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Instructions

1

Let the line be given by the intersection of two planes (see Fig.), for which their general equations are given.A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0. The desired line belongs to both of these planes. Accordingly, we can conclude that all of its points can be found from the solution of the system of these two equations.

2

Let, for example, the planes be giventhe following expressions: 4x-3y4z + 2 = 0 and 3x-y-2z-1 = 0. You can solve this problem in any convenient way for you. Let z = 0, then these equations can be rewritten in the form: 4x-3y = -2 and 3x-y = 1.

3

Accordingly, "y" can be expressed as follows: y = 3x-1. Thus, there will be expressions: 4x-9x + 3 = -2; 5x = 5; x = 1; y = 3-1 = 2. The first point of the straight line is M1 (1, 2, 0).

4

Now suppose that z = 1. From the initial equations we get: 1. 4x-3y-1 + 2 = 0 and 3x-y-2-1 = 0 or 4x-3y = -1 and 3x-y = 3. 2. y = 3x-3, then the first expression will have the form 4x-9x + 9 = -1, 5x = 10, x = 2, y = 6-3 = 3. Proceeding from this, the second point has the coordinates M2 (2, 3, 1).

5

If we draw a straight line through M1 and M2, then the problem will be solved. Nevertheless, it is possible to give a more visual way of finding the position of the desired direct equation - the compilation of the canonical equation.

6

It has the form (x-x0) / m = (y-y0) / n = (z-z0) / p, here{m, n, p} = s are the coordinates of the directing vector of the line. Since two points of the required line are found in the example considered, its directing vector s = M2M2 = {2-1, 3-2, 1-0} = {1, 1, 1}. For M0 (x0, y0, z0) one can take any of the points (M1 or M2). Let this be M1 (1, 2, 0), then the canonical equations of the line of intersection of two planes takes the form: (x-1) = (y-2) = z.