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.

x

^{2}+ y^{2}= c^{2}① (*c*is the hypotenuse of right triangle ABC).c

^{2}+ z^{2}= d^{2}② (*c*is a merely a leg of right triangle ACG;*d*is the hypotenuse).Here’s the “magic.” Substitute for c

^{2}in equation ②:x

^{2}+ y^{2}+ z^{2}= d^{2}. ③ Very neat! I can solve for*d*without actually calculating either*c*o*r**c*^{2}.^{}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.
## No comments:

## Post a Comment