The solution of a task of search of a corner between the parties of some geometrical figure should be begun with the answer to a question: with what figure you deal, that is to be defined a polyhedron before you or a polygon. In stereometry "the flat case" is considered (polygon). Each polygon can be broken into a certain quantity of triangles. Respectively, the solution of this task can be consolidated to search of a corner between the parties of one of triangles, components the figure set to you.
Instruction
1. For a task of each of the parties it is necessary to know its length and one more certain parameter which will set the provision of a triangle on the plane. For this purpose, as a rule, the directed pieces - vectors are used. It should be noted that on the plane many equal vectors can be infinite. The main thing that they had equal length, more precisely the |module and also the direction which is set by an inclination to any axis (in the Cartesian coordinates it is axis 0X). Therefore for convenience vectors it is accepted to set the radius vectors of r= and which beginning is located in a point of the beginning of coordinates with the help.
2. For the solution of the question posed, it is necessary to define a scalar product of vectors and and b (it is designated (a, b)). If a corner between vectors f, then, by definition, the scalar product of two winds is the number equal to the work of modules: (a, b) = |a| |b| cos f (see ris1). In the Cartesian coordinates, if and = {x1, y1 } and b= {x2, y2 }, then (a, b) = h1u2 + h2y1. At the same time scalar square of a vector (and, a)= |a| ^2=x1^2 + x2^2. For b vector – similarly. So, |b| cos f = h1u2 + h2y1. Therefore, cos f = (h1u2 + h2y1) / |b|). This formula is an algorithm of the solution of an objective in "a flat case".
3. Primer1. To find a corner between the parties of a triangle set by vectors of a= {3, 5} and b = {-1, 4}. Proceeding from the theoretical calculations given above it is possible to calculate the required corner. cos f = (x1y2 +x2y1) / |b|)= (-3+20) / (9+25) ^1/2 (1+16) ^1/2=18/6 (17) ^1/2=6/sqrt (17) =1,4552otvet: ф = (1.4552).
4. Now it is necessary to consider a case of a volume figure (polyhedron). In this way of solving the task the corner to honey is perceived by the parties as a corner between edges of a side side of a figure. However, strictly speaking, the basis also is a polyhedron side. Then the solution of an objective comes down to consideration of the first of "a flat case". But vectors will be set already by three coordinates. Often the task option when the parties are not crossed at all remains unaddressed, that is lie on skew lines. In this case the concept of a corner between them is also defined. At a vector task of pieces of straight lines, the way of definition of a corner to honey them is uniform - a scalar product.
5. Example 2. To find a corner f between the parties of any polyhedron set by vectors of a= {3,-5,-2} and b = {3,-4, 6}. As soon as it is found out, that corner will be defined by its cosine, and cos f = (x1kh2 + y1y2+z1z2) / (|a| |b|)= (9+20-12)/(3^2+5^2+2^2) ^1/2(3^2+4^2+6^2) ^1/2=7/sqrt (29)•sqrt(61) =7/sqrt (1769)=0,1664 The answer: õ = (0.1664)