Question

Let \( A (2, 3, 5) \) and \( C(-3, 4, -2) \) be opposite vertices of a parallelogram \( ABCD \). If the diagonal \( \overrightarrow{BD} = \hat{i} + 2 \hat{j} + 3 \hat{k} \), then the area of the parallelogram is equal to

• A. \( \frac{1}{2} \sqrt{410} \)
• B. \( \frac{1}{2} \sqrt{474} \)
• C. \( \frac{1}{2} \sqrt{586} \)
• D. \( \frac{1}{2} \sqrt{306} \)

Answer 1

The Correct Option is B

To determine the area of parallelogram \(ABCD\), given opposite vertices \(A(2, 3, 5)\) and \(C(-3, 4, -2)\), and the diagonal vector \(\overrightarrow{BD} = \hat{i} + 2 \hat{j} + 3 \hat{k}\), the following steps are employed:

1. Determine the vector for diagonal \(\overrightarrow{AC}\) using the coordinates of \(A\) and \(C\): \[ \overrightarrow{AC} = (-3 - 2) \hat{i} + (4 - 3) \hat{j} + (-2 - 5) \hat{k} = -5 \hat{i} + \hat{j} - 7 \hat{k} \]
2. The area of a parallelogram is half the magnitude of the cross product of its diagonals.
3. Compute the cross product \(\overrightarrow{AC} \times \overrightarrow{BD}\) using a determinant: \[ \overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix} \]
4. Evaluate the determinant: \[ = \hat{i}(1 \cdot 3 - (-7) \cdot 2) - \hat{j}(-5 \cdot 3 - (-7) \cdot 1) + \hat{k}(-5 \cdot 2 - 1 \cdot 1) \] \[ = \hat{i}(3 + 14) - \hat{j}(-15 + 7) + \hat{k}(-10 - 1) \] \[ = 17\hat{i} + 8\hat{j} - 11\hat{k} \]
5. Calculate the magnitude of the resulting cross product vector: \[ \|\overrightarrow{AC} \times \overrightarrow{BD}\| = \sqrt{17^2 + 8^2 + (-11)^2} \] \[ = \sqrt{289 + 64 + 121} \] \[ = \sqrt{474} \]
6. Apply the area formula: \[ \text{Area} = \frac{1}{2} \|\overrightarrow{AC} \times \overrightarrow{BD}\| = \frac{1}{2} \sqrt{474} \]

The calculated area of the parallelogram is \(\frac{1}{2} \sqrt{474}\).