Straight lines are called crossed if they are not crossed and are not parallel. This concept of spatial geometry. The objective is solved by methods of analytical geometry by finding of distance between straight lines. At the same time length of a perpendicular, mutual for two straight lines, is calculated.
Instruction
1. Starting the solution of this task, it is necessary to make sure that the straight lines which are really crossed. For this purpose use the following data. Two straight lines in space can be parallel (then they can be placed in one plane), crossed (lie in one plane) and crossed (do not lie in one plane).
2. Let straight lines of L1 and L2 be set by the parametrical equations (see fig. 1a). Here τ – parameter in the system of the equations of a straight line L2. If straight lines are crossed, then they have one point of intersection which coordinates are reached in the systems of the equations of figure 1a at certain values of parameters t and τ. Thus, if the system of the equations (see fig. 1b) rather unknown t and τ has the decision, and only, then straight lines of L1 and L2 are crossed. If this system has no decision, then straight lines are crossed or parallel. Then for decision-making compare the directing vectors of straight lines of s1= {m1, n1, p1 } and s2= {m2, n2, p2 } If straight lines the crossed, then these vectors not collinear and their coordinates {m1, n1, p1 } and {m2, n2, p2 } cannot be proportional.
3. After check start the solution of a task. Its illustration – figure 2. It is required to find d distance between skew lines. Place straight lines in the parallel planes β and α. Then the required distance is equal to length of the general perpendicular to these planes. N normal to the planes β and α has the direction of this perpendicular. Take on each straight line on a point of M1 and M2. The distance of d is equal to an absolute value of a projection of a vector of M2M1 to direction N. For the directing vectors of straight lines of L1 and L2 at the same time it is fair that s1 | | β, and s2 | | α. Therefore you look for a vector of N as the vector work [s1, s2]. Now remember rules of finding of the vector work and calculation of length of a projection in a coordinate form and you can start the solution of specific objectives. At the same time hold to the following plan.
4. The statement of the problem begins a task of the equations of straight lines. As a rule, it is the initial equations (if is not present – lead them to a canonical form). L1: (x-x1) / m1=(y-y1)/n1=(z-z1)/p1; L2: (x-x2) / m2=(y-y2)/n2=(z-z2)/p2. Take M1 (x1, y1, z1), M2 (x2, y2, z2) and find M2M1= vector {x1-x2, y1-y2, z1-z2 }. Write down s1= vectors {m1, n1, p1 }, s2= {m2, n2, p2 }. Find a normal of N as the vector work s1 and s2, N= [s1, s2]. Having received N= {A, B, C }, required distance of d find as an absolute value of a projection of a vector of M2M1 to the N.d= direction | Pr (N) M2M1= (A(x1-x2) + B(y1-y2) + C (z1-z2)) / √ (A^2+B^2+C^2).