Find the volume of the parallelepiped whose vertices are
\( A(3,2,-1), B(-2,2,-3), C(3,5,-2) \) and \( D(-2,5,4) \).
\( A(3,2,-1), B(-2,2,-3), C(3,5,-2) \) and \( D(-2,5,4) \).
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Solution and Explanation
Step 1: Find Three Vectors
\[ \mathbf{AB} = (-2 - 3, 2 - 2, -3 + 1) = (-5,0,-2). \] \[ \mathbf{AC} = (3 - 3, 5 - 2, -2 + 1) = (0,3,-1). \] \[ \mathbf{AD} = (-2 - 3, 5 - 2, 4 + 1) = (-5,3,5). \]
Step 2: Compute the Volume
\[ V = \left| \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) \right|. \]
\[ = (18,-5,15). \] \[ V = \left| (-5,0,-2) \cdot (18,-5,15) \right|. \] \[ = |(-5 \times 18) + (0 \times -5) + (-2 \times 15)|. \] \[ = |-90 - 30| = |120|. \]
Final Answer: \( V = 120 \).
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