Pyramid and vectors

Problem

We consider the pyramid ABCDS with A(0|0|0), B(4|4|2), C(8|0|2), D(4|-4|0), and S(1|1|-4). Its base is a parallelogram.

1. Prove that the parallelogram ABCD is a rectangle.

2. The edge [AS] is normal to the base ABCD. The area of the base is \(24\sqrt{2}\). Determine the volume of the pyramid.

Solution of part a

ABCD forms a rectangle if starting from one of the vertices the angle between the shortest vectors to the othe vertices is a right angle.

It follows that the vectors from A to B and from A to D are orthogonal to each other. The point C lies diagonally opposite of A. Therfore, the parallelogram is indeed a rectangle. Since this solution depends on the information that ABCD is a parallelogram, we check alos the other three inner angles.

Solution of part b

Since the vector from A to S is normal to the base, its length \(h\) equals the height of the pyramid. The area of the base is given as \(A=24\sqrt{2}\). We first briefly check the latter result.

The height of the pyramid is obtained as

Then, volume takes on the value \(V=\frac{h}{3}A=48\). This result can be confirmed directly with the help of Sage.