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#1 2006-07-20 04:42:57

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

3-D angles from axes

Say you have a line that passes through the origin, and it is not parallel to any of the
3 axes x,y, or z.    You are given a point (X,Y,Z) that the line points through.
What is the name of the 2-D x-y graph that contains all z's crunched together, where true overrides false.
Not a plane intersecting the 3-D graph, but more of a shadow of the 3-D graph onto the x-y plane, so
essentially Z just goes away and the x,y points are still valid.
It would be like looking at the 3-D graph in many parallel lines with the z-axis.
In drafting, they must have a name for this "shadow" or perspective.

I would be interested in as many different mathematical names for this 3-D projection onto 2-D.

Thank you for your ideas smile


igloo myrtilles fourmis

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#2 2006-07-20 05:39:22

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: 3-D angles from axes

Also interested in what you call the actual angle between each axis and the line.
This angle would be in a plane that only has one axis in it, not two.

Perhaps the study of "vectors" would have some of these names??

Also interested in the projected angles onto the x-y plane, x-z plane, and y-z plane.

Any ideas are great even if they are made up.  Thanks. smile


igloo myrtilles fourmis

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#3 2006-07-20 06:39:35

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: 3-D angles from axes

You have two angles in 3d, the angle between x and y which is normally refered to as theta (just like in 2d) and the angle between y and z, normally called phi.

To draw lines in 3d, first, you need a way to do so.  That is, if you try to use any equation:

z = f(x, y)

Such as z = x^2 + y, you get a plane.  And that's true for any f(x, y) that is defined on the reals.

To get a line, we use parametric.  So for 2d, a parametric graph is:

(3t, 2t)

Where x = 3t and y = 2t.  When we plug in 0 for t, we get the point (0, 0).  When we plug in 1 for t, we get (3, 2).  And 2 would be (6, 4).  This is actually the line y = 3/2x.

So in 3d, we could have an equation for the line be:

(3t, 2t, 5t)

And so the 2d "shadow" of it would be:

(3t, 2t, 0) or rather just (3t, 2t).


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2006-07-20 06:40:50

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: 3-D angles from axes

Oh, and if you're looking for more information, the type of math this is called is "Vector geometry".


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-07-21 05:24:27

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: 3-D angles from axes

Thanks for the input, mister.  You are very helpful and kind.
Is phi pronounced like "find" without the "nd" ?
I was thinking of using 3D coordinates to try to plot out the vertices to some of SoapyJoe's figures.
Or maybe I'll try the center points of each side and see how far I get.


igloo myrtilles fourmis

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#6 2006-07-21 07:00:43

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: 3-D angles from axes

Is phi pronounced like "find" without the "nd" ?

Think of the line "Fee-fi-fo-fum, I smell the blood of an Englishman" from Jack and the Beanstalk.  Phi can either be pronounced like the "Fee" or the "Fi" in that.  I like "Fee".


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#7 2006-07-21 22:03:26

ben
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Registered: 2006-07-12
Posts: 106

Re: 3-D angles from axes

John E. Franklin wrote:

Not a plane intersecting the 3-D graph, but more of a shadow of the 3-D graph onto the x-y plane, so
essentially Z just goes away and the x,y points are still valid.
It would be like looking at the 3-D graph in many parallel lines with the z-axis.

Your intuition was right, the construction you are referring to is included in the study of vectors.

But lookee. You want to look at the projection of a 3-vector onto 2-space (that's just me being fancy with your own words). Again, you guessed right - the image of a point, or set of points, or vector, whatever from a higher to a lower dimension is called a projection.

Lookee further. Having projected your 3-vector onto the XY plane, what do you do with this "shadow"? Well personally I would evaluate it in terms of the X and Y axes. So effectively you are asking - what is the projection of your 3-vector along each of the X, Y and Z axes. This is an interesting construction that goes by the name of inner (or scalar, or dot) product.

Here's something to think about. Suppose you have a vector that casts no shadow on either the X or Y axes? What would you say about that. Here's hint: on a bright sunny day, under what circumstances can you not see your own shadow?

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#8 2006-07-22 04:17:49

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: 3-D angles from axes

Do you mean "casts no shadow on the x-y plane" ?
Wouldn't a line parallel to the z-axis still cast one tiny dot on the x-y plane whereever the (x,y) of (x,y,z) is?
Maybe you mean something else?
Inner product, dot product, scalar product...
I'll read about it, thanks!


igloo myrtilles fourmis

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#9 2006-07-22 04:36:55

ben
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Registered: 2006-07-12
Posts: 106

Re: 3-D angles from axes

John E. Franklin wrote:

Wouldn't a line parallel to the z-axis still cast one tiny dot on the x-y plane whereever the (x,y) of (x,y,z) is?

Yup! Good work. We call that "tiny dot" the zero projection of z on both X and Y. So you see that if the projection of z on both x and y is zero, then z is perpendicular to both.

Now, as we like to invent words to make ourselves look clever, we say that z is orthogonal to x and y (actually I'm kidding. If I told you the real reason we don't use "perpendicular",  it would scare you witless!)

Wanna see it? I mentioned the "dot product" (I hate that term). A lot of texts use z·x to mean the projection of z along x (hence the dot in dot product). And if z·x = 0, as you rightly guessed, z and x are mutually perpendicular or, harrumph, orthogonal.

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#10 2006-07-25 10:18:18

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: 3-D angles from axes

Thanks for the info, ben.
I tried out the scalar product in 2-D with unit vectors and the inverse cosine computes the angle just
as if I used inverse tangents and subtracted the 2 angles.  Love the formula I found.
cos theta = X1X2 + Y1Y2, where X2^2 + Y2^2 = 1


igloo myrtilles fourmis

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#11 2006-07-25 15:55:52

All_Is_Number
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Registered: 2006-07-10
Posts: 258

Re: 3-D angles from axes

ben wrote:

If I told you the real reason we don't use "perpendicular",  it would scare you witless!

Do tell. I'm not that easily frightened.  cool


You can shear a sheep many times but skin him only once.

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#12 2006-07-26 01:46:09

ben
Member
Registered: 2006-07-12
Posts: 106

Re: 3-D angles from axes

All_Is_Number wrote:
ben wrote:

If I told you the real reason we don't use "perpendicular",  it would scare you witless!

Do tell. I'm not that easily frightened.  cool

Hmm. To do it justice would require a detour into the theory of abstract linear vector spaces. Let me just say this.

In a nice, homely 3-dimensional vector space we will say that the inner product x·y can be defined as |x| |y| cos θ. We say that x and y are orthogonal when x·y = 0 which implies that θ = 90°. And as John E. correctly said, this simply means that x and y have zero projection along each other.

But our most familiar vector spaces are infinite-dimensional, which makes the issue of mutual perpendicularity problematical. So we turn the deinition of orthogonal round, and say two vectors are orthogonal if they have zero mutual projection, and x·y = 0 defines this condition.

Note that this need not imply they are mutually perpendicular. To see this consider two vectors which are not perpendicular. Suppose x has some projection along y. Then x·y ≠ 0. But if we subtract from y the projection of x, we find that x·y = 0, but we haven't rotated any vector, so they are orthogonal but not perpendicular!

I hope that is clear.

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