Question

In the given figure, ABCD is a parallelogram. If \( \vec{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k} \) and \( \vec{DB} = 3\hat{i} - 6\hat{j} + 2\hat{k} \), then find \( \vec{AD} \) and hence find the area of parallelogram ABCD.

Hint:

To find the area of a parallelogram using vectors, compute the cross product of two adjacent sides, and the magnitude of this vector gives the area. Use \(|\vec{AB} \times \vec{AD}|\) for the solution.

Answer 1

To determine \( \vec{AD} \), the vector addition \( \vec{AD} = \vec{AB} + \vec{DB} \) is employed. Given \( \vec{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k} \) and \( \vec{DB} = 3\hat{i} - 6\hat{j} + 2\hat{k} \), \( \vec{AD} \) is calculated by summing these vectors: \[ \vec{AD} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (3\hat{i} - 6\hat{j} + 2\hat{k}). \]

Upon simplification, we obtain: \[ \vec{AD} = (2 + 3)\hat{i} + (-4 - 6)\hat{j} + (5 + 2)\hat{k} = 5\hat{i} - 10\hat{j} + 7\hat{k}. \]

The area of parallelogram ABCD is equivalent to the magnitude of the cross product of vectors \( \vec{AB} \) and \( \vec{AD} \): \[ \text{Area} = |\vec{AB} \times \vec{AD}|. \]

The cross product of \( \vec{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k} \) and \( \vec{AD} = 5\hat{i} - 10\hat{j} + 7\hat{k} \) is computed using a determinant: \[ \vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 5 & -10 & 7 \end{vmatrix}. \]

Expanding the determinant yields: \[ \vec{AB} \times \vec{AD} = \hat{i} \begin{vmatrix} -4 & 5 \\ -10 & 7 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 5 \\ 5 & 7 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -4 \\ 5 & -10 \end{vmatrix}. \]

The individual 2x2 determinants are calculated as follows: \[ \hat{i} = (-4)(7) - (5)(-10) = -28 + 50 = 22, \] \[ \hat{j} = (2)(7) - (5)(5) = 14 - 25 = -11, \] \[ \hat{k} = (2)(-10) - (-4)(5) = -20 + 20 = 0. \] Consequently, \[ \vec{AB} \times \vec{AD} = 22\hat{i} + 11\hat{j} + 0\hat{k}. \]

The magnitude of this cross product is: \[ |\vec{AB} \times \vec{AD}| = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605}. \]

Answer: The area of parallelogram ABCD is \( \sqrt{605} \), as determined by the magnitude of the cross product. \bigskip