Calculating each term:
\[ 34^2 = 1156, \quad 4^2 = 16, \quad 24^2 = 576 \]

Adding these together:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{1156 + 16 + 576} = \sqrt{1748} \]

### Step 5: Simplify the square root
Calculating \( \sqrt{1748} \):
\[ \sqrt{1748} \approx 41.8 \]

### Conclusion
Thus, the area of the parallelogram determined by the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is approximately \( 41.8 \).