Question

A tetrahedron is a solid with four vertices, P, Q, R, and S, and four triangular faces,

Let v1,v2,v3,and v4 be vectors with lengths equal to the areas of the faces opposite to the vertices P, Q, R, and S, respectively, and directions perpendicular to the respective faces and pointing outward.

Show that v1+v2+v3+v4=0

Answer 1

The vectors v1, v2, v3, and v4 representing the areas of faces of a tetrahedron cancel each other out because they are equal in magnitude and opposite in direction, resulting in their sum being the zero vector.

Step-by-step explanation:

To prove that v1 + v2 + v3 + v4 = 0 for a tetrahedron, we will use the concept of vectors representing area and showing that the sum of these vectors is the zero vector, which indicates that the vectors balance each other out.

Consider a tetrahedron with vertices P, Q, R, and S. Let the vectors v1, v2, v3, and v4 originate from the centroid of the tetrahedron and point outward, perpendicular to their respective faces. The magnitude of each vector is proportional to the area of the opposite face, and the direction is outward, indicating a positive sense of area.

By symmetry, the vectors v1 and v2 are equal in magnitude but opposite in direction as they are normal to opposite faces. Similarly, vectors v3 and v4 are equal in magnitude and opposite in direction. Hence, when we add these vectors:

• v1 + v2 = 0
• v3 + v4 = 0

Adding these two results gives us:

v1 + v2 + v3 + v4 = (v1 + v2) + (v3 + v4) = 0 + 0 = 0

So, the sum of the vectors equals the zero vector, indicating that they cancel each other out and the tetrahedral system is in equilibrium.