The area ( in sq. units ) of a parallelogram whose adjacent sides are given by the vectors `vec( i) + hat(k) ` and ` 2 hat(i) + hat(j) + hat(k)` is

Options

A. `sqrt(2)`
B. `sqrt(3)`
C. 3
D. 4

Answer

To find the area of the parallelogram formed by the vectors \( \vec{A} = \hat{i} + \hat{k} \) and \( \vec{B} = 2\hat{i} + \hat{j} + \hat{k} \), 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**:
- Let \( \vec{A} = \hat{i} + \hat{k} \) which can be represented as \( (1, 0, 1) \).
- Let \( \vec{B} = 2\hat{i} + \hat{j} + \hat{k} \) which can be represented as \( (2, 1, 1) \).

2. **Set Up the Cross Product**:
- The cross product \( \vec{A} \times \vec{B} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \vec{A} \) and \( \vec{B} \):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix} \]

3. **Calculate the Determinant**:
- Expanding the determinant:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} \]
- Calculating the 2x2 determinants:
\[ = \hat{i} (0 \cdot 1 - 1 \cdot 1) - \hat{j} (1 \cdot 1 - 2 \cdot 1) + \hat{k} (1 \cdot 1 - 0 \cdot 2) \]
\[ = \hat{i} (-1) - \hat{j} (-1) + \hat{k} (1) \]
\[ = -\hat{i} + \hat{j} + \hat{k} \]

4. **Find the Magnitude of the Cross Product**:
- The magnitude of the vector \( \vec{A} \times \vec{B} = -\hat{i} + \hat{j} + \hat{k} \) is calculated as follows:
\[ |\vec{A} \times \vec{B}| = \sqrt{(-1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

5. **Conclusion**:
- Therefore, the area of the parallelogram is \( \sqrt{3} \) square units.