Find the relative position of the line and the plane. Plane in space - necessary information. The formulas are correct for the cube
Stereometry
Mutual arrangement of straight lines and planes
In space
Parallelism of lines and planes
Two lines in space are called parallel , if they lie in the same plane and do not intersect.
A straight line and a plane are called parallel , if they do not intersect.
The two planes are called parallel , if they do not intersect.
Lines that do not intersect and do not lie in the same plane are called interbreeding .
Sign of parallelism between a line and a plane. If a line that does not belong to a plane is parallel to some line in this plane, then it is parallel to the plane itself.
Sign of parallel planes. If two intersecting lines of one plane are respectively parallel to two lines of another plane, then these planes are parallel.
Sign of crossing lines. If one of two lines lies in a plane, and the other intersects this plane at a point not belonging to the first line, then these lines intersect.
Theorems on parallel lines and parallel planes.
1. Two lines parallel to a third line are parallel.
2. If one of two parallel lines intersects a plane, then the other line also intersects this plane.
3. Through a point outside a given line, you can draw a line parallel to the given one, and only one.
4. If a line is parallel to each of two intersecting planes, then it is parallel to their line of intersection.
5. If two parallel planes are intersected by a third plane, then the lines of intersection are parallel.
6. Through a point not lying in a given plane, you can draw a plane parallel to the given one, and only one.
7. Two planes parallel to the third are parallel to each other.
8. Segments of parallel lines contained between parallel planes are equal.
Angles between straight lines and planes
The angle between a straight line and a plane the angle between a straight line and its projection onto a plane is called (the angle in Fig. 1).
Angle between intersecting lines is the angle between intersecting lines parallel to the given intersecting lines.
Dihedral angle is a figure formed by two half-planes with a common line. Half-planes are called edges , straight – edge dihedral angle.
Linear angle dihedral angle is the angle between half-lines belonging to the faces of the dihedral angle, emanating from one point on the edge and perpendicular to the edge (the angle in Fig. 2).
The degree (radian) measure of a dihedral angle is equal to the degree (radian) measure of its linear angle.
Perpendicularity of lines and planes
Two straight lines are called perpendicular if they intersect at right angles.
A straight line intersecting a plane is called perpendicular this plane if it is perpendicular to any line in the plane passing through the point of intersection of this line and the plane.
The two planes are called perpendicular , if intersecting, they form right dihedral angles.
Sign of perpendicularity of a line and a plane. If a line intersecting a plane is perpendicular to two intersecting lines in this plane, then it is perpendicular to the plane.
Sign of perpendicularity of two planes. If a plane passes through a line perpendicular to another plane, then these planes are perpendicular.
Theorems on perpendicular lines and planes.
1. If a plane is perpendicular to one of two parallel lines, then it is also perpendicular to the other.
2. If two lines are perpendicular to the same plane, then they are parallel.
3. If a line is perpendicular to one of two parallel planes, then it is also perpendicular to the other.
4. If two planes are perpendicular to the same line, then they are parallel.
Perpendicular and oblique
Theorem. If a perpendicular and inclined lines are drawn from one point outside the plane, then:
1) oblique ones having equal projections are equal;
2) of the two inclined ones, the one whose projection is larger is greater;
3) equal obliques have equal projections;
4) of the two projections, the one that corresponds to the larger oblique one is larger.
Three Perpendicular Theorem. In order for a straight line lying in a plane to be perpendicular to an inclined one, it is necessary and sufficient that this straight line be perpendicular to the projection of the inclined one (Fig. 3).
Theorem on the area of the orthogonal projection of a polygon onto a plane. The area of the orthogonal projection of a polygon onto a plane is equal to the product of the area of the polygon and the cosine of the angle between the plane of the polygon and the projection plane.
Construction.
1. On a plane a we conduct a direct A.
3. In plane b through the point A let's make a direct b, parallel to the line A.
4. A straight line has been built b parallel to the plane a.
Proof. Based on the parallelism of a straight line and a plane, a straight line b parallel to the plane a, since it is parallel to the line A, belonging to the plane a.
Study. The problem has an infinite number of solutions, since the straight line A in the plane a is chosen randomly.
Example 2. Determine at what distance from the plane the point is located A, if straight AB intersects the plane at an angle of 45º, the distance from the point A to the point IN belonging to the plane is equal to cm?
Solution. Let's make a drawing (Fig. 5):
AC– perpendicular to the plane a, AB– inclined, angle ABC– angle between straight line AB and plane a. Triangle ABC– rectangular because AC– perpendicular. The required distance from the point A to the plane - this is the leg AC right triangle. Knowing the angle and hypotenuse cm, we will find the leg AC:
Answer: 3 cm.
Example 3. Determine at what distance from the plane of an isosceles triangle is a point located 13 cm from each of the vertices of the triangle if the base and height of the triangle are equal to 8 cm?
Solution. Let's make a drawing (Fig. 6). Dot S away from the points A, IN And WITH at the same distance. So, inclined S.A., S.B. And S.C. equal, SO– the common perpendicular of these inclined ones. By the theorem of obliques and projections AO = VO = CO.
Dot ABOUT– the center of a circle circumscribed about a triangle ABC. Let's find its radius:
Where Sun– base;
AD– the height of a given isosceles triangle.
Finding the sides of a triangle ABC from a right triangle ABD according to the Pythagorean theorem:
Now we find OB:
Consider a triangle SOB: S.B.= 13 cm, OB= = 5 cm. Find the length of the perpendicular SO according to the Pythagorean theorem:
Answer: 12 cm.
Example 4. Given parallel planes a And b. Through the point M, which does not belong to any of them, straight lines are drawn A And b that cross a at points A 1 and IN 1 and the plane b– at points A 2 and IN 2. Find A 1 IN 1 if it is known that MA 1 = 8 cm, A 1 A 2 = 12 cm, A 2 IN 2 = 25 cm.
Solution. Since the condition does not say how the point is located relative to both planes M, then two options are possible: (Fig. 7, a) and (Fig. 7, b). Let's look at each of them. Two intersecting lines A And b define a plane. This plane intersects two parallel planes a And b along parallel lines A 1 IN 1 and A 2 IN 2 according to Theorem 5 about parallel lines and parallel planes.
Triangles MA 1 IN 1 and MA 2 IN 2 are similar (angles A 2 MV 2 and A 1 MV 1 – vertical, corners MA 1 IN 1 and MA 2 IN 2 – internal crosswise lying with parallel lines A 1 IN 1 and A 2 IN 2 and secant A 1 A 2). From the similarity of triangles follows the proportionality of the sides:
Option a):
Option b):
Answer: 10 cm and 50 cm.
Example 5. Through the point A plane g a direct line was drawn AB, forming an angle with the plane a. Via direct AB a plane is drawn r, forming with the plane g corner b. Find the angle between the projection of a straight line AB to the plane g and plane r.
Solution. Let's make a drawing (Fig. 8). From point IN drop the perpendicular to the plane g. Linear dihedral angle between planes g And r- this is a right angle AD DBC, based on the perpendicularity of a line and a plane, as well as Based on the perpendicularity of planes, a plane r perpendicular to the plane of the triangle DBC, since it passes through the line AD. We construct the desired angle by dropping the perpendicular from the point WITH to the plane r, let's denote it Find the sine of this angle of a right triangle MYSELF. Let us introduce an auxiliary segment a = BC. From a triangle ABC: From a triangle Navy we'll find
Then the desired angle
Answer:
Tasks for independent solution
I level
1.1. Through a point, draw a line perpendicular to two given intersecting lines.
1.2. Determine how many different planes can be drawn:
1) through three different points;
2) through four different points, no three of which lie on the same plane?
1.3. Through the vertices of the triangle ABC lying in one of two parallel planes, parallel lines are drawn intersecting the second plane at points A 1 , IN 1 , WITH 1. Prove the equality of triangles ABC And A 1 IN 1 WITH 1 .
1.4. From the top A rectangle ABCD perpendicular restored AM to its plane.
1) prove that triangles MBC And MDC– rectangular;
2) indicate among the segments M.B., M.C., M.D. And M.A. segment of the greatest and shortest length.
1.5. The faces of one dihedral angle are correspondingly parallel to the faces of the other. Determine the relationship between the values of these dihedral angles.
1.6. Find the value of the dihedral angle if the distance from a point taken on one face to the edge is 2 times greater than the distance from the point to the plane of the second face.
1.7. From a point separated from the plane by a distance, two equal inclined slopes are drawn, forming an angle of 60º. Oblique projections are mutually perpendicular. Find the lengths of the inclined ones.
1.8. From the top IN square ABCD perpendicular restored BE to the plane of the square. Angle of inclination of the plane of the triangle ACE to the plane of the square is equal j, the side of the square is A ACE.
Level II
2.1. Through a point that does not belong to one of the two intersecting lines, draw a line intersecting both given lines.
2.2. Parallel lines A, b And With do not lie in the same plane. Through the point A on a straight line A perpendiculars to straight lines b And With, intersecting them at the points respectively IN And WITH. Prove that the line Sun perpendicular to straight lines b And With.
2.3. Through the top A right triangle ABC a plane is drawn parallel to Sun. Legs of a triangle AC= 20 cm, Sun= 15 cm. The projection of one of the legs onto the plane is 12 cm. Find the projection of the hypotenuse.
2.4. In one of the faces of the dihedral angle equal to 30º there is a point M. The distance from it to the edge of the corner is 18 cm. Find the distance from the projection of the point M to the second face to the first face.
2.5. Ends of the segment AB belong to the faces of a dihedral angle equal to 90º. Distance from points A And IN to the edge are equal respectively AA 1 = 3 cm, BB 1 = 6 cm, distance between points on the edge Find the length of the segment AB.
2.6. From a point located at a distance from the plane A, two inclined ones are drawn, forming angles of 45º and 30º with the plane, and an angle of 90º between themselves. Find the distance between the bases of the inclined ones.
2.7. The sides of the triangle are 15 cm, 21 cm and 24 cm. Point M is removed from the plane of the triangle by 73 cm and is at the same distance from its vertices. Find this distance.
2.8. From the center ABOUT circle inscribed in a triangle ABC, a perpendicular is restored to the plane of the triangle OM. Find the distance from the point M to the sides of the triangle, if AB = BC = 10 cm, AC= 12 cm, OM= 4 cm.
2.9. Distances from point M to the sides and vertex of the right angle are 4 cm, 7 cm and 8 cm respectively. Find the distance from the point M to the plane of a right angle.
2.10. Through the base AB isosceles triangle ABC the plane is drawn at an angle b to the plane of the triangle. Vertex WITH removed from the plane by a distance A. Find the area of the triangle ABC, if the base AB of an isosceles triangle is equal to its height.
Level III
3.1. Rectangle Layout ABCD with the parties A And b bent diagonally BD so that the planes of the triangles BAD And BCD became mutually perpendicular. Find the length of the segment AC.
3.2. Two rectangular trapezoids with angles of 60º lie in perpendicular planes and have a larger common ground. The larger sides are 4 cm and 8 cm. Find the distance between the vertices of the straight lines and the vertices of the obtuse angles of the trapezoids if the vertices of their acute angles coincide.
3.3.Cube given ABCDA 1 B 1 C 1 D 1. Find the angle between the straight line CD 1 and plane BDC 1 .
3.4. On the edge AB Cuba ABCDA 1 B 1 C 1 D 1 point taken R- the middle of this rib. Construct a section of the cube with a plane passing through the points C 1 P.D. and find the area of this section if the edge of the cube is equal to A.
3.5. Through the side AD rectangle ABCD a plane is drawn a so that the diagonal BD makes an angle of 30º with this plane. Find the angle between the plane of the rectangle and the plane a, If AB = A, AD = b. Determine at what ratio A And b the problem has a solution.
3.6. Find the locus of points equidistant from the lines defined by the sides of the triangle.
Prism. Parallelepiped
Prism is a polyhedron whose two faces are equal n-gons (bases) , lying in parallel planes, and the remaining n faces are parallelograms (side faces) . Lateral rib The side of a prism that does not belong to the base is called the side of the prism.
A prism whose lateral edges are perpendicular to the planes of the bases is called direct prism (Fig. 1). If the side edges are not perpendicular to the planes of the bases, then the prism is called inclined . Correct A prism is a right prism whose bases are regular polygons.
Height prism is the distance between the planes of the bases. Diagonal A prism is a segment connecting two vertices that do not belong to the same face. Diagonal section is called a section of a prism by a plane passing through two lateral edges that do not belong to the same face. Perpendicular section is called a section of a prism by a plane perpendicular to the side edge of the prism.
Lateral surface area of a prism is the sum of the areas of all lateral faces. Area full surface is called the sum of the areas of all faces of the prism (i.e. the sum of the areas of the side faces and the areas of the bases).
For an arbitrary prism the following formulas are true::
Where l– length of the side rib;
H- height;
P
Q
S side
S full
S base– area of the bases;
V– volume of the prism.
For a straight prism the following formulas are correct:
Where p– base perimeter;
l– length of the side rib;
H- height.
parallelepiped called a prism whose base is a parallelogram. A parallelepiped whose lateral edges are perpendicular to the bases is called direct (Fig. 2). If the side edges are not perpendicular to the bases, then the parallelepiped is called inclined . A right parallelepiped whose base is a rectangle is called rectangular. A rectangular parallelepiped with all edges equal is called cube
The faces of a parallelepiped that do not have common vertices are called opposite . The lengths of edges emanating from one vertex are called measurements parallelepiped. Since a parallelepiped is a prism, its main elements are defined in the same way as they are defined for prisms.
Theorems.
1. The diagonals of a parallelepiped intersect at one point and are bisected by it.
2. In a rectangular parallelepiped, the square of the length of the diagonal is equal to the sum of the squares of its three dimensions:
3. All four diagonals of a rectangular parallelepiped are equal to each other.
For an arbitrary parallelepiped the following formulas are valid:
Where l– length of the side rib;
H- height;
P– perpendicular section perimeter;
Q– Perpendicular cross-sectional area;
S side– lateral surface area;
S full– total surface area;
S base– area of the bases;
V– volume of the prism.
For a right parallelepiped the following formulas are correct:
Where p– base perimeter;
l– length of the side rib;
H– height of a right parallelepiped.
For a rectangular parallelepiped the following formulas are correct:
Where p– base perimeter;
H- height;
d– diagonal;
a,b,c– measurements of a parallelepiped.
The following formulas are correct for a cube:
Where a– rib length;
d- diagonal of the cube.
Example 1. The diagonal of a rectangular parallelepiped is 33 dm, and its dimensions are in the ratio 2: 6: 9. Find the dimensions of the parallelepiped.
Solution. To find the dimensions of the parallelepiped, we use formula (3), i.e. by the fact that the square of the hypotenuse of a cuboid is equal to the sum of the squares of its dimensions. Let us denote by k proportionality factor. Then the dimensions of the parallelepiped will be equal to 2 k, 6k and 9 k. Let us write formula (3) for the problem data:
Solving this equation for k, we get:
This means that the dimensions of the parallelepiped are 6 dm, 18 dm and 27 dm.
Answer: 6 dm, 18 dm, 27 dm.
Example 2. Find the volume of an inclined triangular prism, the base of which is an equilateral triangle with a side of 8 cm, if the side edge is equal to the side of the base and inclined at an angle of 60º to the base.
Solution . Let's make a drawing (Fig. 3).
In order to find the volume of an inclined prism, you need to know the area of its base and height. The area of the base of this prism is the area of an equilateral triangle with a side of 8 cm. Let us calculate it:
The height of a prism is the distance between its bases. From the top A 1 of the upper base, lower the perpendicular to the plane of the lower base A 1 D. Its length will be the height of the prism. Consider D A 1 AD: since this is the angle of inclination of the side edge A 1 A to the base plane, A 1 A= 8 cm. From this triangle we find A 1 D:
Now we calculate the volume using formula (1):
Answer: 192 cm 3.
Example 3. The lateral edge of a regular hexagonal prism is 14 cm. The area of the largest diagonal section is 168 cm 2. Find the total surface area of the prism.
Solution. Let's make a drawing (Fig. 4)
The largest diagonal section is a rectangle A.A. 1 DD 1 since diagonal AD regular hexagon ABCDEF is the largest. In order to calculate the lateral surface area of the prism, it is necessary to know the side of the base and the length of the side edge.
Knowing the area of the diagonal section (rectangle), we find the diagonal of the base.
Since then
Since then AB= 6 cm.
Then the perimeter of the base is:
Let's find the area of the lateral surface of the prism:
The area of a regular hexagon with side 6 cm is:
Find the total surface area of the prism:
Answer:
Example 4. The base of a right parallelepiped is a rhombus. The diagonal cross-sectional areas are 300 cm2 and 875 cm2. Find the area of the lateral surface of the parallelepiped.
Solution. Let's make a drawing (Fig. 5).
Let us denote the side of the rhombus by A, diagonals of a rhombus d 1 and d 2, parallelepiped height h. To find the area of the lateral surface of a right parallelepiped, it is necessary to multiply the perimeter of the base by the height: (formula (2)). Base perimeter p = AB + BC + CD + DA = 4AB = 4a, because ABCD- rhombus H = AA 1 = h. That. Need to find A And h.
Let's consider diagonal sections. AA 1 SS 1 – a rectangle, one side of which is the diagonal of a rhombus AC = d 1, second – side edge AA 1 = h, Then
Similarly for the section BB 1 DD 1 we get:
Using the property of a parallelogram such that the sum of the squares of the diagonals is equal to the sum of the squares of all its sides, we obtain the equality We obtain the following:
Let us express from the first two equalities and substitute them into the third. We get: then
1.3. In an inclined triangular prism, a section is drawn perpendicular to the side edge equal to 12 cm. In the resulting triangle, two sides with lengths cm and 8 cm form an angle of 45°. Find the lateral surface area of the prism.
1.4. The base of a right parallelepiped is a rhombus with a side of 4 cm and an acute angle of 60°. Find the diagonals of the parallelepiped if the length of the side edge is 10 cm.
1.5. The base of a right parallelepiped is a square with a diagonal equal to cm. The lateral edge of the parallelepiped is 5 cm. Find the total surface area of the parallelepiped.
1.6. The base of an inclined parallelepiped is a rectangle with sides 3 cm and 4 cm. A side edge equal to cm is inclined to the plane of the base at an angle of 60°. Find the volume of the parallelepiped.
1.7. Calculate the surface area of a rectangular parallelepiped if two edges and a diagonal emanating from one vertex are 11 cm, cm and 13 cm, respectively.
1.8. Determine the weight of a stone column in the shape of a rectangular parallelepiped with dimensions of 0.3 m, 0.3 m and 2.5 m, if the specific gravity of the material is 2.2 g/cm 3.
1.9. Find the diagonal cross-sectional area of a cube if the diagonal of its face is equal to dm.
1.10. Find the volume of a cube if the distance between two of its vertices that do not lie on the same face is equal to cm.
Level II
2.1. The base of the inclined prism is an equilateral triangle with side cm. The side edge is inclined to the plane of the base at an angle of 30°. Find the cross-sectional area of the prism passing through the side edge and the height of the prism if it is known that one of the vertices of the upper base is projected onto the middle of the side of the lower base.
2.2. The base of the inclined prism is an equilateral triangle ABC with a side equal to 3 cm. Vertex A 1 is projected into the center of triangle ABC. Rib AA 1 makes an angle of 45° with the base plane. Find the lateral surface area of the prism.
2.3. Calculate the volume of an inclined triangular prism if the sides of the base are 7 cm, 5 cm and 8 cm, and the height of the prism is equal to the smaller height of the base triangle.
2.4. The diagonal of a regular quadrangular prism is inclined to the side face at an angle of 30°. Find the angle of inclination to the plane of the base.
2.5. The base of a straight prism is an isosceles trapezoid, the bases of which are 4 cm and 14 cm, and the diagonal is 15 cm. The two lateral faces of the prism are squares. Find the total surface area of the prism.
2.6. The diagonals of a regular hexagonal prism are 19 cm and 21 cm. Find its volume.
2.7. Find the measurements of a rectangular parallelepiped whose diagonal is 8 dm and forms angles of 30° and 40° with its side faces.
2.8. The diagonals of the base of a right parallelepiped are 34 cm and 38 cm, and the areas of the side faces are 800 cm 2 and 1200 cm 2. Find the volume of the parallelepiped.
2.9. Determine the volume of a rectangular parallelepiped in which the diagonals of the side faces emerging from one vertex are 4 cm and 5 cm and form an angle of 60°.
2.10. Find the volume of a cube if the distance from its diagonal to an edge that does not intersect with it is mm.
Level III
3.1. In a regular triangular prism, a section is drawn through the side of the base and the middle of the opposite side edge. The base area is 18 cm 2, and the diagonal of the side edge is inclined to the base at an angle of 60°. Find the cross-sectional area.
3.2. At the base of the prism lies a square ABCD, all of whose vertices are equidistant from the vertex A 1 of the upper base. The angle between the side edge and the base plane is 60°. The side of the base is 12 cm. Construct a section of the prism with a plane passing through vertex C, perpendicular to edge AA 1 and find its area.
3.3. The base of a straight prism is an isosceles trapezoid. The diagonal cross-sectional area and the area of the parallel side faces are respectively equal to 320 cm 2, 176 cm 2 and 336 cm 2. Find the lateral surface area of the prism.
3.4. The area of the base of a right triangular prism is 9 cm 2, the area of the side faces is 18 cm 2, 20 cm 2 and 34 cm 2. Find the volume of the prism.
3.5. Find the diagonals of a rectangular parallelepiped, knowing that the diagonals of its faces are 11 cm, 19 cm and 20 cm.
3.6. The angles formed by the diagonal of the base of a rectangular parallelepiped with the side of the base and the diagonal of the parallelepiped are equal to a and b, respectively. Find the lateral surface area of the parallelepiped if its diagonal is d.
3.7. The area of the section of the cube that is a regular hexagon is equal to cm 2. Find the surface area of the cube.
A straight line may or may not belong to a plane. It belongs to a plane if at least two of its points lie on the plane. Figure 93 shows the Sum plane (axb). Straight l belongs to the Sum plane, since its points 1 and 2 belong to this plane.
If a line does not belong to the plane, it can be parallel to it or intersect it.
A line is parallel to a plane if it is parallel to another line lying in that plane. In Figure 93 there is a straight line m || Sum, since it is parallel to the line l belonging to this plane.
A straight line can intersect a plane at different angles and, in particular, be perpendicular to it. The construction of lines of intersection of a straight line and a plane is given in §61.
Figure 93 - A straight line belonging to a plane
A point in relation to the plane can be located in the following way: belong to it or not belong to it. A point belongs to a plane if it is located on a straight line located in this plane. Figure 94 shows a complex drawing of the Sum plane defined by two parallel lines l And p. There is a line in the plane m. Point A lies in the Sum plane, since it lies on the line m. Dot IN does not belong to the plane, since its second projection does not lie on the corresponding projections of the line.
Figure 94 - Complex drawing of a plane defined by two parallel lines
Conical and cylindrical surfaces
Conical surfaces include surfaces formed by the movement of a rectilinear generatrix l along a curved guide m. The peculiarity of the formation of a conical surface is that in this case one point of the generatrix is always motionless. This point is the vertex of the conical surface (Figure 95, A). The determinant of a conical surface includes the vertex S and guide m, at the same time l"~S; l"^ m.
Cylindrical surfaces are those formed by a straight generatrix / moving along a curved guide T parallel to the given direction S(Figure 95, b). A cylindrical surface can be considered as a special case of a conical surface with a vertex at infinity S.
The determinant of a cylindrical surface consists of a guide T and directions S forming l, while l" || S; l"^m.
If the generators of a cylindrical surface are perpendicular to the projection plane, then such a surface is called projecting. In Figure 95, V a horizontally projecting cylindrical surface is shown.
On cylindrical and conical surfaces, given points are constructed using generators passing through them. Lines on surfaces, such as a line A to figure 95, V or horizontal h in figure 95, a, b, are constructed using individual points belonging to these lines.
Figure 95 - Conical and cylindrical surfaces
Torso surfaces
A torso surface is a surface formed by a rectilinear generatrix l, touching during its movement in all its positions some spatial curve T, called return edge(Figure 96). The return edge completely defines the torso and is a geometric part of the surface determinant. The algorithmic part is the indication of the tangency of the generators to the return edge.
A conical surface is a special case of a torso, which has a return edge T degenerated into a point S- the top of the conical surface. A cylindrical surface is a special case of a torso, whose return edge is a point at infinity.
Figure 96 – Torso surface
Faceted surfaces
Faceted surfaces include surfaces formed by the movement of a rectilinear generatrix l along a broken guide m. Moreover, if one point S the generatrix is motionless, a pyramidal surface is created (Figure 97), if the generatrix is parallel to a given direction when moving S, then a prismatic surface is created (Figure 98).
The elements of faceted surfaces are: vertex S(near a prismatic surface it is at infinity), face (part of the plane limited by one section of the guide m and the extreme positions of the generatrix relative to it l) and edge (line of intersection of adjacent faces).
The determinant of a pyramidal surface includes the vertex S, through which the generators and guides pass: l" ~ S; l^ T.
Determinant of a prismatic surface other than a guide T, contains direction S, to which all generators are parallel l surfaces: l||S; l^ t.
Figure 97 - Pyramid surface
Figure 98 - Prismatic surface
Closed faceted surfaces formed by a certain number (at least four) of faces are called polyhedra. Among the polyhedra there is a group regular polyhedra, in which all faces are regular and congruent polygons, and the polyhedral angles at the vertices are convex and contain the same number of faces. For example: hexahedron - cube (Figure 99, A), tetrahedron - regular quadrilateral (Figure 99, 6) octahedron - polyhedron (Figure 99, V). Crystals have the shape of various polyhedra.
Figure 99 - Polyhedra
Pyramid- a polyhedron, the base of which is an arbitrary polygon, and the side faces are triangles with a common vertex S.
In a complex drawing, a pyramid is defined by projections of its vertices and edges, taking into account their visibility. The visibility of an edge is determined using competing points (Figure 100).
Figure 100 – Determining edge visibility using competing points
Prism- a polyhedron whose base is two identical and mutually parallel polygons, and the side faces are parallelograms. If the edges of the prism are perpendicular to the plane of the base, such a prism is called a straight prism. If the edges of a prism are perpendicular to any projection plane, then its lateral surface is called projecting. Figure 101 shows a comprehensive drawing of a right quadrangular prism with a horizontally projecting surface.
Figure 101 - Complex drawing of a right quadrangular prism with a horizontally projecting surface
When working with a complex drawing of a polyhedron, you have to build lines on its surface, and since a line is a collection of points, you need to be able to build points on the surface.
Any point on a faceted surface can be constructed using a generatrix passing through this point. In the figure there are 100 in the face ACS point built M using generatrix S-5.
Helical surfaces
Helical surfaces include surfaces created by the helical movement of a rectilinear generatrix. Ruled helical surfaces are called helicoids.
A straight helicoid is formed by the movement of a rectilinear generatrix i along two guides: helix T and its axes i; while forming l intersects the screw axis at a right angle (Figure 102, a). Straight helicoid is used to create spiral staircases, augers, as well as power threads in machine tools.
An inclined helicoid is formed by moving the generatrix along a screw guide T and its axes i so that the generator l crosses the axis i at a constant angle φ, different from a straight line, i.e. in any position the generatrix l parallel to one of the generatrices of the guide cone with an apex angle equal to 2φ (Figure 102, b). Inclined helicoids limit the surfaces of the threads.
Figure 102 - Helicoids
Surfaces of revolution
Surfaces of revolution include surfaces formed by rotating a line l around a straight line i , which is the axis of rotation. They can be linear, such as a cone or cylinder of revolution, and non-linear or curved, such as a sphere. The determinant of the surface of revolution includes the generatrix l and axis i . During rotation, each point of the generatrix describes a circle, the plane of which is perpendicular to the axis of rotation. Such circles of the surface of revolution are called parallels. The largest of the parallels is called equator. Equator determines the horizontal outline of the surface if i _|_ P 1 . In this case, the parallels are the horizontals of this surface.
Curves of a surface of revolution resulting from the intersection of the surface by planes passing through the axis of rotation are called meridians. All meridians of one surface are congruent. The frontal meridian is called the main meridian; it determines the frontal outline of the surface of revolution. The profile meridian determines the profile outline of the surface of rotation.
It is most convenient to construct a point on curved surfaces of revolution using surface parallels. There is 103 point in the figure M built on parallel h4.
Figure 103 – Constructing a point on a curved surface
Surfaces of rotation have found the widest application in technology. They limit the surfaces of most engineering parts.
A conical surface of revolution is formed by rotating a straight line i around the straight line intersecting with it - the axis i(Figure 104, A). Dot M on the surface is constructed using a generatrix l and parallels h. This surface is also called a cone of revolution or a right circular cone.
A cylindrical surface of revolution is formed by rotating a straight line l around an axis parallel to it i(Figure 104, b). This surface is also called a cylinder or a right circular cylinder.
A sphere is formed by rotating a circle around its diameter (Figure 104, V). Point A on the surface of the sphere belongs to the prime meridian f, dot IN- equator h, a point M built on an auxiliary parallel h".
Figure 104 - Formation of surfaces of revolution
A torus is formed by rotating a circle or its arc around an axis lying in the plane of the circle. If the axis is located within the resulting circle, then such a torus is called closed (Figure 105, a). If the axis of rotation is outside the circle, then such a torus is called open (Figure 105, b). An open torus is also called a ring.
Figure 105 – Formation of a torus
Surfaces of revolution can also be formed by other second-order curves. Ellipsoid of rotation (Figure 106, A) formed by rotating an ellipse around one of its axes; paraboloid of rotation (Figure 106, b) - rotation of the parabola around its axis; single-sheet hyperboloid of revolution (Figure 106, V) is formed by rotating a hyperbola around an imaginary axis, and a two-sheet (Figure 106, G) - rotation of the hyperbola around the real axis.
Figure 106 – Formation of surfaces of revolution by second-order curves
In the general case, surfaces are depicted as not limited in the direction of propagation of the generating lines (see Figures 97, 98). To solve specific tasks and obtaining geometric shapes is limited to the cutting planes. For example, to obtain a circular cylinder, it is necessary to limit a section of the cylindrical surface to the cutting planes (see Figure 104, b). As a result, we get its upper and lower bases. If the cutting planes are perpendicular to the axis of rotation, the cylinder will be straight; if not, the cylinder will be inclined.
To obtain a circular cone (see Figure 104, A), it is necessary to trim along the top and beyond. If the cutting plane of the base of the cylinder is perpendicular to the axis of rotation, the cone will be straight; if not, it will be inclined. If both cutting planes do not pass through the vertex, the cone will be truncated.
Using the cut plane, you can get a prism and a pyramid. For example, a hexagonal pyramid will be straight if all its edges have the same slope to the cutting plane. In other cases it will be slanted. If it is completed With using cutting planes and none of them passes through the vertex - the pyramid is truncated.
A prism (see Figure 101) can be obtained by limiting a section of the prismatic surface to two cutting planes. If the cutting plane is perpendicular to the edges of, for example, an octagonal prism, it is straight; if not perpendicular, it is inclined.
By choosing the appropriate position of the cutting planes, you can obtain different shapes of geometric figures depending on the conditions of the problem being solved.
In planimetry, the plane is one of the main figures, therefore, it is very important to have a clear understanding of it. This article was created to cover this topic. First, the concept of a plane, its graphical representation is given and the designations of planes are shown. Next, the plane is considered together with a point, a straight line or another plane, and options arise from the relative position in space. In the second and third and fourth paragraphs of the article, all the options for the relative position of two planes, a straight line and a plane, as well as points and planes are analyzed, the basic axioms and graphic illustrations are given. In conclusion, the main methods of defining a plane in space are given.
Page navigation.
Plane - basic concepts, symbols and image.
The simplest and most basic geometric shapes in three-dimensional space there are a point, a line and a plane. We already have an idea of a point and a line on a plane. If we place a plane on which points and lines are depicted in three-dimensional space, then we get points and lines in space. The idea of a plane in space allows us to obtain, for example, the surface of a table or wall. However, a table or wall has finite dimensions, and the plane extends beyond its boundaries to infinity.
Points and lines in space are designated in the same way as on a plane - in large and small Latin letters, respectively. For example, points A and Q, lines a and d. If two points lying on a line are given, then the line can be denoted by two letters corresponding to these points. For example, straight line AB or BA passes through points A and B. Planes are usually denoted by small Greek letters, for example, planes, or.
When solving problems, it becomes necessary to depict planes in a drawing. A plane is usually depicted as a parallelogram or an arbitrary simple closed region.
The plane is usually considered together with points, straight lines or other planes, and problems arise. various options their relative position. Let's move on to their description.
The relative position of the plane and the point.
Let's start with the axiom: there are points in every plane. From it follows the first option for the relative position of the plane and the point - the point can belong to the plane. In other words, a plane can pass through a point. To indicate that a point belongs to a plane, the symbol “” is used. For example, if the plane passes through point A, then you can briefly write .
It should be understood that on a given plane in space there are infinitely many points.
The following axiom shows how many points in space must be marked so that they define a specific plane: through three points that do not lie on the same line, a plane passes, and only one. If three points lying in a plane are known, then the plane can be denoted by three letters corresponding to these points. For example, if a plane passes through points A, B and C, then it can be designated ABC.
Let us formulate another axiom, which gives the second version of the relative position of the plane and the point: there are at least four points that do not lie in the same plane. So, a point in space may not belong to the plane. Indeed, by virtue of the previous axiom, a plane passes through three points in space, and the fourth point may or may not lie on this plane. When writing briefly, use the symbol “”, which is equivalent to the phrase “does not belong”.
For example, if point A does not lie in the plane, then use the short notation.
Straight line and plane in space.
Firstly, a straight line can lie in a plane. In this case, at least two points of this line lie in the plane. This is established by the axiom: if two points of a line lie in a plane, then all points of this line lie in the plane. To briefly record the belonging of a certain line to a given plane, use the symbol “”. For example, the notation means that straight line a lies in the plane.
Secondly, a straight line can intersect a plane. In this case, the straight line and the plane have one single common point, which is called the point of intersection of the straight line and the plane. When writing briefly, I denote the intersection with the symbol “”. For example, the notation means that straight line a intersects the plane at point M. When a plane intersects a certain straight line, the concept of an angle between the straight line and the plane arises.
Separately, it is worth focusing on a straight line that intersects the plane and is perpendicular to any straight line lying in this plane. Such a line is called perpendicular to the plane. To briefly record perpendicularity, use the symbol “”. For a more in-depth study of the material, you can refer to the article perpendicularity of a straight line and a plane.
Of particular importance when solving problems related to the plane is the so-called normal vector of the plane. A normal vector of a plane is any non-zero vector lying on a line perpendicular to this plane.
Thirdly, a straight line may be parallel to the plane, that is, it may not have common points in it. When writing concurrency briefly, use the symbol “”. For example, if line a is parallel to the plane, then we can write . We recommend that you study this case in more detail by referring to the article parallelism of a line and a plane.
It should be said that a straight line lying in a plane divides this plane into two half-planes. The straight line in this case is called the boundary of the half-planes. Any two points of the same half-plane lie on the same side of a line, and two points of different half-planes lie on opposite sides of the boundary line.
Mutual arrangement of planes.
Two planes in space can coincide. In this case they have at least three points in common.
Two planes in space can intersect. The intersection of two planes is a straight line, which is established by the axiom: if two planes have a common point, then they have a common straight line on which all the common points of these planes lie.
In this case, the concept of an angle between intersecting planes arises. Of particular interest is the case when the angle between the planes is ninety degrees. Such planes are called perpendicular. We talked about them in the article perpendicularity of planes.
Finally, two planes in space can be parallel, that is, have no common points. We recommend that you read the article parallelism of planes to get a complete understanding of this option for the relative arrangement of planes.
Methods for defining a plane.
Now we will list the main ways to define a specific plane in space.
Firstly, a plane can be defined by fixing three points in space that do not lie on the same straight line. This method is based on the axiom: through any three points that do not lie on the same line, there is a single plane.
If a plane is fixed and specified in three-dimensional space by indicating the coordinates of its three different points that do not lie on the same straight line, then we can write the equation of the plane passing through the three given points.
The next two methods of defining a plane are a consequence of the previous one. They are based on corollaries of the axiom about a plane passing through three points:
- a plane passes through a line and a point not lying on it, and only one (see also the article equation of a plane passing through a line and a point);
- a single plane passes through two intersecting lines (we recommend that you familiarize yourself with the material in the article: equation of a plane passing through two intersecting lines).
The fourth way to define a plane in space is based on defining parallel lines. Recall that two lines in space are called parallel if they lie in the same plane and do not intersect. Thus, by indicating two parallel lines in space, we will determine the only plane in which these lines lie.
If in three-dimensional space relative to a rectangular coordinate system a plane is specified in the indicated way, then we can create an equation for a plane passing through two parallel lines.
In the know high school In geometry lessons, the following theorem is proven: through a fixed point in space there passes a single plane perpendicular to a given line. Thus, we can define a plane if we specify the point through which it passes and a line perpendicular to it.
If a rectangular coordinate system is fixed in three-dimensional space and a plane is specified in the indicated way, then it is possible to construct an equation for a plane passing through a given point perpendicular to a given straight line.
Instead of a line perpendicular to the plane, you can specify one of the normal vectors of this plane. In this case, it is possible to write
Direct can belong to the plane, be her parallel or cross plane. A line belongs to a plane if two points belonging to the line and the plane have the same elevations. The corollary that follows from what has been said: a point belongs to a plane if it belongs to a line lying in this plane.
A line is parallel to a plane if it is parallel to a line lying in this plane.
A straight line intersecting a plane. To find the point of intersection of a straight line with a plane, it is necessary (Fig. 3.28):
1) draw an auxiliary plane through a given straight line m T;
2) build a line n intersection of a given plane Σ with an auxiliary plane T;
3) mark the intersection point R, given straight line m with the line of intersection n.
Consider the problem (Fig. 3.29). The straight line m is defined on the plan by a point A 6 and an inclination angle of 35°. An auxiliary vertical plane is drawn through this line T, which intersects the plane Σ along the line n (B 2 C 3). Thus, one moves from the relative position of a straight line and a plane to the relative position of two straight lines lying in the same vertical plane. This problem is solved by constructing profiles of these straight lines. Intersection of lines m And n on the profile determines the desired point R. Point elevation R determined by the vertical scale scale.
Straight line perpendicular to the plane. A straight line is perpendicular to a plane if it is perpendicular to any two intersecting lines of this plane. Figure 3.30 shows a straight line m, perpendicular to the plane Σ and intersecting it at point A. On the plan, the projection of the line m and the horizontal planes are mutually perpendicular (a right angle, one side of which is parallel to the projection plane, is projected without distortion. Both lines lie in the same vertical plane, therefore the positions of such lines are inverse in magnitude to each other: l m = l/l u. But l uΣ = lΣ, then l m = l/lΣ, that is, the position of the straight line m is inversely proportional to the position of the plane. The falls of a straight line and a plane are directed in different directions.
3.4. Projections with numerical marks. Surfaces
3.4.1.Polyhedra and curved surfaces. Topographic surface
In nature, many substances have a crystalline structure in the form of polyhedra. A polyhedron is a collection of flat polygons that do not lie in the same plane, where each side of one of them is also a side of the other. When depicting a polyhedron, it is enough to indicate the projections of its vertices, connecting them in a certain order with straight lines - projections of the edges. In this case, it is necessary to indicate visible and invisible edges in the drawing. In Fig. Figure 3.31 shows a prism and a pyramid, as well as finding the marks of points belonging to these surfaces.
A special group of convex polygons is the group of regular polygons in which all faces are equal regular polygons and all polygonal angles are equal. There are five types of regular polygons.
Tetrahedron- a regular quadrilateral, bounded by equilateral triangles, has 4 vertices and 6 edges (Fig. 3.32 a).
Hexahedron- regular hexagon (cube) - 8 vertices, 12 edges (Fig. 3.32b).
Octahedron- a regular octahedron, bounded by eight equilateral triangles - 6 vertices, 12 edges (Fig. 3.32c).
Dodecahedron- regular dodecahedron limited to twelve regular pentagons, connected in three near each vertex.
It has 20 vertices and 30 edges (Fig. 3.32 d).
Icosahedron- a regular twenty-sided triangle, limited by twenty equilateral triangles, connected by five near each vertex. 12 vertices and 30 edges (Fig. 3.32 d).
When constructing a point lying on the face of a polyhedron, it is necessary to draw a straight line belonging to this face and mark the projection of the point on its projection.
Conical surfaces are formed by moving a rectilinear generatrix along a curved guide so that in all positions the generatrix passes through a fixed point - the vertex of the surface. Conical surfaces general view on the plan they are depicted as a guide horizontal and a vertex. In Fig. Figure 3.33 shows the location of a point mark on the surface of a conical surface.
A straight circular cone is represented by a series of concentric circles drawn at equal intervals (Fig. 3.34a). Elliptical cone with a circular base - a series of eccentric circles (Fig. 3.34 b)
Spherical surfaces. A spherical surface is classified as a surface of revolution. It is formed by rotating a circle around its diameter. On the plan, a spherical surface is defined by the center TO and the projection of one of its horizontal lines (the equator of the sphere) (Fig. 3.35).
Topographic surface. A topographic surface is classified as a geometrically irregular surface, since it does not have a geometric law of formation. To characterize a surface, determine the position of its characteristic points relative to the projection plane. In Fig. 3.3 b a gives an example of a section of a topographic surface, which shows the projections of its individual points. Although such a plan makes it possible to get an idea of the shape of the depicted surface, it is not very clear. To give the drawing greater clarity and thereby make it easier to read, projections of points with identical marks are connected by smooth curved lines, which are called horizontals (isolines) (Fig. 3.36 b).
The horizontal lines of a topographic surface are sometimes defined as the lines of intersection of this surface with horizontal planes spaced at the same distance from each other (Fig. 3.37). The difference in elevations between two adjacent horizontal lines is called the section height.
The smaller the difference in elevation between two adjacent horizontal lines, the more accurate the image of a topographic surface is. On plans, contour lines are closed within the drawing or outside it. On steeper slopes, the surface projections of the contour lines come closer together; on flat slopes, their projections diverge.
The shortest distance between the projections of two adjacent horizontal lines on the plan is called the lay. In Fig. 3.38 through point A several straight line segments are drawn on the topographic surface AB, AS And AD. They all have different angles of incidence. The segment has the greatest angle of incidence AC, the location of which is of minimal importance. Therefore, it will be a projection of the line of incidence of the surface at a given location.
In Fig. 3.39 shows an example of constructing a projection of the line of incidence through a given point A. From point A 100, as if from the center, draw an arc of a circle touching the nearest horizontal line at the point At 90. Dot At 90, horizontal h 90, will belong to the fall line. From point At 90 draw an arc tangent to the next horizontal line at the point From 80, etc. From the drawing it is clear that the line of incidence of the topographic surface is a broken line, each link of which is perpendicular to the horizontal, passing through the lower end of the link, which has a lower elevation.
3.4.2.Intersection of a conical surface with a plane
If a cutting plane passes through the vertex of a conical surface, then it intersects it along straight lines forming the surface. In all other cases, the section line will be a flat curve: a circle, an ellipse, etc. Let us consider the case of a conical surface intersecting a plane.
Example 1. Construct the projection of the intersection line of a circular cone Φ( h o , S 5) with a plane Ω parallel to the generatrix of the conical surface.
A conical surface with a given plane location intersects along a parabola. Having interpolated the generatrix t we build horizontal lines of a circular cone - concentric circles with a center S 5. Then we determine the intersection points of the same horizontals of the plane and the cone (Fig. 3.40).
3.4.3. Intersection of a topographic surface with a plane and a straight line
The case of the intersection of a topographic surface with a plane is most often encountered in solving geological problems. In Fig. 3.41 gives an example of constructing the intersection of a topographic surface with the plane Σ. The curve I'm looking for m are determined by the intersection points of the same horizontal planes and the topographic surface.
In Fig. 3.42 gives an example of constructing a true view of a topographic surface with a vertical plane Σ. The required line m is determined by points A, B, C... intersection of the horizontals of the topographic surface with the cutting plane Σ. On the plan, the projection of the curve degenerates into a straight line coinciding with the projection of the plane: m≡ Σ. The profile of the curve m is constructed taking into account the location of the projections of its points on the plan, as well as their elevations.
3.4.4. Surface of equal slope
A surface of equal slope is a ruled surface, all straight lines of which make a constant angle with the horizontal plane. Such a surface can be obtained by moving a straight circular cone with an axis perpendicular to the plane of the plan, so that its top slides along a certain guide, and the axis remains vertical in any position.
In Fig. Figure 3.43 shows a surface of equal slope (i=1/2), the guide of which is a spatial curve A, B, C, D.
Graduation of the plane. As examples, consider the slope planes of the roadway.
Example 1. Longitudinal slope of the roadway i=0, slope slope of the embankment i n =1:1.5, (Fig. 3.44a). It is required to draw horizontal lines every 1 m. The solution comes down to the following. We draw the scale of the slope of the plane perpendicular to the edge of the roadway, mark points at a distance equal to an interval of 1.5 m taken from the linear scale, and determine marks 49, 48 and 47. Through the obtained points we draw the contours of the slope parallel to the edge of the road.
Example 2. Longitudinal slope of the road i≠0, slope of the embankment i n =1:1.5, (Fig. 3.44b). The plane of the roadway is graded. The slope of the roadway is graded as follows. At the point with the vertex 50.00 (or another point) we place the vertex of the cone, describe a circle with a radius equal to the interval of the embankment slope (in our example l= 1.5m). The elevation of this horizontal line of the cone will be one less than the elevation of the vertex, i.e. 49m. We draw a series of circles, we get contour lines 48, 47, tangent to which we draw contours of the embankment slope from the edge points with marks 49, 48, 47.
Graduation of surfaces.
Example 3. If the longitudinal slope of the road is i = 0 and the slope of the embankment is i n = 1: 1.5, then the contour lines of the slopes are drawn through the points of the slope scale, the interval of which is equal to the interval of the slopes of the embankment (Fig. 3.45a). The distance between two projections of adjacent horizontal lines in the direction of the general norm (slope scale) is the same everywhere.
Example 4. If the longitudinal slope of the road is i≠0, and the slope of the embankment is i n =1:1.5, (Fig. 3.45b), then the contour lines are constructed in the same way, except that the slope contours are drawn not in straight lines, but in curves.
3.4.5. Determination of the excavation limit line
Since most soils are unable to maintain vertical walls, slopes (artificial structures) have to be built. The slope imparted by a slope depends on the soil.
In order to give a section of the earth's surface the appearance of a plane with a certain slope, you need to know the line of limits for excavation and excavation works. This line, limiting the planned area, is represented by the lines of intersection of the slopes of embankments and excavations with a given topographic surface.
Since every surface (including flat ones) is depicted using contours, the line of intersection of surfaces is constructed as a set of intersection points of contours with the same marks. Let's look at examples.
Example 1. In Fig. 3.46 shows an earthen structure in the shape of a truncated quadrangular pyramid, standing on a plane N. Upper base ABCD pyramid has a mark 4m and side sizes 2×2.5 m. The side faces (embankment slopes) have a slope of 2:1 and 1:1, the direction of which is shown by arrows.
It is necessary to construct a line of intersection of the slopes of the structure with the plane N and among themselves, as well as construct a longitudinal profile along the axis of symmetry.
First, a diagram of slopes, intervals and scales of deposits, and given slopes is constructed. Perpendicular to each side of the site, the scales of the slopes are drawn at specified intervals, after which the projections of the contour lines with the same marks of adjacent faces are the intersection lines of the slopes, which are projections of the side edges of this pyramid.
The lower base of the pyramid coincides with the zero horizontal slopes. If this earthen structure is crossed by a vertical plane Q, in cross-section you will get a broken line - the longitudinal profile of the structure.
Example 2. Construct a line of intersection of the pit slopes with a flat slope and with each other. Bottom ( ABCD) the pit is a rectangular area with an elevation of 10 m and dimensions of 3x4 m. The axis of the site makes an angle of 5° with the south-north line. The slopes of the excavations have the same slopes of 2:1 (Fig. 3.47).
The line of zero works is established according to the site plan. It is constructed at the points of intersection of the same-named projections of the horizontal lines of the surfaces under consideration. At the points of intersection of the contours of the slopes and the topographic surface with the same marks, the line of intersection of the slopes is found, which are projections of the side edges of a given pit.
In this case, the side slopes of the excavations are adjacent to the bottom of the pit. Line abcd– the desired intersection line. Aa, Bb, Cs, Dd– the edges of the pit, the lines of intersection of the slopes with each other.
4. Questions for self-control and tasks for independent work on the topic "Rectangular projections"
Dot
4.1.1. The essence of the projection method.
4.1.2. What is point projection?
4.1.3. What are projection planes called and designated?
4.1.4. What are projection connection lines in a drawing and how are they located in the drawing in relation to the projection axes?
4.1.5. How to construct the third (profile) projection of a point?
4.1.6. Construct three projections of points A, B, C on a three-picture drawing, write down their coordinates and fill out the table.
4.1.7. Construct the missing projection axes, x A =25, y A =20. Construct a profile projection of point A.
4.1.8. Construct three projections of points according to their coordinates: A(25,20,15), B(20,25,0) and C(35,0,10). Indicate the position of the points in relation to the planes and axes of projections. Which point is closer to the P3 plane?
4.1.9. Material points A and B begin to fall simultaneously. What position will point B be in when point A touches the ground? Determine the visibility of points. Plot points in new position.
4.1.10. Construct three projections of point A, if the point lies in the P 3 plane, and the distance from it to the P 1 plane is 20 mm, to the P 2 plane - 30 mm. Write down the coordinates of the point.
Straight
4.2.1. How can a straight line be defined in a drawing?
4.2.2. Which line is called a straight line general position?
4.2.3. What position can a straight line occupy relative to the projection planes?
4.2.4. In what case does the projection of a line turn to a point?
4.2.5. What is characteristic of a complex straight level drawing?
4.2.6. Determine the relative position of these lines.
a…b a…b a…b
4.2.7. Construct projections of a straight segment AB with a length of 20 mm, parallel to the planes: a) P 2; b) P 1; c) Ox axis. Indicate the angles of inclination of the segment to the projection planes.
4.2.8. Construct projections of segment AB using the coordinates of its ends: A(30,10,10), B(10,15,30). Construct projections of point C dividing the segment in the ratio AC:CB = 1:2.
4.2.9. Determine and record the number of edges of this polyhedron and their position relative to the projection planes.
4.2.10. Through point A, draw a horizontal and a frontal line intersecting straight line m.
4.2.11. Determine the distance between line b and point A
4.2.12. Construct projections of a segment AB with a length of 20 mm passing through point A and perpendicular to the plane a) P 2; b) P 1; c) P 3.
The relative position of two straight lines
The following statements express the necessary and sufficient indications mutual position of two lines in space, given by canonical equations
A) Straight lines cross, i.e. do not lie on the same plane.
b) Lines intersect.
But the vectors are also non-collinear (otherwise their coordinates are proportional).
V) Lines are parallel.
Vectors are collinear, but a vector is not collinear.
G) The straight lines coincide.
All three vectors: , are collinear.
Proof. Let us prove the sufficiency of the indicated signs
A) Consider the vector and direction vectors of the given straight lines
then these vectors are non-coplanar, therefore, these lines do not lie on the same plane.
b) If, then the vectors are coplanar, therefore, these lines lie in the same plane, and since in the case ( b) the direction vectors and these lines are assumed to be non-collinear, then the lines intersect.
V) If the direction vectors and the given lines are collinear, then the lines are either parallel or coincide. In case ( V) the lines are parallel, because By convention, a vector whose beginning is at the point of the first line and its end at the point of the second line is not collinear.
d) If all vectors are collinear, then the lines coincide.
The necessity of signs is proved by contradiction.
Kletenik No. 1007
The following statements give necessary and sufficient conditions for the relative position of the line given by the canonical equations
and the plane defined by the general equation
relative to the general Cartesian coordinate system.
A plane and a line intersect:
The plane and the line are parallel:
The straight line lies on the plane:
Let us first prove the sufficiency of the indicated characteristics. Let us write the equations of this line in parametric form:
Substituting into equation (2 (planes)) the coordinates of an arbitrary point on a given line, taken from formulas (3), we will have:
1. If, then equation (4) has relatively t the only solution:
which means that a given straight line and a given plane have only one common point, i.e. intersect.
2. If, then equation (4) is not satisfied for any value t, i.e. on a given line there is not a single point lying on a given plane, therefore, the given line and the plane are parallel.
3. If, then equation (4) is satisfied for any value t, i.e. all points of a given line lie on a given plane, which means that a given line lies on a given plane.
The sufficient conditions for the relative position of a straight line and a plane that we have derived are also necessary and can be proved immediately by the method of contradiction.
From what has been proved it follows a necessary and sufficient condition that the vector is coplanar to the plane defined by the general equation with respect to the general Cartesian coordinate system.
