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Calculus-Volume of tetrahedron and cross product

by GingerBread27
Tags: calculusvolume, cross, product, tetrahedron
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GingerBread27
#1
Sep15-04, 08:09 PM
P: 109
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.
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HallsofIvy
#2
Sep15-04, 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 |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.
ehild
#3
Sep16-04, 07:16 AM
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P: 10,373
Quote Quote by GingerBread27
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.
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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