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CalculusVolume of tetrahedron and cross product 
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#1
Sep1504, 08:09 PM

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Determine whether the points A = (1, 2, 3), B = (1, 1, 1), C = (1, 0, 2), and
D = (2,2, 0) are coplanar and find the volume of the tetrahedron with vertices ABCD. My professor did this problem in class as a review for an upcoming test and he didn't get the answer that was on the key. He just chuckled and went on but I would like to know how to really do this problem. The answer should be 5/6 and the points are not coplanar. My teacher got the answer of 3. He did the cross product of a and b times c. He got the cross product of a and b to be 3 and multiplying 3 by c resulted in 3. He used the volume of a tetrahedron to be (1/6)ha, h being C and area being a cross b. 


#2
Sep1504, 09:32 PM

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Draw a picture! The area of a parallelgram is "base times height". If you know the lengths of two connected sides, say x and y, and the angle. θ, between them, then the height (measured perpendicular to the base) is y sin(θ). The base is x so the area is xy sin(θ).
One way of defining the cross product of two vectors, u and v, is that it is the vector with length uvsin(θ) (theta is the angle between the two vectors). If one side of a parallelogram is the vector u and the other is v, then the length of the sides are u and v so that the area is uvsin(&theta), exactly the same as the cross product: the area of a parallelogram whose sides are the vectors u and v is exactly the cross product of u and v. 


#3
Sep1604, 07:16 AM

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a.[bxc] You can look after at Mathworld, for example. http://mathworld.wolfram.com/ScalarTripleProduct.html You can calculate the volume of a parallelepiped defined by the three vectors a, b, c. This product is the same as the absolute value of a determinant D, built up from the components of the vectors :  ax ay az   bx by bz   cx cy cz  D = ax by cz + ay bz cx + az bx cy  az by cx  ay bz cx  ax bz cy . Now, the edges od the tetrahedron are not the original vectors A, B, C, D , but the differences, for example with respect to B. a = AB = (0, 1, 2) c = CB = (0, 1, 1) d = DB = (1, 3, 1) If these three vectors are coplanar the points A, B, C, D are in the same plane. In this case the volume of the corresponding parallelepiped is zero. So we calculate the determinant.  0 +1 +2   0 1 +1  = 0 +1 + 0 + 2 +0 + 0 = 3  1 3 1  The volume of the parallelepiped is 3. The volume of the corresponding tetrahedron is one sixth of this value, that is 3/6=1/2. As I understood you, this is the same what your teacher got. That key might be wrong... ehild 


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