vector math/cross product
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I really don't want to ask this question here, but since there's no math forum I haven't much of a choice. Anyway, I have a math question for the gurus again. I'm trying to understand just how a cross product works rather than just do as I'm told kinda thing. So, given this...
| x1 | | x2 | | y1z2 - z1y2 |
| y1 | x | y2 | = | z1x2 - x1z2 |
| z1 | | z2 | | x1y2 - y1x2 |My question is, why is that so? I realize (using the dot product as a reference) that the elements are independent or so I thought. I would've thought that multiplying the two would mean somethign of this nature...
x1 * x2 + y1 * y2, etc.
But I realize that would just be the dot product again. Also, why do I need to subtract at all when multiplying? Can anyone please explain this to me? The book I'm reading did a great job at explaining the dot product, but not the cross product. TIA Jeremy Falcon
Hey Jeremy, Hopefully I can shed some light on your question. Here's a brief tutorial in vector manipulation -- you should be able to find some more detailed discussions elsewhere. Vectors can be written as 3 coordinates -- x, y, and z coordinates. For example, a vector starting at the origin (0,0,0) going to the point (2,3,4) would be writen as <2,3,4> = 2ax + 3ay + 4az ax, ay and az are unit vectors (have a length of 1) and point in the x-, y- and z-directions, respectively. That is, they are defined to be ax = <1,0,0> ay = <0,1,0> az = <0,0,1> Here I'm using the notation that (x,y,z) is a point (in parentheses) and is a vector that has an x-component of x, y-component of y, and z-component of z. Looking at the example above, and using the fact that a constant (scalar) multipled by a vector would behave as you'd expect: 2ax = 2<1,0,0> = <2,0,0> Also, adding vectors behaves as you'd expect, just add up each of the 3 components individually: <1,2,3> + <4,5,6> = <5,7,9> You can see that 2ax + 3ay + 4az = 2<1,0,0> + 3<0,1,0> + 4<0,0,1> = <2,0,0> + <0,3,0> + <0,0,4> = <2,3,4> In linear algebra lingo, the 3 vectos ax, ay, and az for a "basis" for the 3D space you are describing. Any 3 vectors can be used as a basis, provided that 1) their magnitude is 1 and 2) they are not parallel. That's a bit advanced, and we usually just choose the obvious ones of <1,0,0>, <0,1,0>, and <0,0,1> as above. It makes the math easier since each component represents one single direction, which is good. Now that we've gotten that out of the way, let's take a look at multiplication with vectors. You already seem to have a good understanding of the dot product. For the dot product, you take 2 vectors, take the dot product, and get a scalar result (just a single number). On the other hand, when you take a cross product, you take two vectors, take the cross product, and get a vector result. The physical interpretation of this is that you want to find a vector result that is perpendicular to the first 2 vectors. Think of it this way: If you take <1,0,0> x <0,1,0> = ? Basically this is ax x ay. What vector is perpendicular to both the x-direction and the y-direction? az. Try it out with your formula above. Why does this work? We need to look at how the cross p
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Jeremy Falcon wrote:
for the direction I'm headed in (OGL programming)?
Actually Jeremy, you need to answer most of this question for yourself. I believe you will when you are ready. My point here is: Ok, you are going to visually display something. What is to be displayed and why is it important or entertaining to bother displaying what ever it is. When you answer that/those questions, you will start saying, now if I new this, I can answer that. Just one word of advise, do not start using a left handed coordinate system, just because you got tired one night. "Yes I know the voices are not real. But they have some pretty good ideas."
Michael A. Barnhart wrote:
Actually Jeremy, you need to answer most of this question for yourself.
Well, the original idea was making games. But now I see this could open up some other doors for me as well.
Michael A. Barnhart wrote:
Just one word of advise, do not start using a left handed coordinate system, just because you got tired one night.
:laugh: That brings up a question though. Out of curosity, does OGL allow you change to left-handed? I thought it was right-handed, but I'm not a guru. The reason I ask is I'm used to left-handed thanks to apps like 3DS Max and would kinda like to stay that way. Thanks for the reply. Jeremy Falcon
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Jeremy, 1/ I'm on a work trip in the far east at the mo., so don't expect me to reply quickly. 2/ I wrote an article demonstrating using cross products etc to transform points from one coord system to another. Which is handy for an awful lot of reasons. Warping Coordinates with Matrices[^] Debugging through that should help. You can even use my matrix library if you wish. Its not templated etc, but its half decent! Iain.
Iain Clarke wrote:
1/ I'm on a work trip in the far east at the mo., so don't expect me to reply quickly.
:laugh: Understood.
Iain Clarke wrote:
2/ I wrote an article demonstrating using cross products etc to transform points from one coord system to another. Which is handy for an awful lot of reasons.
Oh cool thanks. I guess I should've searched CP first. Jeremy Falcon
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Was simply passing linear algebra. Sorry I am no help too l8 on friday to do math (only trig this late) "Until the day of his death, no man can be sure of his courage" -- Jean Anouilh
:-D It's the thought that counts man. Jeremy Falcon
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This[^] may help.OR it just may really confuse you. cheers, Chris Maunder
CodeProject.com : C++ MVP
It's about the same caliber as the text I was reading, but it did one better by showing a picture. I reckon them things are worth a 1,000 words. Thanks for the link and reply. Jeremy Falcon
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Hey Jeremy, Hopefully I can shed some light on your question. Here's a brief tutorial in vector manipulation -- you should be able to find some more detailed discussions elsewhere. Vectors can be written as 3 coordinates -- x, y, and z coordinates. For example, a vector starting at the origin (0,0,0) going to the point (2,3,4) would be writen as <2,3,4> = 2ax + 3ay + 4az ax, ay and az are unit vectors (have a length of 1) and point in the x-, y- and z-directions, respectively. That is, they are defined to be ax = <1,0,0> ay = <0,1,0> az = <0,0,1> Here I'm using the notation that (x,y,z) is a point (in parentheses) and is a vector that has an x-component of x, y-component of y, and z-component of z. Looking at the example above, and using the fact that a constant (scalar) multipled by a vector would behave as you'd expect: 2ax = 2<1,0,0> = <2,0,0> Also, adding vectors behaves as you'd expect, just add up each of the 3 components individually: <1,2,3> + <4,5,6> = <5,7,9> You can see that 2ax + 3ay + 4az = 2<1,0,0> + 3<0,1,0> + 4<0,0,1> = <2,0,0> + <0,3,0> + <0,0,4> = <2,3,4> In linear algebra lingo, the 3 vectos ax, ay, and az for a "basis" for the 3D space you are describing. Any 3 vectors can be used as a basis, provided that 1) their magnitude is 1 and 2) they are not parallel. That's a bit advanced, and we usually just choose the obvious ones of <1,0,0>, <0,1,0>, and <0,0,1> as above. It makes the math easier since each component represents one single direction, which is good. Now that we've gotten that out of the way, let's take a look at multiplication with vectors. You already seem to have a good understanding of the dot product. For the dot product, you take 2 vectors, take the dot product, and get a scalar result (just a single number). On the other hand, when you take a cross product, you take two vectors, take the cross product, and get a vector result. The physical interpretation of this is that you want to find a vector result that is perpendicular to the first 2 vectors. Think of it this way: If you take <1,0,0> x <0,1,0> = ? Basically this is ax x ay. What vector is perpendicular to both the x-direction and the y-direction? az. Try it out with your formula above. Why does this work? We need to look at how the cross p
Awesome reply! Thanks a lot. Jeremy Falcon
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Michael A. Barnhart wrote:
Actually Jeremy, you need to answer most of this question for yourself.
Well, the original idea was making games. But now I see this could open up some other doors for me as well.
Michael A. Barnhart wrote:
Just one word of advise, do not start using a left handed coordinate system, just because you got tired one night.
:laugh: That brings up a question though. Out of curosity, does OGL allow you change to left-handed? I thought it was right-handed, but I'm not a guru. The reason I ask is I'm used to left-handed thanks to apps like 3DS Max and would kinda like to stay that way. Thanks for the reply. Jeremy Falcon
Jeremy Falcon wrote:
does OGL allow you change to left-handed?
For the raw display no, however how you build relative coordinate systems to place objects, yes you can do both. But it is a mind trip to do so and not violate physics somewhere, so the best advise is to not try (even by accident.)
Jeremy Falcon wrote:
I'm used to left-handed thanks to apps like 3DS Max
I am not familuar with 3DS so I could be mistaken but I would suspect it to be right handed. What throws people and some text improperly state being left handed, when in reality you a travesing a path from the aboslute reference plane to the viewing coordiante system. This is really not left handed but performing an inverse on the tranformation which locates the viewing position to the absolute reference. Think of a stage with props placed on it and the camera dolly moving around. Yes a nit but I feel a more accurate statement of what is physically happening. The fact you can do this I think is very cool. Well just my 2 cents. Have fun. "Yes I know the voices are not real. But they have some pretty good ideas."
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Jeremy Falcon wrote:
does OGL allow you change to left-handed?
For the raw display no, however how you build relative coordinate systems to place objects, yes you can do both. But it is a mind trip to do so and not violate physics somewhere, so the best advise is to not try (even by accident.)
Jeremy Falcon wrote:
I'm used to left-handed thanks to apps like 3DS Max
I am not familuar with 3DS so I could be mistaken but I would suspect it to be right handed. What throws people and some text improperly state being left handed, when in reality you a travesing a path from the aboslute reference plane to the viewing coordiante system. This is really not left handed but performing an inverse on the tranformation which locates the viewing position to the absolute reference. Think of a stage with props placed on it and the camera dolly moving around. Yes a nit but I feel a more accurate statement of what is physically happening. The fact you can do this I think is very cool. Well just my 2 cents. Have fun. "Yes I know the voices are not real. But they have some pretty good ideas."
Michael A. Barnhart wrote:
For the raw display no, however how you build relative coordinate systems to place objects
I was afraid of that. I suppose it wouldn't be too much overhead, but I'm anal about speed. :laugh:
Michael A. Barnhart wrote:
I am not familuar with 3DS so I could be mistaken but I would suspect it to be right handed.
In 3DSMax +X is pointing to the right. And from what I've been told that's left-handed because it matches the direction the thumb is pointed on your left hand. Of course, there could be a way to swap Max to use right-handed for all I know. This is just the default.
Michael A. Barnhart wrote:
Well just my 2 cents. Have fun.
Thanks for the info! Jeremy Falcon
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Jeremy Falcon wrote:
does OGL allow you change to left-handed?
For the raw display no, however how you build relative coordinate systems to place objects, yes you can do both. But it is a mind trip to do so and not violate physics somewhere, so the best advise is to not try (even by accident.)
Jeremy Falcon wrote:
I'm used to left-handed thanks to apps like 3DS Max
I am not familuar with 3DS so I could be mistaken but I would suspect it to be right handed. What throws people and some text improperly state being left handed, when in reality you a travesing a path from the aboslute reference plane to the viewing coordiante system. This is really not left handed but performing an inverse on the tranformation which locates the viewing position to the absolute reference. Think of a stage with props placed on it and the camera dolly moving around. Yes a nit but I feel a more accurate statement of what is physically happening. The fact you can do this I think is very cool. Well just my 2 cents. Have fun. "Yes I know the voices are not real. But they have some pretty good ideas."
Out of curiosty, do you prefer right or left handed? I mean is there are reason why (like most people tend to use one). I heard DX was left-handed (don't know for sure though), so my guess would be not everyone in the industry goes the same way. Anyway just curious as you seem to have more experience in this than I do. Jeremy Falcon
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Out of curiosty, do you prefer right or left handed? I mean is there are reason why (like most people tend to use one). I heard DX was left-handed (don't know for sure though), so my guess would be not everyone in the industry goes the same way. Anyway just curious as you seem to have more experience in this than I do. Jeremy Falcon
Jeremy Falcon wrote:
Out of curiosty, do you prefer right or left handed?
I prefer right handed. In my Finite Element background, right hand is universally used. "Yes I know the voices are not real. But they have some pretty good ideas."
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Michael A. Barnhart wrote:
For the raw display no, however how you build relative coordinate systems to place objects
I was afraid of that. I suppose it wouldn't be too much overhead, but I'm anal about speed. :laugh:
Michael A. Barnhart wrote:
I am not familuar with 3DS so I could be mistaken but I would suspect it to be right handed.
In 3DSMax +X is pointing to the right. And from what I've been told that's left-handed because it matches the direction the thumb is pointed on your left hand. Of course, there could be a way to swap Max to use right-handed for all I know. This is just the default.
Michael A. Barnhart wrote:
Well just my 2 cents. Have fun.
Thanks for the info! Jeremy Falcon
Jeremy Falcon wrote:
In 3DSMax +X is pointing to the right.
Ok, and I assume +Y is up. So it is right hand if +Z is towards you (Screen to viewer) and left if +Z is awar from you. In my work (well past life) the viewer being at a positive distance from the object is common and is right hand. To late in the day to spell very well. "Yes I know the voices are not real. But they have some pretty good ideas." -- modified at 21:20 Saturday 1st July, 2006
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You're thinking too linearly. Start with a 2D cross product and draw both vectors and the resulting cross product vector on paper. You should have an appropriate "ah ha" moment. Marc Pensieve Some people believe what the bible says. Literally. At least [with Wikipedia] you have the chance to correct the wiki -- Jörgen Sigvardsson
a 2D cross product? there is no such thing.
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a 2D cross product? there is no such thing.
Arjang Assadi wrote:
a 2D cross product? there is no such thing.
See? What do I know? Yet my suggestion to get out the graph paper helped. :) Marc Pensieve Some people believe what the bible says. Literally. At least [with Wikipedia] you have the chance to correct the wiki -- Jörgen Sigvardsson
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Jeremy Falcon wrote:
In 3DSMax +X is pointing to the right.
Ok, and I assume +Y is up. So it is right hand if +Z is towards you (Screen to viewer) and left if +Z is awar from you. In my work (well past life) the viewer being at a positive distance from the object is common and is right hand. To late in the day to spell very well. "Yes I know the voices are not real. But they have some pretty good ideas." -- modified at 21:20 Saturday 1st July, 2006
It's left handed then. Now I just gotta decide to stick with one. Jeremy Falcon
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Jeremy Falcon wrote:
Out of curiosty, do you prefer right or left handed?
I prefer right handed. In my Finite Element background, right hand is universally used. "Yes I know the voices are not real. But they have some pretty good ideas."
Thanks for the info. Jeremy Falcon
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Jeremy Falcon wrote:
In 3DSMax +X is pointing to the right.
Ok, and I assume +Y is up. So it is right hand if +Z is towards you (Screen to viewer) and left if +Z is awar from you. In my work (well past life) the viewer being at a positive distance from the object is common and is right hand. To late in the day to spell very well. "Yes I know the voices are not real. But they have some pretty good ideas." -- modified at 21:20 Saturday 1st July, 2006
Nevermind, I decided to actually open up Max and test and it's right handed. :laugh: If Max is right and so is OGL then I think I have a winner! Jeremy Falcon
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Nevermind, I decided to actually open up Max and test and it's right handed. :laugh: If Max is right and so is OGL then I think I have a winner! Jeremy Falcon
Jeremy Falcon wrote:
I have a winner!
Have fun. "Yes I know the voices are not real. But they have some pretty good ideas."
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I really don't want to ask this question here, but since there's no math forum I haven't much of a choice. Anyway, I have a math question for the gurus again. I'm trying to understand just how a cross product works rather than just do as I'm told kinda thing. So, given this...
| x1 | | x2 | | y1z2 - z1y2 |
| y1 | x | y2 | = | z1x2 - x1z2 |
| z1 | | z2 | | x1y2 - y1x2 |My question is, why is that so? I realize (using the dot product as a reference) that the elements are independent or so I thought. I would've thought that multiplying the two would mean somethign of this nature...
x1 * x2 + y1 * y2, etc.
But I realize that would just be the dot product again. Also, why do I need to subtract at all when multiplying? Can anyone please explain this to me? The book I'm reading did a great job at explaining the dot product, but not the cross product. TIA Jeremy Falcon
The cross product of two vectors A and B is a third vector (A x B) that is perpendicular to BOTH A and B. The magnitude of (A x B) is equal to the product of the magnitudes of the first two vectors and the sine of the angle between A and B: |A x B| = |A| |B| sin(/_ A,B) The perpendicularity condition determines the (two-headed) line in the direction of (A x B). The magnitude condition narrows down (A x B) to one of two vectors. The precise choice of (A x B) follows the right hand rule: 1.) Point your RIGHT index finger along A. 2.) Point your RIGHT middle finger along B. 3.) Lift your RIGHT thumb perpendicular to the plane of your index and middle fingers. It will then point along (A x B). (If you use your LEFT hand, you will get -(A x B) NOT (A x B). Whether to use a RIGHT hand rule or a LEFT hand rule is essentially just an arbitrary choice, but it needs to be used consistently. The right hand rule for moving currents and magnetic field in electricity basically come from the right hand rule for cross products.) The origin of the cross product actually comes from looking at tensor products: If A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k then AB = A1*B1*i*i + A1*B2*i*j + ... + A3*B2*k*j + A3*B3*k*k where i, j, and k are unit vectors in the X, Y, and Z directions, respectively. (The tensor product is usually represented as a square matrix.) The dot product is the trace (the sum of the ii part, the jj part, and the kk part) and the cross product is the antisymmetric part of the tensor product. The i component of (A x B) is the jk component of the tensor minus the kj component, etc. In matrix terms, look at AB - BA: The diagonal terms are all zero, the off-diagonal terms are the components of (A x B). As to when you would use it? It depends on how often you need to find a line perpendicular to two other lines. John Morgan Center for Health Statistics Arkansas Department of Health and Human Services
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The cross product of two vectors A and B is a third vector (A x B) that is perpendicular to BOTH A and B. The magnitude of (A x B) is equal to the product of the magnitudes of the first two vectors and the sine of the angle between A and B: |A x B| = |A| |B| sin(/_ A,B) The perpendicularity condition determines the (two-headed) line in the direction of (A x B). The magnitude condition narrows down (A x B) to one of two vectors. The precise choice of (A x B) follows the right hand rule: 1.) Point your RIGHT index finger along A. 2.) Point your RIGHT middle finger along B. 3.) Lift your RIGHT thumb perpendicular to the plane of your index and middle fingers. It will then point along (A x B). (If you use your LEFT hand, you will get -(A x B) NOT (A x B). Whether to use a RIGHT hand rule or a LEFT hand rule is essentially just an arbitrary choice, but it needs to be used consistently. The right hand rule for moving currents and magnetic field in electricity basically come from the right hand rule for cross products.) The origin of the cross product actually comes from looking at tensor products: If A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k then AB = A1*B1*i*i + A1*B2*i*j + ... + A3*B2*k*j + A3*B3*k*k where i, j, and k are unit vectors in the X, Y, and Z directions, respectively. (The tensor product is usually represented as a square matrix.) The dot product is the trace (the sum of the ii part, the jj part, and the kk part) and the cross product is the antisymmetric part of the tensor product. The i component of (A x B) is the jk component of the tensor minus the kj component, etc. In matrix terms, look at AB - BA: The diagonal terms are all zero, the off-diagonal terms are the components of (A x B). As to when you would use it? It depends on how often you need to find a line perpendicular to two other lines. John Morgan Center for Health Statistics Arkansas Department of Health and Human Services
Thanks for the reply. Jeremy Falcon
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Arjang Assadi wrote:
a 2D cross product? there is no such thing.
See? What do I know? Yet my suggestion to get out the graph paper helped. :) Marc Pensieve Some people believe what the bible says. Literally. At least [with Wikipedia] you have the chance to correct the wiki -- Jörgen Sigvardsson
Damn, that was meant to be a joke. I was expecting an answer like if the vectors are on a mobius strip then each is on a 1 dimensional surface and crossing them give a 2D vector! Mark, your suggestion was good and that shows you know about what matters. Regards and Cheers :-D Arjang