line and plane intersection

the same as in the above example, can be calculated applying simpler method. You can edit the visual size of a plane, but it is still only cosmetic. This free Autolisp program calculates and draws a point at the intersection of a line and a plane. Take a look at the graph below. This note will illustrate the algorithm for finding the intersection of a line and a plane using two possible formulations for a plane. One should first test for the most frequent case of a unique intersect point, namely that , since this excludes all the other cases. Fortunately, after all that doom and gloom, you can use 3D coordinates for specifying points in TikZ. As it is fundamentally a 2D-package, it doesn't know how to compute the intersection of the line and plane and so doesn't know when to stop drawing the line. in a point which can be determined by solving the four simultaneous equations. Finally, if the line intersects the plane in a single point, determine this point of intersection. u.z : -u.z); // test if the two planes are parallel if ((ax+ay+az) < SMALL_NUM) { // Pn1 and Pn2 are near parallel // test if disjoint or coincide Vector v = Pn2.V0 - Pn1.V0; if (dot(Pn1.n, v) == 0) // Pn2.V0 lies in Pn1 return 1; // Pn1 and Pn2 coincide else return 0; // Pn1 and Pn2 are disjoint } // Pn1 and Pn2 intersect in a line // first determine max abs coordinate of cross product int maxc; // max coordinate if (ax > ay) { if (ax > az) maxc = 1; else maxc = 3; } else { if (ay > az) maxc = 2; else maxc = 3; } // next, to get a point on the intersect line // zero the max coord, and solve for the other two Point iP; // intersect point float d1, d2; // the constants in the 2 plane equations d1 = -dot(Pn1.n, Pn1.V0); // note: could be pre-stored with plane d2 = -dot(Pn2.n, Pn2.V0); // ditto switch (maxc) { // select max coordinate case 1: // intersect with x=0 iP.x = 0; iP.y = (d2*Pn1.n.z - d1*Pn2.n.z) / u.x; iP.z = (d1*Pn2.n.y - d2*Pn1.n.y) / u.x; break; case 2: // intersect with y=0 iP.x = (d1*Pn2.n.z - d2*Pn1.n.z) / u.y; iP.y = 0; iP.z = (d2*Pn1.n.x - d1*Pn2.n.x) / u.y; break; case 3: // intersect with z=0 iP.x = (d2*Pn1.n.y - d1*Pn2.n.y) / u.z; iP.y = (d1*Pn2.n.x - d2*Pn1.n.x) / u.z; iP.z = 0; } L->P0 = iP; L->P1 = iP + u; return 2;}//===================================================================, James Foley, Andries van Dam, Steven Feiner & John Hughes, "Clipping Lines" in Computer Graphics (3rd Edition) (2013), Joseph O'Rourke, "Search and Intersection" in Computational Geometry in C (2nd Edition) (1998), © Copyright 2012 Dan Sunday, 2001 softSurfer, For computing intersections of lines and segments in 2D and 3D, it is best to use the parametric equation representation for lines. So you have to tell it. Then, coordinates of the point of intersection (x, y, 0) must satisfy equations of the given planes. Let this point be the intersection of the intersection line and the xy coordinate plane. The plane determined by the points , , and and the line passing through the points and intersect in a point which can be determined by solving the four simultaneous equations Here's the question. The line is contained in the plane, i.e., all points of the line are in its intersection with the plane. Surface to choose for the Stokes' theorem for intersection of sphere and plane. P = 0 where n3 = n1 x n2 and d3 = 0 (meaning it passes through the origin). Here are cartoon sketches of each part of this problem. Do a line and a plane always intersect? 0 : t0; // clip to min 0 t1 = t1>1? u.x : -u.x); float ay = (u.y >= 0 ? Explore anything with the first computational knowledge engine. 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