Pythagoras in 3-D

Figure 1. Box ABCDEFGH

For some curious reason I want to know the distance from the bottom left front corner (point A) to top right back corner (point G) in the box above. I also want to know the angle made by the line AG with the bottom of the box. 

 

I will start by unfolding the box to make the flat shape below. Mathematicians call an unfolded shape a net. When the box is unfolded some single points (such as G and H) get represented as two distinct points (G and G’; H and H’).

Figure 2. A net of the same box.

ABCD is a rectangle forming the bottom of the box. B is obviously a right angle and and are both acute angles. The diagonal AC cuts the rectangle ABCD into two congruent (or equal) right-angled triangles, one of which is ABC (shaded light green above).

What may not be so obvious is that ACG’ (shaded light blue) must also be a right angled triangle! See Figure 3 below. It appears that ACG’ is a single straight line. If that is so, external angle ACG’ is 180° and consequently  +  would equal 180°, also. Since we already know that  is acute, it appears that must be an obtuse angle. 

Figure 3. A "zoomed-in" view of Figure 2 with extraneous detail erased.

We all know the saying, “A picture is worth a thousand words.” But folk wisdom is often contradictory and you probably have also heard, “You can’t believe everything you see.” Remember that ABCDEFGH is a 3-dimensional box. AB is the length of the box. BC is the breadth of the box. Length and breadth are dimensions, at 90° to each other. Together they form a horizontal plane. Now look back at Figure 1. Line CG (or equivalently CG’) is the vertical height of the box, so it must make a 90° angle to the horizontal plane! In other words, angle BCG’ is a 90° angle (between the horizontal and the vertical). Since AC is also horizontal, it follows that ACG’ (the angle also labeled ) must also be a 90° angle. How can I assert that AC is also horizontal? It’s simple: any line drawn on a horizontal plane will be a horizontal line, when considered with respect to the third dimension.

 

A glance at the diagram below might be helpful. The x-, y- and z-axes are all mutually perpendicular. Representing 3-D space on a 2-D plane requires that some 90 degree angles appear other than they really are.

Figure 4. Reminding you of 3-D Cartesian axes.

By now you may believe I have wandered rather too far from the main point. But now its time to wrap up.

Figure 5. The triangles are drawn so that angle B and angle C can both be seen to be right angles, although the three dimensional look has been lost.

x² + y² = c² ①                               (c is the hypotenuse of right triangle ABC).

c² + z² = d² ②                               (c is a merely a leg of right triangle ACG; d is the hypotenuse).

Here’s the “magic.” Substitute for c² in equation ②:

 

x² + y² + z² = d². ③          Very neat!  I  can solve for d without actually calculating either c or c².

As a bonus, I can do some trigonometry. The angle between the base of the box and the line AG is  and can be found using any of the inverse sine, cosine or tangent functions, for example  = tan^-1 (z÷c).

In conclusion, my curiosity was satisfied because I looked at the problem from a different perspective. If I actually make a box ABCDEFGH it is obvious that angle ACG is 90°.  When I drew the net, I lost one dimension and the 90° angle became a 180° angle!  Was it wrong to draw the net?  No - the purpose of the net was to show that line segment AC has two roles to play. It can be regarded as the hypotenuse of one triangle but is merely a leg of another triangle.  This dual role led to the substitution that produced equation ③, which extends the 2-D Pythagorean equation into 3-D.