How about a 3D-embedded triangle?

[STxAxTIC]

I'll make this pretty simple.

Consider a 3D space that embeds a triangle having vertices:

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(1,1,0)
(1,0,1)
(0,1,1)
Now, consider these "random" single points:

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(.66,.66,.66)
(.46,.66,.86)
(.83,.89,.26)
(.55,.55,.55)
(.75,.79,.46)
(.79,.89,.66)
Question: Which two of the points given are not in the plane of the triangle, contained within its sides?

Hints and/or Comments:
The rest of the points do live in the plane of the triangle.
I generated all valid points with the same equation, and then simply invented two nonsense ones.
The triangle given is extremely simple but don't get too comfortable with this one. Your thought needs to work for any triangle oriented in any way in 3D.

Bah, that's enough.

Oh and if your name is MasterGy, I assume you've got this. No need to flex but you are 1000% welcome.
Same with Luke and a few others who can't be bothered with such vector trivialities.

Oh and I'll attach an image of a triangle with a line touching it, and then you can imagine that line could have stopped anywhere.

[Qwerkey]

Is this not simply rotating axes * so that the triangle lies in, say, the xy plane and then seeing which points lie inside the triangle (as already covered)?  Or is it somehow more complicated than that?
* A standard mathematical 3D manipulation for those with the skill?

[STxAxTIC]

Is this not simply rotating axes * so that the triangle lies in, say, the xy plane and then seeing which points lie inside the triangle (as already covered)

That's a completely legitimate interpretation of this problem. In "real life", you would center your view of the triangle to see what occurs inside its boundaries... but then you'd need a side view to make sure the line actually touches the triangle. Ideally you can do that in one nice calculation...

There may be other ways using CLEAR, SLEEP, PRINT, LINE, etc., Maybe even _MEM ;-)

[STxAxTIC]

Ooookay, buzzer has sounded, pencils down, pass in your work and seeya Monday.

Here is the answer. Of the points provided, sum across each row using the calculator of your choice. You will find:

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sum(<.66,.66,.66>) = 1.98
sum(<.46,.66,.86>) = 1.98
sum(<.83,.89,.26>) = 1.98
sum(<.55,.55,.55>) = 1.65
sum(<.75,.79,.46>) = 2.00
sum(<.79,.89,.66>) = 2.34

Four of these numbers are very close to 2.0, but two of them are way off. Evidently, there is something about the equation

x + y + z = 2

... I'll stop here and divert to the PDF linked below. I'll attach a picture below for the problem setup you should be picturing.

The full solution is at the bottom on this link:

http://barnes.x10host.com/homedata/Vectors%20and%20Trigonometry.pdf

... just look for the same picture. Also - talking to you @bplus - if you scroll upward in that document you'll see most of the math problems we've hit over the last year or so.

[Qwerkey]

My first reaction is: a linear addition of values from three orthogonal axes has no meaning at all (it's always square/add/square root), so what on earth is he up to now?  I'll have to study this after class (tomorrow)!

[bplus]

Also - talking to you @bplus

As a matter of fact, bplus IS working on this problem independently (of vectors of course) but is still trying to check/settle an issue with 2D math concerning the "fix"ing of angles.

It might be an interesting alternative to vectors, it might fail in 3D but does it even work in 2D, should know in about 5 minutes (that's programmer's time not real time ;-))

BTW screw this:

Ooookay, buzzer has sounded, pencils down, pass in your work and seeya Monday.

It's a cool problem to work on. :)

[STxAxTIC]

Lol don't take me seriously, I thought people gave up.

[Qwerkey]

Qwerkey, you do talk rubbish: x + y+ z = constant is just the equation of a (plane) surface as STxAxTIC was using.
Last year, the American Institute of Mathematical Sciences invited me to give a lecture tour entitled "Qwerkey Talks Nonsense".  They've invited me back this year with the talks entitled "Qwerkey Talks More Nonsense".  And I'm working on next year's  "Qwerkey Talks Even More Nonsense".  Why do they let me anywhere near this site?

[bplus]

Code: QB64: [Select]

1. _TITLE "STx How about a 3D embedded triangle?"
2.  
3. ' I am translating Ashish idea to 3D
4.  
5. DEFDBL A-Z
6. TYPE xyz
7.     x AS DOUBLE
8.     y AS DOUBLE
9.     z AS DOUBLE
10. END TYPE
11.  
12. 'given 3d tri
13. '(1,1,0)
14. '(1,0,1)
15. '(0,1,1)
16. DIM tri(2) AS xyz
17. tri(0).x = 1: tri(0).y = 1: tri(0).z = 0
18. tri(1).x = 1: tri(1).y = 0: tri(1).z = 1
19. tri(2).x = 0: tri(2).y = 1: tri(2).z = 1
20.  
21. 'given test cases   6 points in 3d tri or not? 2 are made up the remaider are in which is which?
22. '(.66,.66,.66)
23. '(.46,.66,.86)
24. '(.83,.89,.26)
25. '(.55,.55,.55)
26. '(.75,.79,.46)
27. '(.79,.89,.66)
28. DIM test(5) AS xyz
29. test(0).x = .66: test(0).y = .66: test(0).z = .66 'on/in
30. test(1).x = .46: test(1).y = .66: test(1).z = .86
31. test(2).x = .83: test(2).y = .89: test(2).z = .26
32. test(3).x = .55: test(3).y = .55: test(3).z = .55 '
33. test(4).x = .75: test(4).y = .79: test(4).z = .46
34. test(5).x = .79: test(5).y = .89: test(5).z = .66 '
35.  
36. triArea = HeroArea(tri(0), tri(1), tri(2))
37. PRINT "Main Triangle Area "; triArea
38. FOR i = 0 TO 5
39.     tArea1 = HeroArea(tri(0), tri(1), test(i))
40.     'PRINT tArea1,
41.     tArea2 = HeroArea(tri(1), tri(2), test(i))
42.     'PRINT tArea2,
43.     tArea3 = HeroArea(tri(2), tri(0), test(i))
44.     'PRINT tArea3,
45.     sum = tArea1 + tArea2 + tArea3
46.     PRINT "Sum of sub tri"; sum; "  differance to Main"; sum - triArea
47.     IF sum <= triArea + .001 THEN
48.         PRINT "Test point"; i; " is on/in the triangle."
49.     ELSE
50.         PRINT "Test point"; i; " is NOT on/in the triangle."
51.     END IF
52.     PRINT
53. NEXT
54. SLEEP
55.  
56. FUNCTION HeroArea (p1 AS xyz, p2 AS xyz, p3 AS xyz)
57.     a = dist3D(p1, p2)
58.     b = dist3D(p2, p3)
59.     c = dist3D(p3, p1)
60.     s = (a + b + c) / 2
61.     HeroArea = SQR(s * (s - a) * (s - b) * (s - c))
62. END FUNCTION
63.  
64. FUNCTION dist3D (p1 AS xyz, p2 AS xyz)
65.     dist3D = SQR((p1.x - p2.x) ^ 2 + (p1.y - p2.y) ^ 2 + (p1.z - p2.z) ^ 2)
66. END FUNCTION
67.  

This should work with any Triangle in 3D.

[STxAxTIC]

Nice, seems legit! I banged it around a little and it seems solid. Nice work, very!

[bplus]

Thanks, I had a whole 'nother angle in mind literally, but then I looked up 3D angles, uh-oh, dot product, LOL.

I was wondering if Ashish would get something up first, it's the same idea applied to 3D.

[Ashish]

@bplus
I had used the *exact* the same idea you have used for solving this problem. @STxAxTIC know this as I discussed this with him at Discord. However, when I run
my program, it doesn't print any point which are NOT in the triangle.

Oh I see, you have added .001 in your condition. ;)
Nice!

[bplus]

@bplus
I had used the *exact* the same idea you have used for solving this problem. @STxAxTIC know this as I discussed this with him at Discord. However, when I run
my program, it doesn't print any point which are NOT in the triangle.

Oh I see, you have added .001 in your condition. ;)
Nice!

Yes, rounding errors makes everything a little more than it is (when squaring stuff allot to have abs values), but @Ashish you encountered that in your clever 2D solution.

Yesterday I spent afternoon reworking my method for 2D solution from points Interior to an angle thinking it might be applicable to 3D as Ashish sol'n based on Triangle Areas. I had a hell of time with blunders or something, prolonged senior moment? Finally I resorted to OPTION _EXPLICIT and had the whole matter cleared up in 5 minutes literally it seems. Then I looked up calculating 3D angles and get this:
  [ You are not allowed to view this attachment ]  

So dang, back to vector math so what is  A . B ?
https://www.mathsisfun.com/algebra/vectors-dot-product.html
It is ax * bx + ay * by + az * bz, not so bad

OK, so what is |A| ?  well it just SQR( A . A )

IF COS(theta) = that mess on right THEN ARC-COS(the mess on right) = the dang angle between the two arms A and B.

OK, so now I am going to try my Interior Angle approach as I finally got it worked to my satisfaction in 2D.
Just for some practice with what I call doorway problems that open up new vistas or new realm of skills.

I learned SIN and COS in math classes but I didn't really learn it until I wanted to draw a hexagon on the computer, that was a doorway problem and I am thinking Angles in 3D might also be one.

So what new blunders await us today :-))

[bplus]

Blunder

OK the Interior of Angles in 3D are cones.

Update: wait... no, the 2 rays of an angle do form a plane or a boring slide that no water-park would ever purchase ;-))

[TempodiBasic]

Hi
it is very impressive how coding can be useful in math! I never have imagined this strong connection between math and programming.
I like math but after a while my brain stops to think if still I am on an abstract formula.... so I have little resource to  use for complex math solutions...nevertheless my brain works better and stay longer active if there is a geometric question. I like so much Logical geometry!
when I read this kind of thread, the only thing that I can do is to follow your discussions and your solutions...with admiration.