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Calculus-Volume of tetrahedron and cross product

  1. Sep 15, 2004 #1
    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. jcsd
  3. Sep 15, 2004 #2

    HallsofIvy

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    Science Advisor

    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 |u||v|sin(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 |u||v|sin(&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.
     
  4. Sep 16, 2004 #3

    ehild

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    Have you learnt about scalar triple product? it is
    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 = A-B = (0, 1, 2)

    c = C-B = (0, -1, 1)

    d = D-B = (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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