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

Answer

To find the area of the parallelogram whose diagonals are given by the vectors \( \mathbf{d_1} = -2\hat{i} + 4\hat{j} + 4\hat{k} \) and \( \mathbf{d_2} = -4\hat{i} - 2\hat{k} \), we can use the formula:

\[ \text{Area} = \frac{1}{2} \|\mathbf{d_1} \times \mathbf{d_2}\| \]

### Step 1: Calculate the cross product \( \mathbf{d_1} \times \mathbf{d_2} \)

The cross product of two vectors \( \mathbf{d_1} \) and \( \mathbf{d_2} \) 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:

\[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & 4 \\ -4 & 0 & -2 \end{vmatrix} \]

### Step 2: Calculate the determinant

Expanding the determinant, we have:

\[ \mathbf{d_1} \times \mathbf{d_2} = \hat{i} \begin{vmatrix} 4 & 4 \\ 0 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 4 \\ -4 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 4 \\ -4 & 0 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 4 & 4 \\ 0 & -2 \end{vmatrix} = (4)(-2) - (4)(0) = -8 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} -2 & 4 \\ -4 & -2 \end{vmatrix} = (-2)(-2) - (4)(-4) = 4 + 16 = 20 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} -2 & 4 \\ -4 & 0 \end{vmatrix} = (-2)(0) - (4)(-4) = 0 + 16 = 16 \]

Putting it all together:

\[ \mathbf{d_1} \times \mathbf{d_2} = -8\hat{i} - 20\hat{j} + 16\hat{k} \]

### Step 3: Calculate the magnitude of the cross product

Now, we find the magnitude of the vector \( \mathbf{d_1} \times \mathbf{d_2} \):

\[ \|\mathbf{d_1} \times \mathbf{d_2}\| = \sqrt{(-8)^2 + (-20)^2 + (16)^2} \]

Calculating each term:

\[ = \sqrt{64 + 400 + 256} = \sqrt{720} = 12\sqrt{5} \]

### Step 4: Calculate the area of the parallelogram

Now, we can use the magnitude to find the area:

\[ \text{Area} = \frac{1}{2} \|\mathbf{d_1} \times \mathbf{d_2}\| = \frac{1}{2} \times 12\sqrt{5} = 6\sqrt{5} \]

### Final Answer

The area of the parallelogram is:

\[ \boxed{6\sqrt{5}} \]