Enormous numbers of stones! Compteted and beautiful. The engineers are dead some 2000 years. Here we use vectors to look at their work. This is a simple problem that solves nicely with vectors.

Pharaoh's Engineers The GREAT PYRAMID of Egypt was constructed to precise proportions. A hypothesis is the pyramid was constructed to fit inside an imaginary hemisphere with each of the pyramid corners and the peak touching the hemisphere. The sketch shows this hypothesized pyramid geometry. Determine angle 0BC, the angle each of the pyramid faces makes with the plane of earth plane (angle 0BC). The actual, measured value is 51°.

♦  Vectors work well in three dimensions but physical stituations do not come with coordinates and unit vectors. We take the center of the pyramid at its base as our origin. We apply X, Y, and Z coordinates in accord with the pyramid sides then indicate the vector triple, I, J and K as our basis.

We seek to determine the angle OBC. The usual approach is to write a vector to some point the path of which includes the sought angle. Next write a second vector to the same point and equate them. The peak and corners of the pyramid contact the hemisphere. We write the position of the peak of the pyramid in two independent ways.

• A vector represented as OC extends from the origin vertically to the peak. The notation OC means a vector starting at "O" and extending to (and directed toward) "C."

• We see another path to the peak that is the sum of three vectors. Our working equation is the equality of these two representations of position of the same point, the peak of the pyramid.

OA + AB + BC = OC

• The above is an implicit vector statement. To proceed we make as many of the components explicit in terms of dimensions and the unit vector basis as we are able.

OA = r (cos 45 I + sin 45 J),

The letter, " r " represents the radius of the hemisphere.

Since we do not know the length of vector AB we write it as:

AB = |AB|J

The notation, |AB|, means length of the vector AB and its direction is that of the unit vector I.

The vector BC has an unknown length. Its direction is written using the unknown angle, α. The vectors are written then equated.

r (cos 45 I+sin 45 J) + |AB|J + |BC|[cosα(- I) +  sinαK] = r K

The above procedure is typical. At this point, we check it to assure it is correct. Next we regroup the equation in accord with the vector triad.

[r cos45  - |BC|cosα ] I + [|AB| - r sin45  ] J + [ |BC|sinα - r ] K = 0

By its right side we see this equation expresses a vector that is zero. The only way that can happen is if each component magnitude of the vector equals zero. Thus we have:

From the I component we obtain:

r cos45 - |BC| cosα = 0

The J component provides:

|AB| - r sin45 = 0

and for K we have:

|BC| sinα - r = 0

Combining the I and K components we determine:

tan α = √2   whereupon     α = 54°

Thus the hypothesis is quite close!