find the volume of the parallelepiped whose vertices are A(3,2,-1),B(-2,2,-3),C(3,5,-2),D(-2,5,4)
Question
find the volume of the parallelepiped whose vertices are A(3,2,-1),B(-2,2,-3),C(3,5,-2),D(-2,5,4)
Solution
The volume of a parallelepiped is given by the scalar triple product of the vectors formed by its edges.
First, we need to find the vectors AB, AC, and AD.
Vector AB = B - A = (-2-3, 2-2, -3-(-1)) = (-5, 0, -2) Vector AC = C - A = (3-3, 5-2, -2-(-1)) = (0, 3, -1) Vector AD = D - A = (-2-3, 5-2, 4-(-1)) = (-5, 3, 5)
The scalar triple product of three vectors (a, b, c) is given by a . (b x c), where . is the dot product and x is the cross product.
First, we find the cross product of AC and AD:
AC x AD = (3*-5 - -13, -1-5 - 03, 0-5 - 3*-5) = (-15+3, 5, 15) = (-12, 5, 15)
Then, we find the dot product of AB and (AC x AD):
AB . (AC x AD) = -5*-12 + 05 + -215 = 60 - 30 = 30
Therefore, the volume of the parallelepiped is 30 cubic units.
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