SUNY Geneseo Department of Mathematics

3D Coordinate Systems

Wednesday, August 31 - Wednesday, August 3

Math 223 01
Fall 2022
Prof. Doug Baldwin

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Anything You Want to Talk About?

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3D Coordinates

Based on “Three-Dimensional Coordinate Systems,” “Distance in R3,” and “Writing Equations in R3,” from section 1.2.

Questions

Notation?

Our notation for points in 3 dimensions is pretty standard, namely a parenthesized list of 3 numbers, in x - y - z order: (x, y, z) e.g., (2, -4, 19).

For vectors (coming up in about a week) we’ll use either a similar component notation, or one based on standard vectors in the x, y, and z directions:

Vector written as delta X, delta Y, delta Z in brackets, or as delta X I hat plus delta Y J hat plus delta Z K hat

(There will be more on this in readings and classes next week, so don’t worry if it doesn’t make total sense yet. And if it does make sense now, that’s great, of course.)

Key Ideas

3 perpendicular axes meeting at an origin define a 3-dimensional coordinate system.

Distance in 3 dimensions can be found using a formula that extends the 2-dimensional formula.

Examples

Is my coordinate system right-handed or left-handed? What could you change to change its handedness?

It’s right-handed. But one could make it left handed by flipping any one axis to point in the opposite of its current direction, or by exchanging any pair of axes (e.g., make x point in the current y direction and vice versa).

Does it make sense to have a 4-dimensional coordinate system? If so, what can you say about what one would look like?

Yes, it makes sense.

It would consist of 4 mutually perpendicular axes meeting at a common origin.

Analogous to the quadrants of a 2D system and the octants in 3D, a 4 dimensional coordinate system would have 16 “hyper-octants” (I made up the term “hyper-octant,” I don’t think they have a standard name, although “hyper” is often used as a prefix when extending an idea from a low number of dimensions to the analogous one in many dimensions).

What is the distance from point (1,2,3) to (-2,4,6)?

Use the distance formula to work this out:

Using square root of change in X squared plus change in Y squared plus change in Z squared to find a distance

I conjecture that the set of points equidistant from (a,0,0) and (b,0,0) is the plane parallel to the yz plane and passing through x = (a+b)/2. Can you show that this is right (or that it’s not right)?

3 D coordinates with 2 points on X axis and plane perpendicular to that axis halfway between the points

Start with using the distance formula to talk about d1 and d2, and see where you can go from there. This is very much how math happens in the real world, i.e., you’re starting with an idea that seems plausible, and then doing a lot of exploratory work, just seeing where it goes that might eventually confirm (or refute) your idea.

D 1 equals D 2 eventually implies X minus A squared equals X minus B squared, so A equals B

Oops! Our exploration led us to a valid and meaningful result, namely that if the points (a,0,0) and (b,0,0) are actually the same point, then all of space is equidistant from it/them. But we were hoping for insight into a different situation, namely if (a,0,0) and (b,0,0) are distinct points, is the subset of space equidistant from both a certain plane? Think about whether you can back up what we did above and take it in a slightly different direction to get an answer to this question, and we’ll start Friday’s class with your questions or ideas.

Problem Set

Problem set 1, on 3D coordinates, is available.

See the handout for details.

Next

Finish the points equidistant from (a,0,0) and (b,0,0) problem.

Then an introduction to Mathematica, particularly its use to plot surfaces in 3 dimensional space.

No new reading.

But be sure Mathematica is installed on your laptop by Friday, and bring the laptop to class.

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