Possible cases of mutual arrangement of a straight line and a plane. Plane in space - necessary information
The mutual position of a straight line and a plane is determined by the number of common points :
1) if a line has two common points with a plane, then it belongs to this plane,
2) if a line has one common point with a plane, then the line intersects the plane,
3) if the point of intersection of a line with a plane is removed to infinity, then the line and the plane are parallel.
Problems in which the relative position of various geometric shapes relative to each other is determined are called positional problems.
The straight line belonging to the plane was considered earlier.
Line parallel to plane, if it is parallel to some straight line lying in this plane. To construct such a straight line, it is necessary to specify any straight line in the plane and draw the required one parallel to it.
Rice. 1.53 Fig. 1.54 Fig.1.55
Let through the dot BUT(Fig. 1.53) it is necessary to draw a straight line AB, parallel to the plane Q, given by a triangle CDF. To do this, through the frontal projection of the point a / points BUT make a frontal projection a / in / desired line parallel to the frontal projection of any line lying in the plane R, e.g. straight CD (a / in /!!s / d /). Through a horizontal projection a points BUT parallel sd make a horizontal projection aw desired line AB (av11 sd). Straight AB parallel to the plane R, given by a triangle CDF.
Of all the possible positions of a line intersecting a plane, we note the case when the line is perpendicular to the plane. Consider the properties of projections of such a line.
Rice. 1.56 Fig. 1.57
The line is perpendicular to the plane(a special case of the intersection of a straight line with a plane) if it is perpendicular to any straight line lying in the plane. To construct projections of a perpendicular to a plane in general position, this is not enough without transforming the projections. Therefore, an additional condition is introduced: a line is perpendicular to a plane if it is perpendicular to two intersecting principal lines(to construct projections, the projection condition is used right angle). In this case: the horizontal and frontal projections of the perpendicular are perpendicular, respectively, to the horizontal projection of the horizontal and the frontal projection of the front of the given plane general position(Fig. 1.54). When a plane is specified by traces, the projections of the perpendicular are perpendicular, respectively, to the frontal - to the frontal trace, horizontal - to the horizontal trace of the plane (Fig. 1.55).
Intersection of a straight line with a projecting plane. Consider a straight line that intersects a plane when the plane is in a particular position.
A plane perpendicular to the projection plane (the projection plane) is projected onto it as a straight line. On this line (the projection of the plane) there must be a corresponding projection of the point at which some line intersects this plane (Fig. 1.56).
In Figure 1.56, the frontal projection of the point To line intersection AB with a triangle CDE is determined at the intersection of their frontal projections, because triangle CDE projected onto the frontal plane as a straight line. We find the horizontal projection of the point of intersection of the line with the plane (it lies on the horizontal projection of the line). Using the method of competing points, we determine the visibility of the line AB relative to the plane of the triangle CDE on the horizontal projection plane.
Figure 1.59 shows a horizontally projecting plane P and a straight line in general position AB. Because plane R is perpendicular to the horizontal plane of projections, then everything that is in it is projected onto the horizontal plane of projections on its trace, including the point of its intersection with the line AB. Therefore, in the complex drawing we have a horizontal projection of the point of intersection of the line with the plane R. According to the belonging of the point to the straight line, we find the frontal projection of the point of intersection of the straight line AB with a plane R. Determine the visibility of the line on the frontal projection plane.
Rice. 1.58 Fig. 1.59
Figure 1.58 shows a comprehensive drawing of the construction of projections of the point of intersection of the line AB with horizontal level plane G. Frontal plane trace G is its frontal projection. Frontal projection of the point of intersection of the plane G with a straight line AB are determined at the intersection of the frontal projection of the straight line and the frontal trace of the plane. Having a frontal projection of the point of intersection, we find the horizontal projection of the point of intersection of the line AB with plane G.
Figure 1.57 shows a plane in general position, given by a triangle CDE and front projection line AB? intersecting the plane at a point K. Frontal projection of a point - k / matches the points a / and b/ . To construct a horizontal projection of the intersection point, draw through the point K in plane CDE straight line (eg. 1-2 ). Let's construct its frontal projection, and then horizontal. Dot K is the point of intersection of the lines AB and 1-2. That is the point K simultaneously belongs to the line AB and the plane of the triangle and, therefore, is the point of their intersection.
The intersection of two planes. A straight line of intersection of two planes is defined by two points, each of which belongs to both planes, or by one point, which belongs to two planes, and the known direction of the line. In both cases, the task is to find a point common to two planes.
Intersection of projecting planes. Two planes can be parallel to each other or intersect. Consider the cases of mutual intersection of planes.
A straight line obtained at the mutual intersection of two planes is completely determined by two points, each of which belongs to both planes, therefore, it is necessary and sufficient to find these two points belonging to the line of intersection of two given planes.
Therefore, in the general case, to construct a line of intersection of two planes, it is necessary to find any two points, each of which belongs to both planes. These points determine the line of intersection of the planes. To find each of these two points, you usually have to perform special constructions. But if at least one of the intersecting planes is perpendicular (or parallel) to any projection plane, then the construction of the projection of the line of their intersection is simplified.
Rice. 1.60 Fig. 1.61
If the planes are given by traces, then it is natural to look for the points that define the line of intersection of the planes at the points of intersection of the traces of the planes of the same name in pairs: the line passing through these points is common to both planes, i.e. their line of intersection.
Consider special cases of the location of one (or both) of the intersecting planes.
The complex drawing (Fig. 1.60) shows horizontally projecting planes P and Q. Then the horizontal projection of their intersection line degenerates into a point, and the frontal projection into a straight line perpendicular to the axis ox.
The complex drawing (Fig. 1.61) shows the planes of private position: the plane R perpendicular to the horizontal projection plane (horizontal projection plane) and the plane Q- horizontal level plane. In this case, the horizontal projection of their line of intersection will coincide with the horizontal trace of the plane R, and the frontal - with a frontal trace of the plane Q.
In the case of specifying planes by traces, it is easy to establish that these planes intersect: if at least one pair of traces of the same name intersect, then the planes intersect each other.
The foregoing applies to planes defined by intersecting traces. If both planes have traces parallel to each other on the horizontal and frontal planes, then these planes can be parallel or intersect. The mutual position of such planes can be judged by constructing a third projection (third trace). If the traces of both planes on the third projection are also parallel, then the planes are parallel to each other. If the traces on the third plane intersect, then the planes given in space intersect.
The complex drawing (Fig. 1.62) shows front-projecting planes defined by a triangle ABC and DEF. The projection of the line of intersection on the frontal projection plane is a point, i.e. Since the triangles are perpendicular to the frontal projection plane, their line of intersection is also perpendicular to the frontal projection plane. Therefore, the horizontal projection of the line of intersection of triangles ( 12 ) is perpendicular to the axis ox. The visibility of the elements of the triangles on the horizontal projection plane is determined using competing points (3,4).
On the complex drawing (Fig. 1.63), two planes are set: one of which is a triangle ABC general position, the other - a triangle DEF perpendicular to the frontal projection plane, i.e. located in a private position (front-projecting). Frontal projection of the line of intersection of triangles ( 1 / 2 / ) is found based on common points that simultaneously belong to both triangles (everything that is in the front-projecting triangle DEF on the frontal projection will result in a line - its projection onto the frontal plane, including the line of its intersection with the triangle ABC. According to the belonging of the points of intersection to the sides of the triangle ABC, we find the horizontal projection of the line of intersection of the triangles. Using the method of competing points, we determine the visibility of triangle elements on the horizontal plane of projections.
Rice. 1.63 Fig. 1.64
Figure 1.64 shows a complex drawing of two planes defined by a triangle in general position ABC and horizontally projecting plane R, given by traces. Since the plane R- horizontally projecting, then everything that is in it, including the line of its intersection with the plane of the triangle ABC, on the horizontal projection will coincide with its
horizontal track. The frontal projection of the line of intersection of these planes is found from the condition that the points of the element belong to (sides) of the plane in general position.
In the case of specifying planes in general position not by traces, then to obtain the line of intersection of the planes, the point of meeting of the side of one triangle with the plane of another triangle is sequentially found. If planes in general position are not given by triangles, then the line of intersection of such planes can be found by introducing two auxiliary secant planes in turn - projecting (for specifying planes by triangles) or level for all other cases.
Intersection of a line in general position with a plane in general position. Previously, cases of intersection of planes were considered, when one of them was projecting. Based on this, we can find the point of intersection of a line in general position with a plane in general position by introducing an additional projecting mediator plane.
Before considering the intersection of planes in general position, consider the intersection of a line in general position with a plane in general position.
To find the meeting point of a line in general position with a plane in general position, it is necessary:
1) enclose a straight line in an auxiliary projecting plane,
2) find the line of intersection of the given and auxiliary planes,
determine a common point belonging simultaneously to two planes (this is their line of intersection) and a straight line.
Rice. 1.65 Fig. 1.66
Rice. 1.67 Fig. 1.68
The complex drawing (Fig. 1.65) shows a triangle CDE general position and direct AB general position. To find the point of intersection of a line with a plane, we conclude the line AB Q. Let's find the line of intersection ( 12 ) intermediary plane Q and given plane CDE. When constructing a horizontal projection of the intersection line, there is a common point To, simultaneously belonging to two planes and a given line AB. From the belonging of a point to a straight line, we find the frontal projection of the point of intersection of a straight line with a given plane. The visibility of the elements of a straight line on the projection planes is determined using competing points.
Figure 1.66 shows an example of finding the meeting point of a straight line AB, which is a horizontal line (the line is parallel to the horizontal plane of projections) and the plane R, in general position, given by traces. To find the point of their intersection, the line AB lies in the horizontally projecting plane Q. Then proceed as in the above example.
To find the meeting point of a horizontally projecting line AB with a plane in general position (Fig. 1.67), through the meeting point of a straight line with a plane (its horizontal projection coincides with the horizontal projection of the straight line itself) we draw a horizontal line (i.e. we bind the point of intersection of a straight line with a plane to a plane R). Having found the frontal projection of the drawn horizontal in the plane R, mark the frontal projection of the meeting point of the line AB with plane R.
To find the line of intersection of planes in general position, given by traces, it is enough to mark two common points that simultaneously belong to both planes. Such points are the points of intersection of their traces (Fig. 1.68).
To find the line of intersection of planes in general position, given by two triangles (Fig. 1.69), we sequentially find the point
meeting of the side of one triangle with the plane of another triangle. Taking any two sides from any triangle, enclosing them in mediators projecting planes, two points are found that simultaneously belong to both triangles - the line of their intersection.
Figure 1.69 shows a complex drawing of triangles ABC and DEF general position. To find the line of intersection of these planes:
1. We conclude the side Sun triangle ABC into the frontal projection plane S(the choice of planes is completely arbitrary).
2. Find the line of intersection of the plane S and plane DEF – 12 .
3. We mark the horizontal projection of the meeting point (common point of two triangles) To from intersection 12 and Sun and find its frontal projection on the frontal projection of the line Sun.
4. We draw the second auxiliary projecting plane Q across the side D.F. triangle DEF.
5. Find the line of intersection of the plane Q and triangle ABC - 3 4.
6. Mark the horizontal projection of the point L, which is the meeting point of the party D.F. with triangle plane ABC and find its frontal projection.
7. We connect the same-named projections of points To and L. to L- line of intersection of planes in general position, given by triangles ABC and DEF.
8. Using the method of competing points, we determine the visibility of the elements of triangles on the projection planes.
Since the above is also valid for the main lines of parallel planes, we can say that planes are parallel if their traces of the same name are parallel(Fig. 1.71).
Figure 1.72 shows the construction of a plane parallel to the given one and passing through the point BUT. In the first case, through the point BUT a straight line (front) is drawn parallel to a given plane G. Thus, a plane is drawn R containing a line parallel to a given plane G and parallel to it. In the second case, through the point BUT a plane is drawn, given by the main lines from the condition of parallelism of these lines to a given plane G.
Mutually perpendicular planes.If one plane contains
at least one line perpendicular to another plane, then such
planes are perpendicular. Figure 1.73 mutually perpendicular planes are shown. Figure 1.74 shows the construction of a plane perpendicular to the one given through the point BUT, using the condition of perpendicularity of a straight line (in this case, the main lines) to the plane.
In the first case, through the point BUT a frontal is drawn perpendicular to the plane R, its horizontal trace is constructed and a horizontal trace of the plane is drawn through it Q , perpendicular to the horizontal trace of the plane R. Through the resulting vanishing point Q X a frontal trace of the plane is drawn Q perpendicular to the front trace of the plane R.
In the second case, horizontal lines are drawn in the plane of the triangle BE and frontal bf and through a given point BUT we set the plane by intersecting straight lines (main lines) perpendicular to the plane of the triangle. To do this, draw through the point BUT horizontal and frontal. The horizontal projection of the horizontal of the desired plane ( N) we draw perpendicular to the horizontal projection of the horizontal of the triangle, the frontal projection of the front of the new plane ( M) is perpendicular to the frontal projection of the front of the triangle.
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.
Line and 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 of a straight line and a plane. If a line not belonging to a plane is parallel to some line in that plane, then it is also 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 intersecting lines. If one of the two lines lies in a plane, and the other intersects this plane at a point that does not belong to the first line, then these lines intersect.
Theorem of 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 intersects this plane.
3. Through a point outside a given line, one can draw a line parallel to the given line, 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, one can draw a plane parallel to the given one, and only one.
7. Two planes parallel to a third are parallel to each other.
8. Segments of parallel lines enclosed between parallel planes are equal.
Angles between lines and planes
Angle between line and plane the angle between the line and its projection onto the plane is called (the angle in Fig. 1).
Angle between skew lines is the angle between intersecting lines parallel respectively to the given skew lines.
dihedral angle A figure formed by two half-planes with a common straight line is called. Half planes are called faces , straight line edge dihedral angle.
Linear angle dihedral angle is the angle between the 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
The two lines are called perpendicular if they intersect at right angles.
A line that intersects 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.
A sign of perpendicularity of a straight line and a plane. If a line intersecting a plane is perpendicular to two intersecting lines in that 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 oblique lines are drawn from one point outside the plane, then:
1) inclined, 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 slope is larger.
Three perpendiculars 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 an orthogonal projection of a polygon onto a plane is equal to the product of the area of the polygon times the cosine of the angle between the plane of the polygon and the projection plane.
Construction.
1. On the plane a draw a straight line a.
3. In plane b through a point BUT let's draw a straight line b, parallel to the line a.
4. Built a straight line b parallel to the plane a.
Proof. On the basis of 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 line a in plane a is chosen arbitrarily.
Example 2 Determine how far a point is from a plane BUT if straight AB intersects the plane at an angle of 45º, the distance from the point BUT to the point AT, 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- the angle between the line AB and plane a. Triangle ABC- rectangular as AC- perpendicular. Desired distance from a point BUT to the plane - this is the leg AC right triangle. Knowing the angle and hypotenuse cm, we find the leg AC:
Answer: 3 cm
Example 3 Determine how far from the plane isosceles triangle find a point 13 cm away from each of the vertices of the triangle if the base and height of the triangle are 8 cm each?
Solution. Let's make a drawing (Fig. 6). Dot S away from points BUT, AT and FROM to the same distance. So inclined SA, SB and SC equal, SO- the common perpendicular of these inclined. By the oblique and projection theorem AO = BO = CO.
Dot O- the center of a circle circumscribed about a triangle ABC. Let's find its radius:
where Sun- base;
AD is the height of the given isosceles triangle.
Finding the sides of a triangle ABC from a right triangle ABD according to the Pythagorean theorem:
Now we find OV:
Consider a triangle SOB: SB= 13 cm, OV= = 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 dot M, which does not belong to any of them, straight lines are drawn a and b, which cross a at points BUT 1 and AT 1 , and the plane b- at points BUT 2 and AT 2. Find BUT 1 AT 1 if it is known that MA 1 = 8 cm, BUT 1 BUT 2 = 12 cm, BUT 2 AT 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 consider each of them. Two intersecting lines a and b define a plane. This plane intersects two parallel planes a and b along parallel lines BUT 1 AT 1 and BUT 2 AT 2 according to Theorem 5 on parallel lines and parallel planes.
triangles MA 1 AT 1 and MA 2 AT 2 are similar (angles BUT 2 MV 2 and BUT 1 MV 1 - vertical, corners MA 1 AT 1 and MA 2 AT 2 - internal cross lying with parallel lines BUT 1 AT 1 and BUT 2 AT 2 and secant BUT 1 BUT 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 dot BUT plane g direct AB forming an angle with the plane a. Through a straight line AB plane drawn r, forming with the plane g corner b. Find the angle between the projection of the line AB to the plane g and plane r.
Solution. Let's make a drawing (Fig. 8). From a point AT drop a perpendicular to the plane g. Linear dihedral angle between planes g and r is the angle AD DBC, on the basis of the perpendicularity of the line and the plane, since and On the basis of the perpendicularity of the planes, the 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 FROM to the plane r, denote it Find the sine of this angle of a right triangle MYSELF. We introduce an auxiliary segment a = sun. From a triangle ABC: From a triangle Navy find
Then the required angle
Answer:
Tasks for independent solution
I level
1.1. Through a point, draw a line perpendicular to two given skew 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 that intersect the second plane at points BUT 1 , AT 1 , FROM one . Prove that the triangles are equal ABC and BUT 1 AT 1 FROM 1 .
1.4. From the top BUT rectangle ABCD erected perpendicular AM to its plane.
1) prove that the triangles MBC and MDC- rectangular;
2) indicate among the segments MB, MC, MD and MA segment of the greatest and the smallest length.
1.5. The faces of one dihedral angle are respectively parallel to the faces of the other. Determine what is the relationship between the values of these dihedral angles.
1.6. Find the value of the dihedral angle if the distance from the point taken on one face to the edge is 2 times 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 lines are drawn, forming an angle of 60º. The projections of the inclined planes are mutually perpendicular. Find the lengths of the obliques.
1.8. From the top AT square ABCD erected perpendicular BE to the plane of the square. The angle of inclination of the plane of the triangle ACE to the plane of the square is j, the side of the square is a ACE.
II level
2.1. Through a point that does not belong to any of the two intersecting lines, draw a line that intersects both given lines.
2.2. Parallel lines a, b and With do not lie in the same plane. Through the dot BUT on a straight line a drawn perpendicular to the lines b and With, intersecting them respectively at the points AT and FROM. Prove that the line Sun perpendicular to straight lines b and With.
2.3. Through the top BUT right triangle ABC a plane drawn parallel to Sun. Triangle legs AC= 20 cm, Sun\u003d 15 cm. The projection of one of the legs on the plane is 12 cm. Find the projection of the hypotenuse.
2.4. In one of the faces of a 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 on the second edge to the first edge.
2.5. Line ends AB belong to the faces of a dihedral angle equal to 90º. Distance from points BUT and AT up to the edge are equal respectively AA 1 = 3 cm, BB 1 \u003d 6 cm, distance between points on the edge Find the length of the segment AB.
2.6. From a point separated from the plane by a distance a, two inclined ones are drawn, forming angles of 45º and 30º with the plane, and between themselves an angle of 90º. Find the distance between the bases of the slopes.
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 is at the same distance from its vertices. Find this distance.
2.8. From the center O circle inscribed in a triangle ABC, perpendicular to the plane of the triangle OM. Find the distance from a point M to the sides of the triangle, if AB = BC = 10 cm AC= 12 cm, OM= 4 cm.
2.9. Distances from a point M to the sides and vertex of the right angle are respectively 4 cm, 7 cm and 8 cm. Find the distance from the point M to the plane of the right angle.
2.10. Through the base AB isosceles triangle ABC plane drawn at an angle b to the plane of the triangle. Vertex FROM removed from the plane at a distance a. Find the area of a triangle ABC if the base AB of an isosceles triangle is equal to its height.
III level
3.1. Rectangle layout ABCD with the parties a and b folded diagonally BD so that the planes of the triangles bad and BCD become 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 large lateral 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 trapezium if the vertices of their acute angles coincide.
3.3 Cube is given ABCDA 1 B 1 C 1 D one . Find the angle between the line CD 1 and plane bdc 1 .
3.4. on edge AB Cuba ABCDA 1 B 1 C 1 D 1 point taken R is the middle of this edge. Construct a section of a cube by a plane passing through the points C 1 PD and find the area of this section if the edge of the cube is a.
3.5. Across the side AD rectangle ABCD plane drawn a so that the diagonal BD makes an angle of 30 degrees with this plane. Find the angle between the plane of the rectangle and the plane a, if AB = a, AD=b. Determine 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 called a polyhedron whose two faces are equal n-gons (grounds) , lying in parallel planes, and the remaining n faces are parallelograms (side faces) . Side rib prism is the side of the lateral face that does not belong to the base.
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 oblique . correct A prism is a straight prism whose bases are regular polygons.
Height prism is called 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 A section of a prism by a plane passing through two side edges that do not belong to the same face is called. Perpendicular section called the section of the prism by a plane perpendicular to the lateral edge of the prism.
Side surface area prism is the sum of the areas of all side faces. area full surface the sum of the areas of all the faces of the prism is called (i.e., the sum of the areas of the side faces and the areas of the bases).
For an arbitrary prism, the formulas are true:
where l is the length of the side rib;
H- height;
P
Q
S side
S full
S main is the area of the bases;
V is the volume of the prism.
For a straight prism, the following formulas are true:
where p- the perimeter of the base;
l is the length of the side rib;
H- height.
Parallelepiped A prism whose base is a parallelogram is called. 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 oblique . A right parallelepiped whose base is a rectangle is called rectangular. A rectangular parallelepiped in which all edges are 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 the box is a prism, its main elements are defined in the same way as they are defined for prisms.
Theorems.
1. The diagonals of the parallelepiped intersect at one point and bisect 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 cuboid are equal to each other.
For an arbitrary parallelepiped, the following formulas are true:
where l is the length of the side rib;
H- height;
P is the perimeter of the perpendicular section;
Q– Area of perpendicular section;
S side is the lateral surface area;
S full is the total surface area;
S main is the area of the bases;
V is the volume of the prism.
For a right parallelepiped, the following formulas are true:
where p- the perimeter of the base;
l is the length of the side rib;
H is the height of the right parallelepiped.
For a rectangular parallelepiped, the following formulas are true:
where p- the perimeter of the base;
H- height;
d- diagonal;
a,b,c– measurements of a parallelepiped.
The correct formulas for a cube are:
where a is the length of the rib;
d is the diagonal of the cube.
Example 1 The diagonal of a rectangular cuboid is 33 dm, and its measurements are related as 2:6:9. Find the measurements of the cuboid.
Solution. To find the dimensions of the parallelepiped, we use formula (3), i.e. the fact that the square of the hypotenuse of a cuboid is equal to the sum of the squares of its dimensions. Denote by k coefficient of proportionality. Then the dimensions of the parallelepiped will be equal to 2 k, 6k and 9 k. We write formula (3) for the problem data:
Solving this equation for k, we get:
Hence, the dimensions of the parallelepiped are 6 dm, 18 dm and 27 dm.
Answer: 6 dm, 18 dm, 27 dm.
Example 2 Find volume of oblique triangular prism, which is based on equilateral triangle with a side of 8 cm, if the side edge is equal to the side of the base and is 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 \u200b\u200bits 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's calculate it:
The height of a prism is the distance between its bases. From the top BUT 1 of the upper base we lower the perpendicular to the plane of the lower base BUT 1 D. Its length will be the height of the prism. Consider D BUT 1 AD: since this is the angle of inclination of the side rib BUT 1 BUT to the base plane BUT 1 BUT= 8 cm. From this triangle we find BUT 1 D:
Now we calculate the volume using formula (1):
Answer: 192 cm3.
Example 3 The lateral edge of a regular hexagonal prism is 14 cm. The area of \u200b\u200bthe 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 AA 1 DD 1 , since the diagonal AD regular hexagon ABCDEF is the largest. In order to calculate the lateral surface area of a prism, it is necessary to know the side of the base and the length of the lateral rib.
Knowing the area of the diagonal section (rectangle), we find the diagonal of the base.
Because , then
Since then AB= 6 cm.
Then the perimeter of the base is:
Find the area of the lateral surface of the prism:
The area of a regular hexagon with a side of 6 cm is:
Find the total surface area of the prism:
Answer:
Example 4 The base of a right parallelepiped is a rhombus. The areas of diagonal sections are 300 cm 2 and 875 cm 2. Find the area of the side surface of the parallelepiped.
Solution. Let's make a drawing (Fig. 5).
Denote the side of the rhombus by a, the diagonals of the rhombus d 1 and d 2 , the height of the box h. To find the lateral surface area of a straight 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.
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 get the equality We get the following:
From the first two equalities, we express and substitute 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 side 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 of 3 cm and 4 cm. The side edge equal to cm is inclined to the base plane at an angle of 60 °. Find the volume of the parallelepiped.
1.7. Calculate the surface area of a cuboid if two edges and a diagonal emanating from the same vertex are 11 cm, cm and 13 cm, respectively.
1.8. Determine the weight of a stone column having 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/cm3.
1.9. Find the area of the diagonal section of a cube if the diagonal of its face is 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 cm.
II level
2.1. The base of an inclined prism is an equilateral triangle with side cm. The lateral edge is inclined to the base plane 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. The vertex A 1 is projected into the center of the triangle ABC. The 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 lower 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 base plane.
2.5. The base of a straight prism is isosceles trapezium, whose bases are 4 cm and 14 cm, and the diagonal is 15 cm. The two side 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 which forms angles of 30° and 40° with the side faces.
2.8. The diagonals of the base of a straight 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 cuboid in which the diagonals of the side faces coming out of 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.
III level
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 cm2, and the diagonal of the side face is inclined to the base at an angle of 60°. Find the sectional area.
3.2. The base of the prism is a square ABCD, all vertices of which are equidistant from the top A 1 of the upper base. The angle between the side edge and the plane of the base is 60°. The side of the base is 12 cm. Construct a section of the prism by a plane passing through the vertex C, perpendicular to the edge AA 1 and find its area.
3.3. The base of a right prism is an isosceles trapezoid. The area of the diagonal section and the area of the 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 straight 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 cuboid, 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 area of the lateral surface of the parallelepiped if its diagonal is d.
3.7. The area of the section of a cube that is regular hexagon, is equal to cm 2. Find the surface area of the cube.
The mutual arrangement of a straight line and a plane in space admits three cases. A line and a plane can intersect at one point. They may be parallel. Finally, a line can lie in a plane. Finding out the specific situation for a straight line and a plane depends on the way they are described.
Assume that the plane π is given general equationπ: Ax + By + Cz + D = 0, and the line L is canonical equations(x - x 0) / l \u003d (y - y 0) / m \u003d (z - z 0) / n. The straight line equations give the coordinates of the point M 0 (x 0; y 0; z 0) on the straight line and the coordinates of the directing vector s = (l; m; n) of this straight line, and the plane equation - the coordinates of its normal vector n = (A; B; C).
If the line L and the plane π intersect, then the direction vector s of the line is not parallel to the plane π. Hence, the normal vector n of the plane is not orthogonal to the vector s, i.e. their dot product is not zero. In terms of the coefficients of the equations of the line and the 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 "parallel" and "the line belongs to the plane", we need to check whether point of a line in a given plane.
Thus, all three cases of the relative position of the line and the plane are separated by checking the corresponding conditions:
If the line L is given by its general equations:
then the relative position of the straight line and the plane π can be analyzed as follows. From the general equations of the straight line and the general equation of the plane, we compose 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 qualifier (6.6)
different from zero. Finally, if system (6.6) has infinitely many solutions, then the line belongs to the plane.
The angle between a line and a plane. The angle φ between the line L: (x - x 0) / l \u003d (y - y 0) / m \u003d (z - z 0) / n and the plane π: Ax + By + Cz + D \u003d 0 is within 0 ° (in case of parallelism) to 90° (in case of perpendicularity of a line and a plane). The sine of this angle is equal to |cosψ|, where ψ is the angle between the direction vector of the line s and the normal vector n of the plane (Fig. 6.4). Calculating the cosine of the angle between two vectors in terms of their coordinates (see (2.16)), we obtain
The condition of perpendicularity of a line and a plane is equivalent to the fact that the normal vector of the plane and the direction vector of the line are collinear. In terms of the coordinates of the vectors, this condition is written as a double equality
In planimetry, the plane is one of the main figures, therefore, it is very important to have a clear idea of \u200b\u200bit. This article was created to cover this topic. First, the concept of a plane is given, its graphic representation and plane designations are shown. Further, the plane is considered together with a point, a straight line or another plane, while options arise from the relative position in space. In the second, third and fourth paragraphs of the article, all variants of the mutual arrangement of two planes, a straight line and a plane, as well as a point and a plane, are analyzed, the main axioms and graphic illustrations are given. In conclusion, the main ways of specifying a plane in space are given.
Page navigation.
Plane - basic concepts, notation and image.
The simplest and basic geometric shapes in three-dimensional space are a point, a line and a plane. We already have an idea of a point and a line in the plane. If we place a plane on which points and lines are depicted in three-dimensional space, then we will get points and lines in space. The idea of a plane in space allows you to get, for example, the surface of a table or wall. However, a table or wall has finite dimensions, and the plane extends beyond their boundaries to infinity.
Points and lines in space are denoted in the same way as on a plane - in capital and small Latin letters, respectively. For example, points A and Q, lines a and d. If two points are given that lie on a line, then the line can be denoted by two letters corresponding to these points. For example, the 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 the drawing. The plane is usually depicted as a parallelogram or an arbitrary simple closed area.
The plane is usually considered together with points, straight lines or other planes, and various variants of their mutual arrangement arise. We turn to their description.
Mutual arrangement of a plane and a point.
Let's start with an axiom: there are points in every plane. From it follows the first variant of the mutual arrangement of the plane and the point - the point may belong to the plane. In other words, a plane can pass through a point. To indicate the belonging of a point to any plane, the symbol "" is used. For example, if the plane passes through point A, then you can briefly write .
It should be understood that there are infinitely many points on a given plane in space.
The following axiom shows how many points in space must be marked in order for them to define a particular plane: through three points that do not lie on one straight line, a plane passes, and only one. If three points are known that lie in a plane, then the plane can be denoted by three letters corresponding to these points. For example, if the plane passes through points A, B and C, then it can be designated ABC.
Let us formulate one more axiom, which gives the second variant of the mutual arrangement 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 of space, and the fourth point may or may not lie on this plane. When shorthand, the symbol "" is used, which is equivalent to the phrase "does not belong."
For example, if point A does not lie in the plane, then a short notation is used.
Line and plane in space.
First, a 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. For a short record of belonging to a certain line of a given plane, use the symbol "". For example, the entry means that the line a lies in the plane.
Second, the line can intersect the plane. In this case, the line and the plane have one single common point, which is called the point of intersection of the line and the plane. With a short record, the intersection is denoted by the symbol "". For example, the entry means that the line a intersects the plane at the point M. When a certain line intersects a plane, the concept of an angle between a line and a plane arises.
Separately, it is worth dwelling on a line that intersects a plane and is perpendicular to any line lying in this plane. Such a line is called perpendicular to the plane. For a short record of perpendicularity, the symbol "" is used. For a deeper study of the material, you can refer to the article perpendicularity of a straight line and a plane.
Of particular importance in 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 can be parallel to a plane, that is, not have common points in it. When shorthand for parallelism, the symbol "" is used. For example, if the line a is parallel to the plane, then you can write . We recommend that you study this case in more detail by referring to the article parallelism of a straight 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 the 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 common points of these planes lie.
In this case, the concept of the 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 picture of this variant of the relative position of the planes.
Plane definition methods.
Now we list the main ways to set a specific plane in space.
First, 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 straight line, there is only one plane.
If a plane is fixed and given in three-dimensional space by specifying the coordinates of its three different points that do not lie on the same straight line, then we can write the equation of a plane passing through three given points.
The next two ways of specifying a plane are a consequence of the previous one. They are based on the consequences of the axiom about a plane passing through three points:
- a plane passes through a line and a point not lying on it, moreover, 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 of the article the equation of a plane passing through two intersecting lines).
The fourth way to define a plane in space is based on the definition of parallel lines. Recall that two lines in space are called parallel if they lie in the same plane and do not intersect. Thus, by specifying two parallel lines in space, we determine the only plane in which these lines lie.
If in three-dimensional space with respect to a rectangular coordinate system a plane is given in the indicated way, then we can compose an equation for a plane passing through two parallel lines.
I know high school in geometry lessons, the following theorem is proved: a single plane passes through a fixed point in space, perpendicular to a given line. Thus, we can define a plane if we specify a 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 given in the indicated way, then it is possible to compose an equation for a plane passing through a given point perpendicular to a given straight line.
Instead of a straight line perpendicular to a plane, one of the normal vectors of this plane can be specified. In this case, it is possible to write
Mutual arrangement of two straight lines
The following assertions express the necessary and sufficient signs mutual arrangement of two straight lines in space given by canonical equations
a) The lines intersect, i.e. do not lie on the same plane.
b) The lines intersect.
But the vectors and are not collinear (otherwise their coordinates are proportional).
in) The lines are parallel.
The vectors and are collinear, but the vector is not collinear with them.
G) The lines coincide.
All three vectors: , are collinear.
Proof. Let us prove the sufficiency of the indicated criteria
a) Consider the vector and direction vectors of given 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 of and these lines are assumed to be non-collinear, then the lines intersect.
in) If the direction vectors and given lines are collinear, then the lines are either parallel or coincide. When ( in) the lines are parallel, because by condition, the vector, the beginning of which is at the point of the first line, and the end - at the point of the second line, is not collinear and.
d) If all vectors and are collinear, then the lines coincide.
The necessity of features 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 given by the general equation
relative to a common Cartesian coordinate system.
A plane and a line intersect:
Plane and line are parallel:
The line lies on the plane:
Let us first prove the sufficiency of the indicated criteria. We write the equations of this straight line in parametric form:
Substituting into equation (2 (plane)) the coordinates of an arbitrary point of this line, taken from formulas (3), we will have:
1. If, then equation (4) has relatively t only decision:
which means that the given line and the given plane have only one common point, i.e. intersect.
2. If, then equation (4) is not satisfied for any value t, i.e. there is no point on a given line that lies on a given plane, therefore, the given line and 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, so the given line lies on a given plane.
The sufficient conditions for the mutual position of the line and the plane that we have derived are both necessary and can be proved immediately by contradiction.
From what has been proved follows the necessary and sufficient condition the fact that the vector is coplanar to the plane given by the general equation with respect to the general Cartesian coordinate system.