Find the area of the parallelogram whose sides are represented by `2hati + 4hatj - 6hatk and hati +2hatk.`

Explanation

To find the area of the parallelogram whose sides are represented by the vectors \( \mathbf{A} = 2\hat{i} + 4\hat{j} - 6\hat{k} \) and \( \mathbf{B} = \hat{i} + 2\hat{k} \), we can use the formula for the area of a parallelogram formed by two vectors, which is given by the magnitude of the cross product of the two vectors.

### Step-by-Step Solution:

1. **Identify the vectors:**
\[ \mathbf{A} = 2\hat{i} + 4\hat{j} - 6\hat{k} \]
\[ \mathbf{B} = \hat{i} + 2\hat{k} \]

2. **Set up the cross product:**
The area of the parallelogram is given by:
\[ \text{Area} = |\mathbf{A} \times \mathbf{B}| \]
To compute the cross product \( \mathbf{A} \times \mathbf{B} \), we can use 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 \( \mathbf{B} \):
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -6 \\ 1 & 0 & 2 \end{vmatrix} \]

3. **Calculate the determinant:**
Expanding the determinant:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 4 & -6 \\ 0 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -6 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 4 \\ 1 & 0 \end{vmatrix} \]

- For \( \hat{i} \):
\[ \begin{vmatrix} 4 & -6 \\ 0 & 2 \end{vmatrix} = (4)(2) - (0)(-6) = 8 \]
- For \( \hat{j} \):
\[ \begin{vmatrix} 2 & -6 \\ 1 & 2 \end{vmatrix} = (2)(2) - (1)(-6) = 4 + 6 = 10 \]
- For \( \hat{k} \):
\[ \begin{vmatrix} 2 & 4 \\ 1 & 0 \end{vmatrix} = (2)(0) - (1)(4) = -4 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 8\hat{i} - 10\hat{j} - 4\hat{k} \]

4. **Find the magnitude of the cross product:**
The magnitude of \( \mathbf{A} \times \mathbf{B} \) is calculated as follows:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(8)^2 + (-10)^2 + (-4)^2} \]
\[ = \sqrt{64 + 100 + 16} = \sqrt{180} \]

5. **Calculate the area:**
Thus, the area of the parallelogram is:
\[ \text{Area} = \sqrt{180} \]

### Final Answer:
The area of the parallelogram is \( \sqrt{180} \).