If the diagonals of a parallelogram are represented by the vectors ` 3hati + hatj -2hatk and hati + 3hatj -4hatk`, then its area in square units , is

Options

A. ` 5sqrt3`
B. `6sqrt3`
C. `sqrt42`
D. `2sqrt42`

Explanation

To find the area of a parallelogram given its diagonals represented by the vectors \( \mathbf{d_1} = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \mathbf{d_2} = \hat{i} + 3\hat{j} - 4\hat{k} \), we can follow these steps:

### Step 1: Write down the diagonal vectors
We have:
\[ \mathbf{d_1} = 3\hat{i} + \hat{j} - 2\hat{k} \]
\[ \mathbf{d_2} = \hat{i} + 3\hat{j} - 4\hat{k} \]

### Step 2: Calculate the cross product of the diagonals
The area \( A \) of the parallelogram can be calculated using the formula:
\[ A = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| \]
To find the cross product \( \mathbf{d_1} \times \mathbf{d_2} \), we can set up the determinant:
\[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 3 & -4 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \mathbf{d_1} \times \mathbf{d_2} = \hat{i} \begin{vmatrix} 1 & -2 \\ 3 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -2 \\ 1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. For \( \hat{i} \):
\[ 1 \cdot (-4) - (-2) \cdot 3 = -4 + 6 = 2 \]
2. For \( -\hat{j} \):
\[ 3 \cdot (-4) - (-2) \cdot 1 = -12 + 2 = -10 \quad \Rightarrow \quad 10\hat{j} \]
3. For \( \hat{k} \):
\[ 3 \cdot 3 - 1 \cdot 1 = 9 - 1 = 8 \]

Putting it all together:
\[ \mathbf{d_1} \times \mathbf{d_2} = 2\hat{i} + 10\hat{j} + 8\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now we need to find the magnitude:
\[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \sqrt{2^2 + 10^2 + 8^2} = \sqrt{4 + 100 + 64} = \sqrt{168} \]
This can be simplified:
\[ \sqrt{168} = \sqrt{4 \cdot 42} = 2\sqrt{42} \]

### Step 5: Calculate the area of the parallelogram
Finally, we find the area:
\[ A = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \frac{1}{2} \cdot 2\sqrt{42} = \sqrt{42} \]

### Final Answer
The area of the parallelogram is:
\[ \sqrt{42} \text{ square units} \]