CodePudding user response:
13.If N and O M collision, have the Ox - Nx/ag=Oy - Ny/Vy=Oz - Nz/Vz, Ox - Mx/Wx=Oy - My Wy=Oz - Mz/Wz.
Solvable Oy=Ny + (Mx - Nx) VyWy/(VxWy - VyWx). Make sure Oy, insist on VxWy!=VyWx, also can solve other similar relations,
Can be used as reference
CodePudding user response:
Question 13:M N collided need to meet (1) two trajectory coplanar, and (2) time on the intersection of identity,
The positive solution to this problem:
(1) ag/Wx=Vy Wy=Vz/Wz is set up, set up parallel, never intersect
(2) if (1) was not judge whether different planes, judge condition is that the linear N (Mx, My, Mz) and point of the plane of the normal vector and linear M vertical or not? If vertical two straight lines coplanar intersection, otherwise, in different planes are never intersect in different planes, calculated as follows:
N (1) and (Mx, My, Mz) plane normal vector: a vector (Nx - Mx, Ny - My, Nz - Mz) cross-product vector (Vx, Vy, Vz), the calculation results for (Qx, Qy, Qz), (Qx, Qy, Qz) dot (Vx, Vy, Vz)=QxVx + QyVy + QzVz=? 0, if is not equal to zero, then face two straight lines, never intersect,
(3) if (2) the result equals 0, then two straight lines coplanar intersection,
Set a time T, two straight lines intersect:
Nx=WxT VxT + + Mx
VyT + My + Ny=WyT
VzT + Nz=WzT + Mz
Need to meet: (Nx - Mx)/(Wx - ag)=(Ny - My)/(Wy - Vy)=(Nz - Mz)/(Wz - Vz), the method is very stupid, but well understood