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 Posted: Sun, 6th Oct 2013 10:21 Post subject: Help with geometry problem |
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Got a test tomorrow and so im doing old test problems. Got one I just cant seem to understand, dont know where to begin:
We have four known points, A = (;1 ;2 ;1) B = (3 ;3 ;3) C = (2; 4; 5) and D = (7; 4; 4)
We have a new point E which is located on the line such that the distance between AC and CE are equal. I need to find the coordinates for point E.
Anyone wanna shed some light on this?
Dont mess with God, he can impregnate your girlfriend/wife without taking his pants off!
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Frant
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| Posted: Sun, 6th Oct 2013 12:18 Post subject: |
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If I knew what kind of system (2D? 3D?) you're using and what ; means I might help since I don't recognize that way of describing coordinates.
If it's a 3D-matrix with x.y.z coordinates it should be a breeze, but I haven't read any higher math in school, I think math from a programmers perspective, ie. Euclidean space.
C = (2; 4; 5)
Is that in X Y Z format? Because I see three different kinds of values.. ;N N; and N
Doesn't make sense in a Euclidean space.
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
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| Posted: Sun, 6th Oct 2013 12:23 Post subject: |
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Its the x, y and z coordinates in the room (space). sorry my course is in swedish so im not familiar with the exact expressions used in english...
So in the room, space we have a line and we need to find the coordinates for point E which is somewhere on the line.
Dont mess with God, he can impregnate your girlfriend/wife without taking his pants off!
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Frant
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| Posted: Sun, 6th Oct 2013 12:27 Post subject: |
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Ok, vad står ;N N; och N för?
X can be + (to the right of the center point in the 3D-matrix) or - (to the left of that point).
Same with Y and Z, ie. relative to an imaginary 0,0,0.
If you shift C to become 0,0,0 (ie move the matrix so your relative space centers on C it should make it easier. Then you use the X, Y and Z values and add/subtract to all the other values to get their relative positions along the line. Then you get A's relative distance and can put E on the other side of the "C"enter. When all that is done, "transpose" the values back, ie. reverse the add/subtractions and you have E.
If I understand the problem correctly.
That's the long way of doing things.
The short way of doing things.. Get the relative values between A and C, then shift the symbols (minus to plus and plus to minus) and you should have a point E on the other side of C that is equally as far away from C as A is from C. The distance between A & E = 2(AC).
I guess (without actually knowing the school math) that it becomes something like:
C - A <- C = E ? (where <- means inverting the symbols of each XYZ value of the result)
It could also be as simple as A = E but I doubt that's what the question is after.
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
Last edited by Frant on Sun, 6th Oct 2013 12:35; edited 3 times in total
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| Posted: Sun, 6th Oct 2013 12:59 Post subject: |
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The problem in swedish:
5. Punkterna A = (1; 2; 1 ) B = (3; 3; 3) C = (2 ; 4; 5 ) och D = (7; 4; 4)
är givna i ett ONH-system.
a) Beräkna AB × AC
b) Beräkna volymen av tetraedern ABCD
c) Ange på parameterfri form ekvationen för den räta linje L som går genom A och B
d) Punkten E ligger på linjen L och är belägen så att sträckorna AC och CE är
lika långa. Bestäm koordinaterna för punkten E.
Its the d) problem im not sure on.
The answer is E = (19/3; 14/3; 19/3)
Dont mess with God, he can impregnate your girlfriend/wife without taking his pants off!
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| Posted: Sun, 6th Oct 2013 13:02 Post subject: |
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jag testade att addera vektorerna OA + AC + CE = OE där OA är vektorn från origo till punkten A, men fick inte rätt svar...
Dont mess with God, he can impregnate your girlfriend/wife without taking his pants off!
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| Posted: Sun, 6th Oct 2013 13:29 Post subject: |
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you will now be known as Dr Jackson.
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Frant
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| Posted: Sun, 6th Oct 2013 14:20 Post subject: |
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Okay, then it's a different system (ONH) you're using, hence I'm of no help since I think "simple Euclidean quantized matrix" with decimal/hexadecimal X, Y and Z values. My longer version was basically by "shifting" origo to C. But obviously your problem is entirely focused on the FORM of calculating it and using the correct form of formulas and systems to reach the right answer, not just get the E-coordinates. To me origo = 0,0,0
I'm familiar with Cartesian and Euclidean space coordinates. I have no idea what ONH stands for but apparently you need to calculate areas, angles and stuff to get the answer.
Good luck and break a leg. There are enough of people here that KNOW math that can help you.
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
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| Posted: Sun, 6th Oct 2013 14:22 Post subject: |
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thnx for the effort Frant!
Dont mess with God, he can impregnate your girlfriend/wife without taking his pants off!
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| Posted: Sun, 6th Oct 2013 20:17 Post subject: |
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a class mate helped me with the problem. thnx anyways guys
Dont mess with God, he can impregnate your girlfriend/wife without taking his pants off!
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Frant
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| Posted: Mon, 7th Oct 2013 12:43 Post subject: |
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| Acer wrote: | | a class mate helped me with the problem. thnx anyways guys |
Can you run it down for me? I'm interested in learning how you did it.
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
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| Posted: Tue, 8th Oct 2013 08:31 Post subject: |
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Well I might as well write how I solved it. Bold = vector, *=dot product, || = lenght. This is obviously euclidean 3d space. Working in the plane spanned by AB and AC we can easily draw a picture of the setup. http://tinypic.com/r/13za8eg/5
Now the answer can be read off as
E =A + 2[ ( B - A )*( C - A ) ] ( B - A )/|B - A|^2
There is no formulas besides the dot product which essentially is a projection operator. The only trick is to work in a smart plane.
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Frant
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| Posted: Tue, 8th Oct 2013 09:21 Post subject: |
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C - A = 1,2,4
C + 1,2,4 = E = 3,6,9
That's the way I'd do it and it's simple and obvious.
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
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| Posted: Tue, 8th Oct 2013 09:28 Post subject: |
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But that is trivially not on the line through A and B.
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Frant
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| Posted: Tue, 8th Oct 2013 10:24 Post subject: |
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It should be a point on the line that is equally far from C as A is.
But obviously that solution is only based on the original premise (as far as I can understand the weird form of coordinate-notation).
If it's a straight line in a euclidean space it's a sound and simple solution according to the basic premise, find the coordinates for E that is equally far from C as the distance between A and C.
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
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| Posted: Tue, 8th Oct 2013 10:29 Post subject: |
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Read the problem in swedish then. It is not any line, it is the line L through A and B. C is not a point on the line.
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Frant
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Location: Your Mom
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