What is the area of the parallelogram having diagonals `3hat(i) + hat(j) + 2hat(k) and hat(i) - 3hat(j) + 4hat(k)` ?

Options

A. `5 sqrt3` square units
B. `4 sqrt5` square units
C. `3 sqrt3` square units
D. `15 sqrt2` square units

Answer

To find the area of the parallelogram given its diagonals, we can use the formula:

\[ \text{Area} = \frac{1}{2} \| \mathbf{D_1} \times \mathbf{D_2} \| \]

where \( \mathbf{D_1} \) and \( \mathbf{D_2} \) are the vectors representing the diagonals of the parallelogram.

### Step 1: Identify the diagonals
The given diagonals are:
\[ \mathbf{D_1} = 3\hat{i} + \hat{j} + 2\hat{k} \]
\[ \mathbf{D_2} = \hat{i} - 3\hat{j} + 4\hat{k} \]

### Step 2: Calculate the cross product \( \mathbf{D_1} \times \mathbf{D_2} \)
To calculate the cross product, 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 \( \mathbf{D_1} \) and \( \mathbf{D_2} \):

\[ \mathbf{D_1} \times \mathbf{D_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & -3 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:

\[ \mathbf{D_1} \times \mathbf{D_2} = \hat{i} \begin{vmatrix} 1 & 2 \\ -3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 1 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & 2 \\ -3 & 4 \end{vmatrix} = (1)(4) - (2)(-3) = 4 + 6 = 10 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & 2 \\ 1 & 4 \end{vmatrix} = (3)(4) - (2)(1) = 12 - 2 = 10 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & 1 \\ 1 & -3 \end{vmatrix} = (3)(-3) - (1)(1) = -9 - 1 = -10 \]

Putting it all together:
\[ \mathbf{D_1} \times \mathbf{D_2} = 10\hat{i} - 10\hat{j} - 10\hat{k} \]

### Step 4: Calculate the magnitude of the cross product
Now, we find the magnitude of the vector:
\[ \|\mathbf{D_1} \times \mathbf{D_2}\| = \sqrt{(10)^2 + (-10)^2 + (-10)^2} = \sqrt{100 + 100 + 100} = \sqrt{300} = 10\sqrt{3} \]

### Step 5: Calculate the area of the parallelogram
Using the formula for the area:
\[ \text{Area} = \frac{1}{2} \|\mathbf{D_1} \times \mathbf{D_2}\| = \frac{1}{2} (10\sqrt{3}) = 5\sqrt{3} \]

Thus, the area of the parallelogram is:
\[ \boxed{5\sqrt{3}} \]