ABCD is a parallelogram with `vec(AC) = hati - 2hatj + hatk` and `vec(BD) = -hati + 2hatj - 5hatk`. Area of this parallelogram is equal to:

Options

A. `sqrt(5)//2` sq. units
B. `2sqrt(5)` sq. units
C. `4sqrt(5)` sq. units
D. `sqrt(5)` sq. units

Answer

To find the area of the parallelogram ABCD given the vectors \(\vec{AC} = \hat{i} - 2\hat{j} + \hat{k}\) and \(\vec{BD} = -\hat{i} + 2\hat{j} - 5\hat{k}\), we can follow these steps:

### Step 1: Identify the vectors
We have:
\[ \vec{AC} = \hat{i} - 2\hat{j} + \hat{k} \]
\[ \vec{BD} = -\hat{i} + 2\hat{j} - 5\hat{k} \]

### Step 2: Find the cross product of the vectors
The area of the parallelogram can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} |\vec{AC} \times \vec{BD}| \]
To find \(\vec{AC} \times \vec{BD}\), we set up a determinant:
\[ \vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ -1 & 2 & -5 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:
\[ \vec{AC} \times \vec{BD} = \hat{i} \begin{vmatrix} -2 & 1 \\ 2 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ -1 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ -1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ (-2)(-5) - (1)(2) = 10 - 2 = 8 \]

2. For \(-\hat{j}\):
\[ (1)(-5) - (1)(-1) = -5 + 1 = -4 \quad \text{(so it becomes +4)} \]

3. For \(\hat{k}\):
\[ (1)(2) - (-2)(-1) = 2 - 2 = 0 \]

Putting it all together:
\[ \vec{AC} \times \vec{BD} = 8\hat{i} + 4\hat{j} + 0\hat{k} = 8\hat{i} + 4\hat{j} \]

### Step 4: Find the magnitude of the cross product
Now we find the magnitude:
\[ |\vec{AC} \times \vec{BD}| = \sqrt{(8)^2 + (4)^2} = \sqrt{64 + 16} = \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} \]

### Step 5: Calculate the area of the parallelogram
Now, substituting back into the area formula:
\[ \text{Area} = \frac{1}{2} \times 4\sqrt{5} = 2\sqrt{5} \]

Thus, the area of the parallelogram ABCD is:
\[ \text{Area} = 2\sqrt{5} \text{ square units} \]

### Final Answer
The area of the parallelogram ABCD is \(2\sqrt{5}\) square units.

---