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
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
Updated On: Jan 13, 2026
- \( \frac{1}{2} \sqrt{410} \)
- \( \frac{1}{2} \sqrt{474} \)
- \( \frac{1}{2} \sqrt{586} \)
- \( \frac{1}{2} \sqrt{306} \)
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The Correct Option is B
Approach Solution - 1
To find the area of the parallelogram \(ABCD\), where \(A(2, 3, 5)\) and \(C(-3, 4, -2)\) are opposite vertices and the diagonal vector \(\overrightarrow{BD} = \hat{i} + 2 \hat{j} + 3 \hat{k}\), we can follow these steps:
- Begin by identifying the vectors representing the diagonals of the parallelogram:
- Vector \(\overrightarrow{AC}\) (\(A\) to \(C\)) is given by: \[ \overrightarrow{AC} = (-3 - 2) \hat{i} + (4 - 3) \hat{j} + (-2 - 5) \hat{k} = -5 \hat{i} + \hat{j} - 7 \hat{k} \]
- Let \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\) be the two diagonals of the parallelogram. The area of the parallelogram is given by the formula for the cross product of these vectors:
- Find the cross product \(\overrightarrow{AC} \times \overrightarrow{BD}\):
- The cross product is calculated using the determinant matrix: \[ \overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix} \]
- Calculate the determinant: \[ = \hat{i}(1 \cdot 3 - (-7) \cdot 2) - \hat{j}(-5 \cdot 3 - (-7) \cdot 1) + \hat{k}(-5 \cdot 2 - 1 \cdot 1) \] \[ = \hat{i}(3 + 14) - \hat{j}(-15 + 7) + \hat{k}(-10 - 1) \] \[ = \hat{i}(17) - \hat{j}(-8) + \hat{k}(-11) \] \[ = 17\hat{i} + 8\hat{j} - 11\hat{k} \]
- Calculate the magnitude of the cross product: \[ \|\overrightarrow{AC} \times \overrightarrow{BD}\| = \sqrt{17^2 + 8^2 + (-11)^2} \] \[ = \sqrt{289 + 64 + 121} \] \[ = \sqrt{474} \]
- Substitute back into the area formula: \[ \text{Area} = \frac{1}{2} \sqrt{474} \]
Thus, the area of the parallelogram is \(\frac{1}{2} \sqrt{474}\).
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Approach Solution -2
The area is given by:
Area = \( \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}| \)
Calculate \( \overrightarrow{AC} = (-5i + j - 7k) \) and \( \overrightarrow{BD} = i + 2j + 3k \) and find the cross product.
Then,
Area = \( \frac{1}{2} \sqrt{474} \)
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