Calculating 3d plane for two 3d vectors

If I have two vector coordinates representing positions on the surface of the earth where the center of the earth is (0,0,0) and the Up vector is (0,0,1); What is the best way to calculate a 3d plane that is running along the two vectors (the 开发者_Go百科direction v2 - v1) but back by a set number of meters (just imagine a virtual clip plane that is behind the two vectors and running parallel to them).

well, you do the cross product of v1 and v2 to get the normal of your plane (don't forget to normalize if you want to), then the 4th element of your plane would just be 0 (because it crosses 0,0,0).

and then you want to project the plane in a certain direction based on the UP vector, not the plane's normal?

in that case I think you would just get the dot product of the normal and the up vector, then multiply the inverse of the dot (1/dot) by the number of units you want to project along the up vector and set that as your 4th element?

to clarify, that creates a plane where the two vectors, and the center of the earth are points on the plane, the plane can then be projected up or down by a certain amount in the UP direction.

Planes are usually described by a normal vector N, and as all points x,y,z fitting the equation Ax + By + Cz + D = 0. (A,B,C) is the normal vector. It doesn't even need to be normalized (unit length) if you choose D. Sounds like you want a plane to which v1 and v2 are parallel (and v2-v1 too). For that, make N perpendicular to v1 and v2 by setting it to the cross product of v2 x v1. Then pick a point in (x,y,z) coordinates that you know the plane should pass through. Plug N and (x,y,z) into the equation and compute D.

1- Find the Normal vector N = V1 X V2

2-Select a point that you want your plane tuch P0 ==>R0

3- All the other points(P==>R) in the plane follow N(R-R0)=0

See the link