Question

In this problem, you will illustrate a geometric fact using vectors associated with 2D regions in 3D space. We start with the four points given as:

A : (0, 0, 0) B : (1, 0, 0) C : (0, 1, 0) D : (0, 0, 1).

These four points define a 3D object which is a tetrahedron (not a regular one, however).

1. Find an equation for the plane through points A, B, C; call this plane P1. Using this, find a normal to P1 which points “outwards” (out of the 3D solid, that is). Find the area of the triangle ABC. Finally, find the “outwards” normal to P1 which has length equal to the area of triangle ABC; call this normal vector ~n1.

2. Now find an equation for the plane through the points A, B, D; call this plane P2. Using this equation, find a normal to P2 which points “outwards.” Find the area of the triangle ABD. Finally, find the “outwards” normal to P2 which has length equal to the area of the triangle ABD; call this normal vector ~n2.

3. Repeat step 2 with the points A, C, D; call the resulting plane P3 and normal ~n3.

4. Repeat step 2 with the points B, C, D; call the resulting plane P4 and normal ~n4.

5. Compute ~n1 + ~n2 + ~n3 + ~n4. You should get that the vector sum is equal to zero. Do you have any idea why this should be true?

How do you compute the area of one of those triangles? One possible way is to use the cross product. We know that k~u × ~vk is equal to the area of the parallelogram formed by ~u and ~v. So, the area of the “triangle” formed by these two vectors should be half of this.

Answer 1