The area (in sq units) of the parallelogram whose diagonals are along the vectors `8hati - 6hatj` and `3hati + 4hatj - 12hatk` is:

Options

A. 65
B. 52
C. 26
D. 20

Answer

To find the area of the parallelogram whose diagonals are represented by the vectors \( \mathbf{d_1} = 8\hat{i} - 6\hat{j} \) and \( \mathbf{d_2} = 3\hat{i} + 4\hat{j} - 12\hat{k} \), we can follow these steps:

### Step 1: Identify the vectors
The diagonals of the parallelogram are given as:
- \( \mathbf{d_1} = 8\hat{i} - 6\hat{j} \)
- \( \mathbf{d_2} = 3\hat{i} + 4\hat{j} - 12\hat{k} \)

### Step 2: Calculate the cross product of the vectors
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 \( \mathbf{d_1} \times \mathbf{d_2} \), we set up the determinant:
\[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & -6 & 0 \\ 3 & 4 & -12 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:
\[ \mathbf{d_1} \times \mathbf{d_2} = \hat{i} \begin{vmatrix} -6 & 0 \\ 4 & -12 \end{vmatrix} - \hat{j} \begin{vmatrix} 8 & 0 \\ 3 & -12 \end{vmatrix} + \hat{k} \begin{vmatrix} 8 & -6 \\ 3 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} -6 & 0 \\ 4 & -12 \end{vmatrix} = (-6)(-12) - (0)(4) = 72 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 8 & 0 \\ 3 & -12 \end{vmatrix} = (8)(-12) - (0)(3) = -96 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 8 & -6 \\ 3 & 4 \end{vmatrix} = (8)(4) - (-6)(3) = 32 + 18 = 50 \]

Putting it all together:
\[ \mathbf{d_1} \times \mathbf{d_2} = 72\hat{i} + 96\hat{j} + 50\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now, we calculate the magnitude:
\[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \sqrt{72^2 + 96^2 + 50^2} \]
Calculating each term:
- \( 72^2 = 5184 \)
- \( 96^2 = 9216 \)
- \( 50^2 = 2500 \)

Adding these:
\[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \sqrt{5184 + 9216 + 2500} = \sqrt{16900} = 130 \]

### Step 5: Calculate the area of the parallelogram
Finally, we find the area:
\[ A = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \frac{1}{2} \times 130 = 65 \text{ square units} \]

### Final Answer:
The area of the parallelogram is \( 65 \) square units.

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