Problem solved : from Three.js / WebGL - transparent planes hiding other planes behind them thanks @Alex Under jsfiddle code var material = new THREE.MeshBasicMaterial({ transparent: true, side: THREE.DoubleSide, fog: false, color: 0x00FF00, opacity: 0.2, depthWrite: false, depthTest: false }); ...

data-structures,geometry,plane

It is possible, at the cost of recalculating the distance to the origin every time you need it.

math,3d,mathematical-optimization,plane

Your code looks good. You can shorten it somewhat to this: void position_from_plane(float r_co[3], const float p[4]) { const float d = -p[3] / (p[0]*p[0] + p[1]*p[1] + p[2]*p[2]); r_co[0] = p[0]*d; r_co[1] = p[1]*d; r_co[2] = p[2]*d; } You could get slightly shorter code if you were to intersect...

If you have your target point P with coordinates r_P = (x,y,z) and a plane with normal n=(nx,ny,nz) you need to define an origin on the plane, as well as two orthogonal directions for x and y. For example if you origin is at r_O = (ox, oy, oz) and...

java,vector,3d,plane,dot-product

Your average method changes the value of a, to make it the same as the average point. So your cube isn't a cube, after you've called average - three of the faces have rotated into new positions. So whatever happens in the loop over collider is wrong.

opengl,textures,glsl,shadow,plane

I would go for the following idea: for each pixel in the range of your light yo will want to do a trace to the light origin and test if there is no black pixel on the path. It will look more or less like that: bool inTheShadow = false;...

Ranges of x and y in which to display planes: X = linspace(-100,100,10); Y = linspace(-100,100,10); [ XX, YY ] = meshgrid(X,Y); Calculating z coordinate for planes: ZZ1 = 1 - XX - YY; ZZ2 = 4/3 - 2/3 * XX + 1/3 *YY; Displaying: figure; hold on; mesh(XX,YY,ZZ1); mesh(XX,YY,ZZ2);...

I assume that the ray is perpendicular to the plane, otherwise your question does not make much sense. If this is the case, then V = +/- N. The points P which lay in the plane all satisfy the equation: < P, N> = <Q0, N> Where <,> denotes the...