you are right about the p = d * n.

but if you only make boxes, why not find the planes from the corners of the box?

also, wouldn''t you have to do 8 intersections?

((+/- 1, +/- 1, +/- 1), 1)
...
...

Well you should consider your pb is simply a multiplication of a vector by the inverse of a 3x3 matrix. In fact you have been reinventing the wheel ! LOL

Let your 3 planes be A, B and C and I their intersection. I belongs to every plane thus :

A.n*I = -A.d
B.n*I = -B.d
C.n*I = -C.d

( means that the projected component of I following A.n is the opposite distance of the plane A to the origin )

Now if you still cant see the matrix involved here :

I.x I.y I.z
-------------------------
| A.n.x | A.n.y | A.n.z | | A.d
| B.n.x | B.n.y | B.n.z | = | B.d = D
| C.n.x | C.n.y | C.n.z | | C.d
-------------------------

M*I = D => I = (1/M)*D if Det(M)!=0 ( intersection could be a line, empty, or even a plane if A=B=C )

Thus you probably have a math lib. Use it U dont need much more code.

Then inverting a 3x3 matrix M seen as 3 row vectors { a=A.n, b=B.n, c=C.n } is simple (note the permutations in the formula ) :

{ b^c, c^a, a^b }/Det(M) and Det(M)=(a^b)*c

Thus :
I.x = (b^c) * D / Det(M)
I.y = (c^a) * D / Det(M)
I.z = (a^b) * D / Det(M)

PS: The inversion formula would have worked for column vectors but it s less convenient as A.n, B.n, C.n are rows. That''s not weird at all because : transpose(1/M) = 1/transpose(M)

OK ???

Charles B

Sorry I supposed you knew standard Math notations. But if U have pbs to solve this issue it''s obvious that you dont To me beeing fluent in maths notations is really important to be quicker and see the inner logics and solve small pbs like that. U should read some books about notations in linear algebra. Hmm do you know operator overloading in C++ ? Some 3D libs use it but it can be confusing in code. I dont like it too much.

x=DotProduct(A,B); <=> x=A*B
A=CrossProduct(B,C); <=> A=B^C

D is a 3D vector (shown as a column matrix) with components :
D{ A.d, B.d, C.d }
It is not a scalar ( not real number, not a float if u prefer ) !
Else dont be confused just because I kept A.d, B.d, C.d. I could haved renamed these components D.x, D.y, D.z but they are not 3D "world" space coordinates. It would have been confusing !

a,b,c are the 3 normals. I have defined them ! They just rename A.n, B.n, C.n, the plane normals, because I am a lazzy writter.

Well the mysterious part now ! LOL

I say that I consider the matrices as 3 rows of 3D vectors. It''s a quick way to explain that M is :

a.x a.y a.z
b.x b.y b.z
c.x c.y c.z

Considering the matrix as columns (which is the common way) would have been stupid here. I would have required new vectors {a.x,b.x,c.x}, ...,{a.z,b.z,c.z}. Instead the rows are well known : a,b,c, the plane normals.

Then let''s call N the inverse of M :

N = 1/M

I give the following formula for matrix inversion :

N = { b^c, c^a, a^b }/Det(M)

OOOPS Shame on me ! ( Supposed to be a brute at maths ! ) My memory failed at this point ! I forgot to say that there is a transposition here ! It means that N is now a matrix of column vectors !!!

Det(M) is a scalar. Multiplying a matrix by a scalar is totally equivalent to uniform scaling in fact. If you prefer N is :

| (b^c).x/Det(M) (c^a).y/Det(M) (a^b).z/Det(M) |
| (b^c).x/Det(M) (c^a).y/Det(M) (a^b).z/Det(M) |
| (b^c).x/Det(M) (c^a).y/Det(M) (a^b).z/Det(M) |

Hmm I am not too lazy this time ! LOL

I said that our starting problem was :

(1) D = M*I

which is a much more compact way of saying :

A.d = A.n*I = a*I = a.x*I.x + a.y*I.y + a.z*I.z
B.d = B.n*I = b*I = b.x*I.x + ...
C.d = C.n*I = c*I = c.x*I.x + ...

( BTW I let a mistake ! Forget the minus signs of the 1st 3 lines and read what LilBudyWizer says, U have to choose which A.d you consider. Personnally I prefer A.d to be the distance origin to plane A in the direction of A.n : O----A.d----->(A)--A.n---> here A.d > 0 )

Then solving your problem, I mean finding I is simply inverting (1) That''s all :

(1) => I = (1/M)*D = N*D, if M can be inversed (Det(M)!=0)

Aren''t maths simplier with powerful notations ?
... As long as u remember the inversion formula by heart or you find the function in your maths packages.

I = (1/Det(M))*( (b^c)*A.d + (c^a)*B.d + (a^b)*C.d )

Which more detailed is :

I.x = ((b^c).x/Det(M))*A.d + ((c^a).y/Det(M))*B.d + ...
I.y = ((b^c).y/Det(M))*A.d
I.z = ((b^c).z/Det(M))*A.d

Last thing if your 3 planes are orthogonal ( your cube example ) then Det(M) = 1. This leads to your result.

But I think the more general result for any 3 planes is more interesting to create a nice function. Call it something like :

int Intersect3Planes( Plane &A, Plane &B, Plane &C, Point &I );

Just add a special test if( fabs(Det-1.0f) < SMALL ) to detect the cube case and avoid the division : 1/Det(M). Computing Det requires only one more dot product ( 3 muls 2 adds ) because (a^b) is required elsewhere.

If Det=0 then you should return that there is no defined point intersection. If you are perfectionnist you could return the degree of the intersection: 3 for a plane (A=B=C), 2 for a line, 1 for a point, 0 for nothing.

Charles B

Another thing your plane equations dont need to be normalized if u divide by Det(M).

For instance :

{ {2,0,0}, 2 } : 2*x=2

is the same plane as

{ {1,0,0}, 1 } : x=1

but your function would fail here and mine would still work

Charles B