Determine the volume of the parallelepiped with vertices at the origin (1,2,3), (-1,1,2), (2,1,4).Question 3Select one:A.7 cu. unitB.12 cu. unitC.10 cu. unitD.9 cu. unitClear my choice
Question
Determine the volume of the parallelepiped with vertices at the origin (1,2,3), (-1,1,2), (2,1,4).Question 3Select one:A.7 cu. unitB.12 cu. unitC.10 cu. unitD.9 cu. unitClear my choice
Solution
To determine the volume of the parallelepiped with vertices at the origin (1,2,3), (-1,1,2), and (2,1,4), we can use the formula for the volume of a parallelepiped.
Step 1: Find the vectors formed by the three given points. We can find two vectors by subtracting the coordinates of the origin from the coordinates of the other two points: Vector A = (-1,1,2) - (1,2,3) = (-2,-1,-1) Vector B = (2,1,4) - (1,2,3) = (1,-1,1)
Step 2: Calculate the cross product of Vector A and Vector B. The cross product of two vectors gives us a vector that is perpendicular to both of the original vectors. The magnitude of this cross product vector represents the area of the parallelogram formed by Vector A and Vector B. We can use the formula for the magnitude of a cross product: |Vector A x Vector B| = |(-2,-1,-1) x (1,-1,1)|
To calculate the cross product, we can use the determinant method: |Vector A x Vector B| = |i j k | | -2 -1 -1 | | 1 -1 1 |
Expanding the determinant, we get: |Vector A x Vector B| = (1 * -1 - (-1) * -1)i - ((-2) * -1 - (-1) * 1)j + ((-2) * 1 - 1 * 1)k = (-1 - 1)i - (2 + 1)j - (2 - 1)k = -2i - 3j - 1k
Step 3: Calculate the magnitude of the cross product vector: |Vector A x Vector B| = sqrt((-2)^2 + (-3)^2 + (-1)^2) = sqrt(4 + 9 + 1) = sqrt(14)
Step 4: Calculate the volume of the parallelepiped. The volume of a parallelepiped is equal to the magnitude of the cross product vector multiplied by the base area. Since the base area is equal to the magnitude of the cross product vector, the volume is given by: Volume = |Vector A x Vector B| * base area = sqrt(14) * sqrt(14) = 14
Therefore, the volume of the parallelepiped is 14 cubic units.
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