The vector `A = 3i-k, b=i+2j` are adjacent sides of a parallelogram . Its area is

Options

A. `(1)/(2) sqrt(17)`
B. `(1)/(2)sqrt(14)`
C. `sqrt(41)`
D. `(1)/(2)sqrt(7)`

Answer

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

### Step-by-Step Solution:

1. **Identify the vectors**:
\[ \mathbf{A} = 3\mathbf{i} - \mathbf{k}, \quad \mathbf{B} = \mathbf{i} + 2\mathbf{j} \]

2. **Set up the cross product**:
The area \( A \) of the parallelogram can be calculated using the formula:
\[ A = |\mathbf{A} \times \mathbf{B}| \]
To compute \( \mathbf{A} \times \mathbf{B} \), we will use the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{A} \) and \( \mathbf{B} \).

3. **Construct the determinant**:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -1 \\ 1 & 2 & 0 \end{vmatrix} \]

4. **Calculate the determinant**:
Expanding this determinant:
\[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} 0 & -1 \\ 2 & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & -1 \\ 1 & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 0 \\ 1 & 2 \end{vmatrix} \]

Now, calculating each of the 2x2 determinants:
- For \( \mathbf{i} \):
\[ \begin{vmatrix} 0 & -1 \\ 2 & 0 \end{vmatrix} = (0)(0) - (-1)(2) = 2 \]
- For \( \mathbf{j} \):
\[ \begin{vmatrix} 3 & -1 \\ 1 & 0 \end{vmatrix} = (3)(0) - (-1)(1) = 1 \]
- For \( \mathbf{k} \):
\[ \begin{vmatrix} 3 & 0 \\ 1 & 2 \end{vmatrix} = (3)(2) - (0)(1) = 6 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 2\mathbf{i} - 1\mathbf{j} + 6\mathbf{k} = 2\mathbf{i} - \mathbf{j} + 6\mathbf{k} \]

5. **Find the magnitude of the cross product**:
The magnitude \( |\mathbf{A} \times \mathbf{B}| \) is calculated as follows:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(2)^2 + (-1)^2 + (6)^2} = \sqrt{4 + 1 + 36} = \sqrt{41} \]

6. **Conclusion**:
The area of the parallelogram is:
\[ A = \sqrt{41} \]