This is a multi-variable prob.
We know that a tetrahedron is a solid which has four vertices and has four triangular faces. For every side of a tetrahedron, you can take a vector whose length then is equal to the area of the face and further where its direction is pointing outward perpendicular to the face. Mark these vectors as v_1, v_2, v_3, and v_4.
1.) Show that v_1 + v_2 + v_3 + v_4 = 0
2.) Suppose that we have that 3 of the faces of the tetrahedron are perpendicular to each other (now it will look as though it could fit in the very corner of a box. Label these faces A,B,C and the area of the 4th face as D. Prove that D^2 = A^2 + B^2 + C^2
HINTS: you will need to use vector products - also, you may want to use a coordinate system such that one vertex will lie at the origin (and then from there consider edges as vectors)
This is apparently our hardest problem for this assignment. I'd appreciate it if someone could point me in the right direction of how to solve these.