Question

The area of the parallelogram whose diagonals are \[ \mathbf{d_1} = \frac{3}{2} \hat{i} + \frac{1}{2} \hat{j} - \hat{k}, \quad \mathbf{d_2} = 2 \hat{i} - 6 \hat{j} + 8 \hat{k} \] is:

• A. \( 5\sqrt{3} \)
• B. \( 5\sqrt{2} \)
• C. \( 25\sqrt{3} \)
• D. \( 25\sqrt{2} \)

Hint:

The area of the parallelogram with diagonals \( \mathbf{d_1} \) and \( \mathbf{d_2} \) is given by \( \frac{1}{2} | \mathbf{d_1} \times \mathbf{d_2} | \).

Answer 1

The Correct Option is A

Step 1: Formula for the area of the parallelogram. The area of the parallelogram formed by two diagonals \( \mathbf{d_1} \) and \( \mathbf{d_2} \) is given by: \[ \text{Area} = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| \] where \( \mathbf{d_1} \) and \( \mathbf{d_2} \) are the diagonals, and \( \times \) represents the cross product. 
Step 2: Calculate the cross product. We need to find the cross product of \( \mathbf{d_1} \) and \( \mathbf{d_2} \): \[ \mathbf{d_1} = \left( \frac{3}{2}, \frac{1}{2}, -1 \right), \quad \mathbf{d_2} = (2, -6, 8) \] The cross product \( \mathbf{d_1} \times \mathbf{d_2} \) is: \[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{1}{2} & -1 \\ 2 & -6 & 8 \end{vmatrix} \] Solving the determinant: \[ \mathbf{d_1} \times \mathbf{d_2} = \hat{i} \left( \frac{1}{2} \times 8 - (-1) \times (-6) \right) - \hat{j} \left( \frac{3}{2} \times 8 - (-1) \times 2 \right) + \hat{k} \left( \frac{3}{2} \times (-6) - \frac{1}{2} \times 2 \right) \] \[ \mathbf{d_1} \times \mathbf{d_2} = \hat{i} \left( 4 - 6 \right) - \hat{j} \left( 12 + 2 \right) + \hat{k} \left( -9 - 1 \right) \] \[ \mathbf{d_1} \times \mathbf{d_2} = -2 \hat{i} - 14 \hat{j} - 10 \hat{k} \] 
Step 3: Find the magnitude of the cross product. The magnitude of the vector \( \mathbf{d_1} \times \mathbf{d_2} \) is: \[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} \] \[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \sqrt{4 + 196 + 100} = \sqrt{300} \] \[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = 10\sqrt{3} \] 
Step 4: Calculate the area. Now, we calculate the area of the parallelogram: \[ \text{Area} = \frac{1}{2} \times 10\sqrt{3} = 5\sqrt{3} \]