The vector `-2hati+4hatj+4hatk and -4hati-4hatk` represent the diagonals BD and AC of a parallelogram ABCD. Then, find the area of the parallelogram.

Answer

To find the area of the parallelogram ABCD given the diagonals BD and AC represented by the vectors \(-2\hat{i} + 4\hat{j} + 4\hat{k}\) and \(-4\hat{i} + 0\hat{j} - 4\hat{k}\), we can follow these steps:

### Step 1: Identify the Diagonal Vectors
Let:
- \(\mathbf{d_1} = -2\hat{i} + 4\hat{j} + 4\hat{k}\) (Diagonal AC)
- \(\mathbf{d_2} = -4\hat{i} + 0\hat{j} - 4\hat{k}\) (Diagonal BD)

### Step 2: Calculate the Cross Product \(\mathbf{d_1} \times \mathbf{d_2}\)
To find the area of the parallelogram, we need to calculate the cross product of the two diagonal vectors. The area \(A\) of the parallelogram is given by:

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

The cross product can be calculated using the determinant of a matrix:

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

### Step 3: Calculate the Determinant
Expanding the determinant:

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

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ 4 \cdot (-4) - 4 \cdot 0 = -16 \]

2. For \(-\hat{j}\):
\[ -((-2) \cdot (-4) - 4 \cdot (-4)) = - (8 + 16) = -24 \]

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

Putting it all together:

\[ \mathbf{d_1} \times \mathbf{d_2} = -16\hat{i} + 24\hat{j} + 16\hat{k} \]

### Step 4: Calculate the Magnitude of the Cross Product
Now we find the magnitude of the cross product:

\[ |\mathbf{d_1} \times \mathbf{d_2}| = \sqrt{(-16)^2 + (24)^2 + (16)^2} \]

Calculating each term:

- \((-16)^2 = 256\)
- \(24^2 = 576\)
- \(16^2 = 256\)

Adding these:

\[ |\mathbf{d_1} \times \mathbf{d_2}| = \sqrt{256 + 576 + 256} = \sqrt{1088} \]

### Step 5: Simplify the Magnitude
We can simplify \(\sqrt{1088}\):

\[ \sqrt{1088} = \sqrt{64 \times 17} = 8\sqrt{17} \]

### Step 6: Calculate the Area of the Parallelogram
Finally, we can find the area of the parallelogram:

\[ A = \frac{1}{2} |\mathbf{d_1} \times \mathbf{d_2}| = \frac{1}{2} \times 8\sqrt{17} = 4\sqrt{17} \]

Thus, the area of the parallelogram ABCD is:

\[ \boxed{4\sqrt{17}} \text{ square units} \]