Location of straight line and plane. The relative position of the straight line and the plane

The straight line belongs to the plane, if it has two common points or one common point and is parallel to any line lying in the plane. Let the plane in the drawing be defined by two intersecting lines. In this plane, it is required to construct two straight lines m and n in accordance with these conditions ( G(a b)) (Fig. 4.5).

Solution. 1. We arbitrarily draw m 2, since the line belongs to the plane, mark the projections of the points of intersection of it with the lines A And b and determine their horizontal projections, draw m 1 through 1 1 and 2 1.

2. Through the point K of the plane we draw n 2 ║m 2 and n 1 ║m 1.

A straight line is parallel to a plane, if it is parallel to any line lying in the plane.

The intersection of a line and a plane. There are three possible cases of location of the straight line and the plane relative to the projection planes. Depending on this, the intersection point of the straight line and the plane is determined.

First case – straight line and plane – projecting position. In this case, the intersection point is available in the drawing (both of its projections); it only needs to be designated.

EXAMPLE In the drawing, a plane is given by traces Σ ( h 0 f 0)– horizontal projecting position – and straight l– frontally projecting position. Determine the point of their intersection (Fig. 4.6).

There is already an intersection point in the drawing - K(K 1 K 2).

Second case– either a straight line or a plane – of the projecting position. In this case, on one of the projection planes the projection of the intersection point already exists; it needs to be designated, and on the second projection plane it must be found by belonging.

EXAMPLES. In Fig. 4.7, and the plane is depicted with traces of a frontally projecting position and a straight line l – general position. The projection of the intersection point K 2 is already available in the drawing, and the projection K 1 must be found based on the belonging of the point K to the straight line l. On
rice. 4.7, b is a general plane, and straight line m is frontally projecting, then K 2 already exists (coincides with m 2), and K 1 must be found from the condition that the point belongs to the plane. To do this, pass through K
straight ( h– horizontal) lying in a plane.

Third case– both a straight line and a plane – in general position. In this case, to determine the point of intersection of the line and the plane, it is necessary to use the so-called intermediary - the projecting plane. To do this, an auxiliary cutting plane is drawn through the straight line. This plane intersects a given plane along a line. If this line intersects a given line, then there is a point of intersection of the line and the plane.

EXAMPLES. In Fig. 4.8 the plane is represented by a triangle ABC - general position - and a straight line l– general position. To determine the intersection point K, it is necessary through l draw a frontally projecting plane Σ, construct a line of intersection of Δ and Σ in the triangle (in the drawing this is segment 1,2), determine K 1 and, by accessory, K 2. Then the line visibility is determined l in relation to the triangle by competing points. On P 1, points 3 and 4 are taken as competing points. The projection of point 4 is visible on P 1, since its Z coordinate is greater than that of point 3, therefore, the projection l 1 from this point to K 1 will be invisible.

On P 2 the competing points are point 1, belonging to AB, and point 5, belonging to l. Point 1 will be visible, since its Y coordinate is greater than that of point 5, and therefore the projection of the straight line l 2 up to K 2 invisible.

Stereometry

Mutual arrangement 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 larger;

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 required angle

Answer:

Tasks for independent decision

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) in three various 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 triangle plane 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 are drawn 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 removed from the plane of the triangle by 73 cm and located 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 top right angle respectively equal to 4 cm, 7 cm and 8 cm. 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 base. 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 straight 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. Total surface area 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 bisect it.

2. B 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 us 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 face 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 ​​parallel side faces are respectively 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.

The relative position of a straight line and a plane in space allows for three cases. A straight line and a plane can intersect at one point. They can be parallel. Finally, a straight line can lie in a plane. Finding out specific situation for a straight line and a plane depends on the method of their description.

Let us assume that the plane π is given by the general equation π: Ax + By + Cz + D = 0, and the line L is given by the canonical equations (x - x 0)/l = (y - y 0)/m = (z - z 0) /n. The equations of a line give the coordinates of the point M 0 (x 0 ; y 0 ; z 0) on the line and the coordinates of the direction vector s = (l; m; n) of this line, and the equation of the plane gives the coordinates of its normal vector n = (A; B; C).

If straight line L and plane π intersect, then the direction vector s of the straight line is not parallel to plane π. This means that the normal vector n of the plane is not orthogonal to the vector s, i.e. their scalar product is not equal to zero. Through the coefficients of the equations of the line and plane, this condition is written as the inequality A1 + Bm + Cn ≠ 0.

If the line and the plane are parallel or the line lies in the plane, then the condition s ⊥ n is satisfied, which in coordinates reduces to the equality Al + Bm + Cn = 0. To separate the cases of “parallel” and “the line belongs to the plane”, you need to check whether point of a straight line in a given plane.

Thus, all three cases of the relative position of a straight line and a plane are separated by checking the corresponding conditions:

If the straight line L is given by its general equations:

then the relative position of the straight line and the π plane can be analyzed as follows. From the general equations of the line and general equation let's create a plane system of three linear equations with three unknowns

If this system has no solutions, then the line is parallel to the plane. If it has a unique solution, then the line and the plane intersect at a single point. The latter is equivalent to system determinant (6.6)

different from zero. Finally, if system (6.6) has infinitely many solutions, then the straight line belongs to the plane.

The angle between a straight line and a plane. The angle φ between the straight line L: (x - x 0)/l = (y - y 0)/m = (z - z 0)/n and the plane π: Ax + By + Cz + D = 0 is within the range of 0 ° (in case of parallelism) up to 90° (in case of perpendicularity to a straight line and a plane). The sine of this angle is equal to |cosψ|, where ψ is the angle between the directing vector of the straight line s and the normal vector n of the plane (Fig. 6.4). Having calculated the cosine of the angle between two vectors through their coordinates (see (2.16)), we obtain

The condition that a line and a plane are perpendicular is equivalent to the fact that the normal vector of the plane and the direction vector of the line are collinear. Through the coordinates of the vectors, this condition is written as a double equality

TICKET 16.

Properties of a pyramid whose dihedral angles are equal.

A) If the lateral faces of a pyramid with its base form equal dihedral angles, then all the heights of the lateral faces of the pyramid are equal (for a regular pyramid these are apothems), and the top of the pyramid is projected into the center of a circle inscribed in the base polygon.

B) A pyramid can have equal dihedral angles at the base when a circle can be inscribed in the polygon of the base.

Prism. Definition. Elements. Types of prisms.

Prism- is a polyhedron, two of whose faces are equal polygons located in parallel planes, and the remaining faces are parallelograms.

Faces that are in parallel planes are called reasons prisms, and the remaining faces - side faces prisms.

Depending on the base of the prism there are:

1) triangular

2) quadrangular

3) hexagonal

A prism with lateral edges perpendicular to its bases is called straight prism.

A right prism is called regular if its bases are regular polygons.

TICKET 17.

Property of diagonals of a rectangular parallelepiped.

All four diagonals intersect at one point and bisect there.

In a rectangular parallelepiped, all diagonals are equal.

In a rectangular parallelepiped, the square of any diagonal is equal to the sum of the squares of its three dimensions.

Drawing the diagonal of the base AC, we obtain triangles AC 1 C and ACB. Both of them are rectangular: the first because the parallelepiped is straight and, therefore, edge CC 1 is perpendicular to the base; the second because the parallelepiped is rectangular and, therefore, a rectangle lies at its base. From these triangles we find:

AC 1 2 = AC 2 + CC 1 2 and AC 2 = AB 2 + BC 2

Therefore, AC 1 2 = AB 2 + BC 2 + CC 1 2 = AB 2 + AD 2 + AA 1 2.

Cases of mutual arrangement of two planes.

PROPERTY 1:

The lines of intersection of two parallel planes with a third plane are parallel.

PROPERTY 2:

Segments of parallel lines enclosed between two parallel planes are equal in length.

PROPERTY 3

Through every point in space that does not lie in a given plane, it is possible to draw a plane parallel to this plane, and only one.

TICKET 18.

Property of opposite faces of a parallelepiped.

The opposite faces of a parallelepiped are parallel and equal.

For example , the planes of parallelograms AA 1 B 1 B and DD 1 C 1 C are parallel, since the intersecting lines AB and AA 1 of the plane AA 1 B 1 are respectively parallel to the two intersecting lines DC and DD 1 of the plane DD 1 C 1. Parallelograms AA 1 B 1 B and DD 1 C 1 C are equal (that is, they can be combined by overlapping), since the sides AB and DC, AA 1 and DD 1 are equal, and the angles A 1 AB and D 1 DC are equal.

Surface areas of a prism, pyramid, regular pyramid.

Correct pyramid: Sfull. =3SASB+Sbas.

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, wherein 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 generatrices 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 cusp 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. From among the polyhedra, a group of regular polyhedra is distinguished, 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 one. If the edges of a prism are perpendicular to any projection plane, then lateral surface it 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 rotation. 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 revolution 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 problems and obtain geometric shapes 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.