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.
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Solution and Explanation
To find \( \vec{AD} \), we use the relationship: \[ \vec{AD} = \vec{AB} + \vec{DB}. \] Given that \( \vec{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k} \) and \( \vec{DB} = 3\hat{i} - 6\hat{j} + 2\hat{k} \), we calculate \( \vec{AD} \): \[ \vec{AD} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (3\hat{i} - 6\hat{j} + 2\hat{k}). \]
Simplifying: \[ \vec{AD} = (2 + 3)\hat{i} + (-4 - 6)\hat{j} + (5 + 2)\hat{k} = 5\hat{i} - 10\hat{j} + 7\hat{k}. \]
The area of parallelogram ABCD is given by 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 as follows: \[ \vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 5 & -10 & 7 \end{vmatrix}. \]
Expanding the determinant:
\[ \vec{AB} \times \vec{AD} = \hat{i} \begin{vmatrix} -4 & 5 \\ -10 & 7 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 5 \\ 5 & 7 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -4 \\ 5 & -10 \end{vmatrix}. \]
Calculating each 2x2 determinant: \[ \hat{i} = (-4)(7) - (5)(-10) = -28 + 50 = 22, \] \[ \hat{j} = (2)(7) - (5)(5) = 14 - 25 = -11, \] \[ \hat{k} = (2)(-10) - (-4)(5) = -20 + 20 = 0. \] Thus, \[ \vec{AB} \times \vec{AD} = 22\hat{i} + 11\hat{j} + 0\hat{k}. \]
The magnitude of the cross product is: \[ |\vec{AB} \times \vec{AD}| = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605}. \]
Answer: The magnitude of the cross product is \( \sqrt{605} \), so the area of parallelogram ABCD is \( \sqrt{605} \). \bigskip
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