In this problem, you will illustrate a geometric fact usingvectors associated with 2D regions in 3D space. We start with thefour 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 (nota regular one, however).

1. Find an equation for the plane through points A, B, C; callthis plane P1. Using this, find a normal to P1 which points“outwards” (out of the 3D solid, that is). Find the area of thetriangle ABC. Finally, find the “outwards” normal to P1 which haslength 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 P2which points “outwards.” Find the area of the triangle ABD.Finally, find the “outwards” normal to P2 which has length equal tothe area of the triangle ABD; call this normal vector ~n2.

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

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

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

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