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Plane Equation

Started by programering Apr 5, 2006 at 12:54 PM 37 replies 6.8k views

Original Post

Apr 5, 2006 at 12:54 PM

I have asked this before in this thread when I didn't understand. As I read it now I definitly think I would deserve to be rated down for that. I was pissing you off, but no article explained that well so I could understand it. Now I understod I had to try harder to understand. So I have checked it out at Wikipedia. But I think I got it now. It was said that the parameter B was pointing as the normal right up at the face. But in what directions are the other two parameters pointing then? Then I got the idea that the parameters are koeffient (like ratio) values that tells how much it differs from the world's xyz axis. Therefore my guess is: A * x + B * y + C * z + D = 0 Is the same as: A * x + B * y + C * z = -D x is the x axis delta from the plane's origin to the actual point y is the y axis delta from the plane's origin to the actual point z is the z axis delta from the plane's origin to the actual point A is the value for how the plane's x axis differs from the world's x axis. B is the value for how the plane's y axis differs from the world's y axis. C is the value for how the plane's z axis differs from the world's z axis. D is the length formed by sqrt(x*x + y*y + z*z) So I just need to be confirmed if this is true. Thanks in advance!

Mushu

Apr 5, 2006 at 1:10 PM

A, B and C are essentially a normal to the plane, < A, B, C >. Remember a normal is a vector - a distance, not a position. D is to establish position, and is the distance between the origin and the plane, along the normal vector.

programering

Apr 6, 2006 at 4:38 AM

Where's the origin and the point to check at the picture fig(3)?
Is the origin where x, y and z meets?
Is the point the black dot in the other end of D?

Is the magnitude of a vector the same as the length?

Is the vectors xyz component the delta (the difference) the displacement between two points.

To get the magnitude, do one write |AB|? where A and B is two points.

Do one get the length by sqrt(x*x + y*y + z*z).

Is it like this then:

A*x + B*x + C*z + D = 0

A,B,C the normal vector direction.
x,y,z the delta between the origin and the point to check.
D the distance between formed by sqrt(x*x + y*y + z*z)

If the point lies in the plane the expression(equation) then will be true, equal zero.

haegarr

Apr 6, 2006 at 5:14 AM

Your understanding of x, y, z, and D is not correct.

I normally explain the whole thing the following way, where a bold uppercase letter denotes a point, and a bold lowercase letter denotes a direction vector.

To understand it totally, you first have to know a consequence of the dot product:
a . b = |a| * |b| * cos( < a,b > )
Read this as follows: The dot product of two direction vectors is proportional to the length of each vector and the cosine of the angle between the both vectors. For simplicity, let us assume that a is of unit length, so it could be seen that the dot product will result in the length of the projection of b onto a, or, in other words, the length of b in parallel direction of a (there is also a perpendicular component, but that doesn't play a role here).

Now, let us define a distance formula like this
d := ( P - O ) . n
where P and O denotes two points, and n a direction vector.

With this, a plane equation in Hessian normal form is given by
P := ( P - O ) . n = 0
where O (from "origin", not a zero!) denotes any point on the plane, and n denotes the plane's normal.

What does the definition express? Look at the dot product inside. It says: The projected length of the difference vector of any point P in the space and the origin of the plane must be 0 for P being part of the plane. Any P set into the equation above that result in a distance not equal to zero is distant of the plane and hence not part of the plane.

Next, resolve the brace to yield in
P . n - O . n = 0
and remember that that P is variable but O and n are fixed parameters for a given plane. Hence the right dot product could be computed as a scalar
- O . n =: D
From the definition of the dot product this is the (negated) length of the projected difference vector between the space origin and the plane origin, or, in other words, the distance of the plane from the space origin! (Notice that negative values may occur since the direction of the normal plays a role; so D is not really a distance!) To be precise, the full form would be
- ( O - 0 ) . n =: D
where the subtracted 0 (a bold null) denotes the space origin, so that the point is transformed into a direction vector.

Furthurmore, if selecting the 3D space as the space of interest, the left dot product could be expressed by components as
P . n = P_x * n_x + P_y * n_y + P_z * n_z

Now choose some other names:
A := n_x
B := n_y
C := n_z
and
x := P_x
y := P_y
z := P_z
and after substitution you yield in the known formula
A x + B y + C z + D = 0

Hence, A, B, C are just the components of the plane's normal, and D is a distance measure of the plane from the space origin. x, y, and z are the components of any space point that is either part of the plane if they fulfil the equation, or else not.

Quote:
Original post by programering
A,B,C the normal vector direction.
x,y,z the delta between the origin and the point to check.
D the distance between formed by sqrt(x*x + y*y + z*z)

With looking at the explanation above, 1st line is true, 2nd and 3rd are false.

Quote:
Original post by programering
To get the magnitude, do one write |AB|? where A and B is two points.

Yes, but normally (at least in Germany) a horizontal bar has to be written over it (unfortunately I don't know how to make them in HTML). The two vertical bars denote the "absolute value" or "lenght" or "magnitude" (it is a notion of the so-called 2-norm). If A and B denote two points, then AB denote the difference vector from A to B, computed as
AB = B - A

EDIT: As Taralieth has pointed out absolute correctly, the dot product shows the cosine (not the sine) between the both vectors. Already corrected above. Thx.

[Edited by - haegarr on April 6, 2006 7:14:28 AM]

Taralieth

Apr 6, 2006 at 7:06 AM

Quote:
Original post by haegarr
Your understanding of x, y, z, and D is not correct.

I normally explain the whole thing the following way, where a bold uppercase letter denotes a point, and a bold lowercase letter denotes a direction vector.

To understand it totally, you first have to know a consequence of the dot product:
a . b = |a| * |b| * sin( < a,b > )
Read this as follows: The dot product of two direction vectors is proportional to the length of each vector and the sine of the angle between the both vectors. For simplicity, let us assume that a is of unit length, so it could be seen that the dot product will result in the length of the projection of b onto a, or, in other words, the length of b in parallel direction of a (there is also a perpendicular component, but that doesn't play a role here).

...

Small correction to Haegarr. The dot product is actually the length of each vector times cosine of the angle in between both vectors. Otherwise I agree with what Haegarr wrote.

Quote:
Original post by haegarr
With this, a plane equation in Hessian normal form is given by
P := ( P - O ) . n = 0
where O (from "origin", not a zero!) denotes any point on the plane, and n denotes the plane's normal.

What does the definition express? Look at the dot product inside. It says: The projected length of the difference vector of any point P in the space and the origin of the plane must be 0 for P being part of the plane. Any P set into the equation above that result in a distance not equal to zero is distant of the plane and hence not part of the plane.

Another way you can think about it is that if P is a point along the plane, then P - O gives you a vector along the plane since both P and O is on the plane. That means that the vector P - O must be perpendicular to the normal of the plane which means the dot product of them must be 0 (i.e. |P|*|O|*cos90 = 0)

TerrorFLOP

Apr 6, 2006 at 9:15 PM

programering

I wrote a Word document which explains planes in depth.

I would've posted it here in this thread, but it's a bit tricky trying to post a few pages of screen shots.

If you'd like to view it, send me a private message and I could email you the doc.

programering

Apr 21, 2006 at 2:47 AM

I have googled for plane equation
to see if I could find articles about it that
explain it well so I would get to understand it better.
Google Search: "plane equation"
And I choosed the article "Polygons and the Plane Equation" in the list,
and that explained well and got me to understand that right now.

The triangle with the vertices A, B and C.

1. First you get the normal of a surface by the
cross product of the two vectors U and V between the
vertices A-B and A-C pointing out from any
vertex in the triangle, in this case the vertex A.
That's A, B and C are the normal vector x, y and z components.
The normal have no position in space, it just specifies the direction.

2. Second you normalize the normal vector to give it a length
of 1 by dividing every A, B and C component by 1.

3. Third you get the D component from the dot product of the
normal and any vertex point of the triangle.

A*x + B*y + C*z + D = 0

A, B and C are the normal vector's x, y and z components.

D is the value of the dot product of the normal and any
vertex point in the triangle (must it be the same vertex
that was the center of the two vectors of between the
vertices A-B and A-C, in this case A?).

x, y and z are the point to check if it is in the plane
(Is it the delta from the center vertex A?).

If (A*x + B*y + C*z + D) > 0 Then the point is above the plane
If (A*x + B*y + C*z + D) = 0 Then the point is in the plane (coplanar)
If (A*x + B*y + C*z + D) < 0 Then the point is below the plane

Is it right now then?

haegarr

Apr 21, 2006 at 3:47 AM

Quote:
Original post by programering
1. First you get the normal of a surface by the
cross product of the two vectors U and V between the
vertices A-B and A-C pointing out from any
vertex in the triangle, in this case the vertex A.

Yes. The principles behind this are: A cross product of two non-colinear (!) and not zero length (!) vectors is a vector perpendicular to both of its input vectors. For the normal of a surface you have to pick the point of which you need to know the normal, compute 2 tangential vectors and the cross product from them. In your case the surface is planar, so that any 2 non-colinear vectors inside the surface are good as tangential vectors, and the normal will furthurmore be the same at all points on the surface.

Quote:
Original post by programering
That's A, B and C are the normal vector x, y and z components.

Yes. (However, notice that A, B, and C are used as positions in space in the image, so it is a little bit confusing here to use the same variable names (you may consider to use the bold kind of writing for vectors in the future)).

Quote:
Original post by programering
The normal have no position in space, it just specifies the direction.

Correct. A normal is a direction vector, and direction vectors have no position. This difference to position vectors (or "points") is important for transformations. That is why I differ also in writing a upper case letter A for a position vector and a lower case u for direction vectors.

Quote:
Original post by programering
2. Second you normalize the normal vector to give it a length
of 1 by dividing every A, B and C component by 1.

Err, no. "Normalization" is the operation to make a vector unit length, that's right. To do so, you compute its length and divide each component by this length:
n' := n / |n| = [ x/|n| y/|n| z/|n| ]
where
|n| := sqrt( x*x + y*y + z*z )
in 3D space.

Notice please that A,B,C are points in space. Points could not sensefully be normalized.

Notice furthur that it isn't sufficient to normalize u and v, since the normal is computed as the cross product of u and v, and there is a sine in-between:
|u x v| := |u| * |v| * |sin( <u,v>)|
So, until you guarantee that the angle between u and v isn't +/-90° (so that the sine becomes +/-1) you have to normalize _after_ the cross product.

You could furthur see from the sine why the both vectors must not be co-linear, or else the angle between will be 0° or 180°, and its sine will be 0 making the cross product disappear.

Quote:
Original post by programering
3. Third you get the D component from the dot product of the
normal and any vertex point of the triangle.

Closely. First, notice the negation, since the D is the negative dot-product of the normal and any point on the surface. Second, (as already mentioned) you could use _any_ point of the surface, so of course also one of the vertices. The dot-product does a projection, so it plays no role what point you use as long as it is on the surface.

Quote:
Original post by programering
A*x + B*y + C*z + D = 0

A, B and C are the normal vector's x, y and z components.

Yes. That fact comes from the definition of the dot-product.

Quote:
Original post by programering
D is the value of the dot product of the normal and any
vertex point in the triangle (must it be the same vertex
that was the center of the two vectors of between the
vertices A-B and A-C, in this case A?).

See above. Absolutely any point on the (planar) surface could be used.

Quote:
Original post by programering
x, y and z are the point to check if it is in the plane
(Is it the delta from the center vertex A?).

It is any point in space.

Quote:
Original post by programering
If (A*x + B*y + C*z + D) > 0 Then the point is above the plane
If (A*x + B*y + C*z + D) = 0 Then the point is in the plane (coplanar)
If (A*x + B*y + C*z + D) < 0 Then the point is below the plane

Yes, but this needs one more definition: Here "above" means "in direction of the normal", and "below" means "against the direction of the normal". If you negate the normal (what will denote the same surface) the meaning of "above" and "below" will just exchange.

Furthurmore I'm not sure about the term "coplanar" w.r.t. points. I think only planes could be co-planar.

Good job :) I think you mostly has understood the topic now.

programering

Apr 21, 2006 at 12:14 PM

Quote:
Original post by haegarr
Good job :) I think you mostly has understood the topic now.

Not quite, there's still things to clear out.

It is of very great value I get to understand this.
Cos I want to understand this.

Quote:
Original post by haegarr
Quote:
Original post by programering
1. First you get the normal of a surface by the
cross product of the two vectors U and V between the
vertices A-B and A-C pointing out from any
vertex in the triangle, in this case the vertex A.

Yes. The principles behind this are: A cross product of two non-colinear (!) and not zero length (!) vectors is a vector perpendicular to both of its input vectors. For the normal of a surface you have to pick the point of which you need to know the normal, compute 2 tangential vectors and the cross product from them. In your case the surface is planar, so that any 2 non-colinear vectors inside the surface are good as tangential vectors, and the normal will furthurmore be the same at all points on the surface.

What's colinear?
What does "co" stand for?
What's tangential? Is it touching?
What does the "surface is planar" mean?

Quote:
Original post by haegarr
Quote:
Original post by programering
That's A, B and C are the normal vector x, y and z components.

Yes. (However, notice that A, B, and C are used as positions in space in the image,
so it is a little bit confusing here to use the same variable names (you may consider to use the bold kind of writing for vectors in the future)).

Do you mean the triangle's vertices A, B and C?

Quote:
Original post by haegarr
Quote:
Original post by programering
The normal have no position in space, it just specifies the direction.

Correct. A normal is a direction vector, and direction vectors have no position. This difference to position vectors (or "points") is important for transformations. That is why I differ also in writing a upper case letter A for a position vector and a lower case u for direction vectors.

So both the (A, B and C) and (x, y and z) vectors are the normal.
A, B and C are the position vector in uppercace.
x, y and z are the direction vector in lowercase.

Quote:
Original post by haegarr
Quote:
Original post by programering
2. Second you normalize the normal vector to give it a length
of 1 by dividing every A, B and C component by 1.

Err, no. "Normalization" is the operation to make a vector unit length, that's right. To do so, you compute its length and divide each component by this length:
n' := n / |n| = [ x/|n| y/|n| z/|n| ]
where
|n| := sqrt( x*x + y*y + z*z )
in 3D space.

void Normalize(Vector *vect){// shall it be integers or real numbers as float and double?   int length = GetLength(vect);   vect->x /= length;   vect->y /= length;   vect->z /= length;}int GetLength(Vector *v){// shall it be integers or real numbers as float and double?   return sqrt(v->x*v->x + v->y*v->y + v->z*v->z);}

Quote:
Original post by haegarr
Notice please that A,B,C are points in space. Points could not sensefully be normalized.

So it is the normal's direction vector by the components x, y and z which get normalized.

Quote:
Original post by haegarr
Notice furthur that it isn't sufficient to normalize u and v, since the normal is computed as the cross product of u and v, and there is a sine in-between:
|u x v| := |u| * |v| * |sin( <u,v>)|
So, until you guarantee that the angle between u and v isn't +/-90° (so that the sine becomes +/-1) you have to normalize _after_ the cross product.

The length of the vector from the cross product of u and v shall equal the length of u and v * eachother * sin(the angle between u and v)

Quote:
Original post by haegarr
Quote:
Original post by programering
x, y and z are the point to check if it is in the plane
(Is it the delta from the center vertex A?).

It is any point in space.

Is it the point in space to be checked if it is in the plane?

Quote:
Original post by haegarr
Quote:
Original post by programering
If (A*x + B*y + C*z + D) > 0 Then the point is above the plane
If (A*x + B*y + C*z + D) = 0 Then the point is in the plane (coplanar)
If (A*x + B*y + C*z + D) < 0 Then the point is below the plane

Yes, but this needs one more definition: Here "above" means "in direction of the normal", and "below" means "against the direction of the normal". If you negate the normal (what will denote the same surface) the meaning of "above" and "below" will just exchange.

Quote:
Original post by haegarr
Furthurmore I'm not sure about the term "coplanar" w.r.t. points. I think only planes could be co-planar.

What does "v.r.t. points" means?

Thanks for helping me out.
It's a bit hard for me to understand, but I do my best.
And I have bad read understanding.
I would have rated you, but it won't have any effect
since I have much lower than you, otherwise I would.

When I finally get to understand things, I will write articles which explains it very well so people like me can understand.
But I seem to be the only one who don´t understand things here.

Illco

Apr 21, 2006 at 1:34 PM

Ok; I'll add my few cents below with some additions. I've tried to avoid specific mathematical definitions as you can find them -- a more natural-language oriented approach seemed more appropriate.

Quote:
What's colinear?

It means the vectors directions are the same, regardless of their length. So for example, two vectors (0,1,0) and (0,100,0) are colinear. That is not the whole story: (0,1,0) and (0,-1,0) are also colinear as their directions are on the same line, and you might abusingly think of the one as the other with negative length.

Quote:
What does "co" stand for?

It's a prefix to indicate the term colinear relates two a pair of vectors rather than a single one. Think in general of co-pilot, co-driver, co-operate, etc.

Quote:
What's tangential? Is it touching?

Yes it's touching. If you have a parabola then the tangent on the top will be horizontal, for example.

In this case, the explanation tries to tell the following. Normally, for an arbitrary curved surface, to compute the normal vector you would require two vectors that are tangents of the surface i.e. touching it's curvature. The tangents are orthogonal to each other as well as to the normal.

In this case, since the surface is planar, it is not required that the two vectors for computing the normal are orthogonal to each other; any two actually different (non-colinear) vectors in the plane will do.

Quote:
What does the "surface is planar" mean?

It means the surface is not curved, but flat. If you would have an infinitely flat plane, you could position such that it fits together with any planar surface. You cannot do this with non-planar surfaces, as the curvature will at some point move away from your infinite plane.

Actually a flat plane is a tautology: planes are always flat whereas surfaces can be curved.

Quote:
Shall it be integers or real numbers as float and double?

In general these computations are done using floating-point numbers, so stick to those for now.

Quote:
What does "w.r.t. points" means?

It means "with respect to points" i.e. how the term coplanar should be interpreted in the context of points.

FYI: coplanar points are points lying in the same plane. So any two points will be coplanar, as will be any three points but not any four points (in R^3, naturally).

Greetz,

Illco

programering

Apr 23, 2006 at 11:10 AM

Because you self know this so well you explain it in a higher knowing level that you understand.
But I need to get to know the basics first.

Quote:
Original post by Illco
Quote:
Original post by programering
What's colinear?

It means the vectors directions are the same, regardless of their length. So for example, two vectors (0,1,0) and (0,100,0) are colinear. That is not the whole story: (0,1,0) and (0,-1,0) are also colinear as their directions are on the same line, and you might abusingly think of the one as the other with negative length.

Good explaination, I get it, thanks!

Quote:
Original post by Illco
Quote:
Original post by programering
What does "co" stand for?

It's a prefix to indicate the term colinear relates two a pair of vectors rather than a single one. Think in general of co-pilot, co-driver, co-operate, etc.

Ok.

Quote:
Original post by Illco
Quote:
Original post by programering
What's tangential? Is it touching?

Yes it's touching. If you have a parabola then the tangent on the top will be horizontal, for example.

parabola? Is it the TV receiver of the satelite?

Quote:
Original post by Illco
In this case, the explanation tries to tell the following. Normally, for an arbitrary curved surface, to compute the normal vector you would require two vectors that are tangents of the surface i.e. touching it's curvature. The tangents are orthogonal to each other as well as to the normal.

What's arbitrary? I couldn't find it in the game dictionary.
So to get the normal of a curved surface you take the cross product from two tangential vectors that touches the surface in both ends.

Quote:
Original post by Illco
Quote:
Original post by programering
What does the "surface is planar" mean?

It means the surface is not curved, but flat. If you would have an infinitely flat plane, you could position such that it fits together with any planar surface. You cannot do this with non-planar surfaces, as the curvature will at some point move away from your infinite plane.

Ok.

Quote:
Original post by Illco
In this case, since the surface is planar, it is not required that the two vectors for computing the normal are orthogonal to each other; any two actually different (non-colinear) vectors in the plane will do.

Google Search: ortogonal vector

Quote:
Original post by Illco
Actually a flat plane is a tautology: planes are always flat whereas surfaces can be curved.

Ok.

Quote:
Original post by Illco
Quote:
Original post by programering
Shall it be integers or real numbers as float and double?

In general these computations are done using floating-point numbers, so stick to those for now.

Ok.

Quote:
Original post by Illco
Quote:
Original post by programering
What does "w.r.t. points" means?

It means "with respect to points" i.e. how the term coplanar should be interpreted in the context of points.

BTW, what does "i.e." stand for?

Quote:
Original post by Illco
FYI: coplanar points are points lying in the same plane. So any two points will be coplanar, as will be any three points but not any four points (in R^3, naturally).

I understand.
FYI ? For Your Information?

I think I get it now :-).

A*x + B*y + C*z + D = 0

A, B, and C are the point of the normal. Uppercase for point vectors.
x, y and z are the direction of the normal. Lowercase for direction vectors.
D is the dot product of the point and direction vectors which returns the
distance for the space origin to the plane.

If (A*x + B*y + C*z + D) < 0, then the point is?
If (A*x + B*y + C*z + D) = 0, then the point is in the plane.
If (A*x + B*y + C*z + D) > 0, then the point is?

nobodynews

Apr 23, 2006 at 4:14 PM

I'm not going to try and answer all the questions about planes because I'm not that solid on the math, but I'll help out by answering some of the definitions you're having problems with.

Quote:
parabola? Is it the TV receiver of the satelite?

A parabola is simply any curve fitting the equation: y = ax^2 + bx + c

Quote:
What's arbitrary? I couldn't find it in the game dictionary

Arbitrary is a general term meaning any one out an infinite number of possibilities. So an arbitrary curve surface would be any curved surface you can think of.

Quote:
BTW, what does "i.e." stand for?

i.e. is a latin abbreviation. It is short for id est, which means 'that is'. So when someone says i.e. they are basically just clarifying a statement they just made. Similarly, the abbreviation e.g. is short for 'exempli gratia' and means 'for example'. From wikipedia:

Quote:
Exempli gratia (e.g.) and id est (i.e.) are commonly confused and misused in colloquial English. The former, exempli gratia, means "for example", and is used before giving examples of something ("I have lots of favorite colors, e.g., blue, green, and hot pink"). The latter, id est, means "that is", and is used before clarifying the meaning of something, when elaborating, specifying, or explaining rather than when giving examples ("I have lots of favorite colors, i.e., I can't decide on just one"). Both "e.g." and "i.e." should generally be followed by a comma, just as "for example" and "that is" would be.

FYI does mean For Your Information.

Zakwayda

Apr 23, 2006 at 8:35 PM

I'm just sort of jumping in here without having read the whole thread, but...

Quote:
Original post by programering
A, B, and C are the point of the normal. Uppercase for point vectors.
x, y and z are the direction of the normal. Lowercase for direction vectors.
D is the dot product of the point and direction vectors which returns the
distance for the space origin to the plane.

The terminology you're using there seems a little confused (it may just be a language issue). The difference between points and vectors can be a little slippery, but in the plane equation, (A, B, C) is a vector, otherwise known as the plane normal, and (x, y, z) is any point. Those points (that is, values for x, y and z) for which the equality holds are on the plane.

Your definition of D is pretty close. In this particular form of the plane equation, it's actually the negative of the dot product of any point known to be on the plane, with the plane normal (the term normal is preferred to 'direction vector' here). If the normal is unit length, D is in fact the (signed) distance from the origin to the plane.

And:

Quote:
If (A*x + B*y + C*z + D) < 0, then the point is behind the plane
If (A*x + B*y + C*z + D) = 0, then the point is in the plane.
If (A*x + B*y + C*z + D) > 0, then the point is in front of the plane

Where 'behind' and 'in front' refer somewhat informally to the negative and positive half-space associated with the plane.

haegarr

Apr 24, 2006 at 2:57 AM

Quote:
Original post by jyk
I'm just sort of jumping in here without having read the whole thread, but...
Quote:
Original post by programering
...
The terminology you're using there seems a little confused (it may just be a language issue). The difference between points and vectors can be a little slippery, but in the plane equation, (A, B, C) is a vector, otherwise known as the plane normal, and (x, y, z) is any point.

Yes, and sorry for that little confusion. Perhaps I have introduced it because I have not mentioned explicitely that I write vectors in bold weight, hence position vectors are written as bold uppercase letters and direction vectors as bold lowercase letters in my replies above (as in any post I've done on GDnet). Scalars, on the other hand, are written in normal weight by me.

Boder

Apr 24, 2006 at 3:49 AM

This thread rocks.

So for the normal that defines the plane (N) and a point (p) with vector (p - 0 = P) and (d) is some scalar distance.

N . P = d

defines some plane.

And ( Nx, Ny, Nz, -d ) is the four-component plane equation.

haegarr

Apr 24, 2006 at 4:20 AM

Quote:
Original post by Boder
So for the normal that defines the plane (N) and a point (p) with vector (p - 0 = P) and (d) is some scalar distance.

N . P = d

defines some plane.

Yes. But the above use of uppercase and lowercase letters for vectors is reverse to my use in previous replies. In the notation of previous replies it would look like
n . p == d w/ p := P - 0
(read it as "vector n dot vector p is identical to scalar d with p being defined as the difference of point P and the origin point 0").

Notice please that both the normal and the distance is needed to define a plane, while the point is a variable. The definition means verbosely "a plane is build of all those points in the 3D space that fulfil the plane equation; any point not fulfilling the plane equation lies outside the plane, either in the "positive halfspace" (w.r.t. the direction of the planes normal) or in the "negative halfspace".

Quote:
Original post by Boder
And ( Nx, Ny, Nz, -d ) is the four-component plane equation.

The tuple defines the parameters of a kind of plane equation. The equation itself is
nx*px + ny*py + nz*pz - d == 0
with px, py, and pz being the variables of the equation.

programering

Apr 24, 2006 at 2:53 PM

I have read some articles here to better understand this.

Is this true?:

One normalize both the normal's
point vector and
the direction vector
to make them unit vectors with
the magnitude/length of 1.

The Cross Product of two vectors
returns a vector orthogonal/perpendicular/
(right angle)/(90 degrees) to both.

[Edited by - programering on April 24, 2006 3:53:40 PM]

Zakwayda

Apr 24, 2006 at 3:12 PM

Quote:
One normalize both the normal's
point vector and
the direction vector
to make them unit vectors with
the magnitude/length of 1.

Where did you get this exactly? Is it an excerpt from an article or tutorial?

programering

Apr 24, 2006 at 3:16 PM

Quote:
Original post by jyk
Quote:
One normalize both the normal's
point vector and
the direction vector
to make them unit vectors with
the magnitude/length of 1.
Where did you get this exactly? Is it an excerpt from an article or tutorial?

A Texture Mapping Technique
Why? Is it wrong?

haegarr

Apr 25, 2006 at 2:56 AM

The following answers may seem nitpicking, but look at all the related problem posts arised here in GDnet, and you will see the sense. I'm not a mathematician, so I may be unprecise but hopefully not wrong in the one or other formulation.

Quote:
Original post by programering
One normalize both the normal's
point vector and
the direction vector
to make them unit vectors with
the magnitude/length of 1.

The distinction between position vectors and direction vectors could be done in two ways: Either using homogeneous co-ordinates or by defining a suitable vector algebra (say rules like "a point minus a point results in a direction"). Using the homogeneous co-ordinates has IMHO the advantage to see some things directly.

If you apply the length formula to a position vector in homogeneous co-ordinates you get
sqrt( x² + y² + z² + 1² )
what actually is the length of the vector in a mathematical sense but not the value you are looking for. What you want to see instead is the "distance of the point from the origin"!

On the other hand, computing the same for a direction vector will yield in
sqrt( x² + y² + z² + 0² )
what is okay.

Computing the length of the difference vector of a point P := (x y z 1) and the origin 0 := (0 0 0 1)
|P - 0| = | ( x-0 y-0 z-0 1-1 ) | = sqrt( x² + y² + z² )
does the trick.

Compare this with the rule I gave as example for a vector algebra rule above: "a point minus a point results in a direction". See that exactly that is done here!!

Now, we could also do the normalization step. Let us define
d := |P| = sqrt( x² + y² + z² + 1 )
and apply that to yield in
P' := P / d = ( x/d y/d z/d 1/d )
The length of this vector is 1. Okay, now let us compute the affine vector from it:
( x y z )
Err, is that a vector of length 1? No, of course not.

In other words: A position vector is not suitable to compute a length from. If you say "I compute the length of a vector" you ever mean "of a direction vector". If you say "I compute the length of a position vector" you are wrong; what you wanted to say is "I compute the distance of a point from the origin" and you mean "I compute the length of the difference vector of the point from the origin".

Similarly, a normal is a direction vector and hence does not have an associated position. So the term "the normal's point vector" from the OP is also not correct. You could convert a direction to a point only by adding the direction to a point (notice that this is just an equivalence transformation of the one rule of the algebra I have given above!).

In summary, IMHO there are 2 mistakes in the cited sentence above.

Quote:
Original post by programering
The Cross Product of two vectors
returns a vector orthogonal/perpendicular/
(right angle)/(90 degrees) to both.

This is true in many cases but not in general. To be true both original vectors
(a) must not be of zero length, and
(b) must not be co-linear.
You have isolated the sentence from its original context, and perhaps the context have decribed a situation where that conditions were fulfilled.

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