Conditions for parallelism and perpendicularity

1 °. Condition of coplanarity of two planes

Let two planes be given:

A 1 x + B 1 y + C 1 z + D 1 = 0, n 1 = {A 1 ; B 1 ; C 1 } ≠ 0 ;(1)

A 2 x + B 2 y + C 2 z + D 2 = 0, n 2 = {A 2 ; B 2 ; C 2 } ≠ 0 .(2)

When are they coplanar (i.e. parallel or the same)? Obviously, this will be if and only if their normal vectors are collinear. Applying the coplanarity criterion, we obtain

Proposition 1. Two planes are coplanar if and only if the vector product of their normal vectors is equal to the zero vector:

[n 1 , n 2 ] = 0 .

2 °. The condition of coincidence of two planes

Proposition 2. Planes (1) and (2) coincide if and only if all four of their coefficients are proportional, i.e., there exists a number λ such that

A 2 = λ A 1 , B 2 = λ B 1 , C 2 = λ C 1 , D 2 = λ D 1 . (3)

Proof. Let conditions (3) be satisfied. Then the equation of the second plane can be written as follows:

λ A 1 x + λ B 1 y + λ C 1 z + λ D 1 = 0.

λ ≠ 0, otherwise it would be A 2 = B 2 = C 2 = D 2 = 0, which contradicts the condition n 2 ≠ 0 ... Therefore, the last equation is equivalent to equation (1), which means that the two planes coincide.

Suppose now, on the contrary, it is known that these planes coincide. Then their normal vectors are collinear, i.e., there exists a number λ such that

A 2 = λ A 1 , B 2 = λ B 1 , C 2 = λ C 1 .

Equation (2) can now be rewritten as:

λ A 1 x + λ B 1 y + λ C 1 z + D 2 = 0.

Multiplying equation (1) by λ, we obtain an equivalent equation for the first plane (since λ ≠ 0):

λ A 1 x + λ B 1 y + λ C 1 z + λ D 1 = 0.

Let's take some point ( x 0 , y 0 , z 0) from the first (and therefore the second) plane and substitute its coordinates into the last two equations; we get the correct equalities:

λ A 1 x 0 + λ B 1 y 0 + λ C 1 z 0 + D 2 = 0 ;

λ A 1 x 0 + λ B 1 y 0 + λ C 1 z 0 + λ D 1 = 0.

Subtracting the bottom from the top, we get D 2 - λ D 1 = 0, i.e. D 2 = λ D 1, QED.

3 °. Condition of perpendicularity of two planes

Obviously, for this it is necessary and sufficient that the normal vectors be perpendicular.

Proposition 3. Two planes are perpendicular if and only if the scalar product of normal vectors is zero:

(n 1 , n 2) = 0 .

Let the equation of the plane be given

Ax + By + Cz + D = 0, n = {A; B; C} ≠ 0 ,

and point M 0 = (x 0 , y 0 , z 0). We derive the formula for the distance from a point to a plane:

Take an arbitrary point Q = (x 1 , y 1 , z 1) lying in this plane. Its coordinates satisfy the plane equation:

Ax 1 + By 1 + Cz 1 + D = 0.

Note now that the required distance d is equal to the absolute value of the vector projection on the direction of the vector n (here we take the projection as a numerical value, not as a vector). Next, we apply the formula to calculate the projection:

A similar formula is valid for the distance d from point M 0 = (x 0 , y 0) plane to a straight line given by the general equation Ax + By + C = 0.

PROBLEMS C2 OF THE UNIFIED STATE EXAM IN MATH FOR FINDING THE DISTANCE FROM A POINT TO A PLANE

Kulikova Anastasia Yurievna

5th year student, department of mat. analysis, algebra and geometry EI KFU, RF, Republic of Tatarstan, Elabuga

Ganeeva Aigul Rifovna

scientific adviser, Ph.D. ped. Sciences, Associate Professor, EI KFU, RF, Republic of Tatarstan, Elabuga

In recent years, tasks for calculating the distance from a point to a plane have appeared in the tasks of the exam in mathematics. In this article, using the example of one problem, various methods of finding the distance from a point to a plane are considered. The most suitable method can be used to solve various problems. Having solved the problem with one method, another method can check the correctness of the result obtained.

Definition. The distance from a point to a plane that does not contain this point is the length of the perpendicular segment dropped from this point onto the given plane.

Task. Given a rectangular parallelepiped ABWITHDA 1 B 1 C 1 D 1 with sides AB=2, BC=4, AA 1 = 6. Find the distance from the point D to plane ASD 1 .

1 way. Using definition... Find the distance r ( D, ASD 1) from point D to plane ASD 1 (fig. 1).

Figure 1. First method

We will carry out DH⊥AS, therefore, by the theorem about three perpendiculars D 1 H⊥AS and (DD 1 H)⊥AS... We will carry out straight DT perpendicular D 1 H... Straight DT lies in the plane DD 1 H, hence DT⊥AC... Hence, DT⊥ASD 1.

ADC find the hypotenuse AS and height DH

From a right triangle D 1 DH find the hypotenuse D 1 H and height DT

Answer: .

Method 2.Volume method (using an auxiliary pyramid). The problem of this type can be reduced to the problem of calculating the height of a pyramid, where the height of the pyramid is the desired distance from a point to a plane. Prove that this height is the desired distance; find the volume of this pyramid in two ways and express this height.

Note that with this method there is no need to construct a perpendicular from a given point to a given plane.

Rectangular parallelepiped - a parallelepiped, all of whose faces are rectangles.

AB=CD=2, BC=AD=4, AA 1 =6.

The desired distance is the height h pyramids ACD 1 D dropped from the top D on the basis ACD 1 (fig. 2).

Let's calculate the volume of the pyramid ACD 1 D two ways.

Calculating, in the first way we take as the base ∆ ACD 1, then

Calculating, in the second way we take as the base ∆ ACD, then

Equating the right-hand sides of the last two equalities, we obtain

Figure 2. Second method

Of right-angled triangles ASD, ADD 1 , CDD 1 find the hypotenuses using the Pythagorean theorem

ACD

Calculate the area of ​​the triangle ASD 1 using Heron's formula

Answer: .

Method 3. Coordinate method.

Let a point be given M(x 0 ,y 0 ,z 0) and the plane α given by the equation ax+by+cz+d= 0 in a rectangular Cartesian coordinate system. Distance from point M to the plane α can be calculated by the formula:

Let's introduce a coordinate system (Fig. 3). Origin of coordinates at a point V;

Straight AB- axis NS, straight Sun- axis y, straight BB 1 - axis z.

Figure 3. The third method

B(0,0,0), A(2,0,0), WITH(0,4,0), D(2,4,0), D 1 (2,4,6).

Let be ax +by+ cz+ d= 0 - plane equation ACD 1 . Substituting the coordinates of the points into it A, C, D 1 we get:

Plane equation ACD 1 will take the form

Answer: .

Method 4. Vector method.

Let us introduce a basis (Fig. 4),.

Figure 4. Fourth method

Consider in space some plane π and an arbitrary point M 0. Let us choose for the plane unit normal vector n with the beginning at some point М 1 ∈ π, and let p (М 0, π) be the distance from point М 0 to the plane π. Then (fig.5.5)

p (М 0, π) = | pr n M 1 M 0 | = | nM 1 M 0 |, (5.8)

since | n | = 1.

If the plane π is given in a rectangular coordinate system by its general equation Ax + By + Cz + D = 0, then its normal vector is a vector with coordinates (A; B; C) and as a unit normal vector one can choose

Let (x 0; y 0; z 0) and (x 1; y 1; z 1) coordinates of points M 0 and M 1. Then the equality Ax 1 + By 1 + Cz 1 + D = 0 holds, since the point M 1 belongs to the plane, and we can find the coordinates of the vector M 1 M 0: M 1 M 0 = (x 0 -x 1; y 0 -y 1; z 0 -z 1). Writing down scalar product nM 1 M 0 in coordinate form and transforming (5.8), we obtain

since Ax 1 + By 1 + Cz 1 = - D. So, to calculate the distance from a point to a plane, you need to substitute the coordinates of the point in the general equation of the plane, and then divide the absolute value of the result by a normalizing factor equal to the length of the corresponding normal vector.

Let there be a plane ... Let's draw a normal
through the origin of coordinates O. Let
- the angles formed by the normal with coordinate axes.
... Let be - the length of the normal segment
before crossing the plane. Assuming that the direction cosines of the normal are known , we derive the equation of the plane .

Let be
) Is an arbitrary point of the plane. The unit normal vector has coordinates. Find the projection of the vector
to normal.

Since the point M belongs to the plane, then

.

This is the equation of a given plane, called normal .

Distance from point to plane

Let the plane be given ,M*
- point in space, d - its distance from the plane.

Definition. By deviation points M * from the plane is called the number ( + d), if M* lies on the other side of the plane where the positive direction of the normal points , and the number (- d), if the point is located on the other side of the plane:

.

Theorem. Let the plane with unit normal given by the normal equation:

Let be M*
- point in space Deviation t. M* from the plane is given by the expression

Proof. Projection t.
* by the normal we denote Q. Point deviation M * from the plane is

.

Rule. To find deviation T. M* from the plane, you need to substitute the coordinates m into the normal equation of the plane. M* ... The distance from a point to a plane is .

Reducing the general equation of the plane to normal form

Let one and the same plane be given by two equations:

General equation,

Normal equation.

Since both equations define the same plane, their coefficients are proportional:

We square the first three equalities and add:

From here we find - normalizing factor:

. (10)

Multiplying the general equation of the plane by the normalizing factor, we get the normal equation of the plane:

Examples of tasks on the topic "Plane".

Example 1. Make a plane equation passing through a given point
(2,1, -1) and parallel to the plane.

Solution... Normal to plane :
... Since the planes are parallel, the normal is also the normal to the desired plane ... Using the equation of the plane passing through a given point (3), we obtain for the plane the equation:

Answer:

Example 2. The base of the perpendicular dropped from the origin to the plane , is the point
... Find the equation of the plane .

Solution... Vector
is normal to the plane ... Point M 0 belongs to the plane. You can use the equation of a plane passing through a given point (3):

Answer:

Example 3. Construct plane passing through the points

and perpendicular to the plane :.

Therefore, for some point M (x, y, z) belonged to the plane , it is necessary that three vectors
were coplanar:

=0.

It remains to reveal the determinant and reduce the resulting expression to the form of the general equation (1).

Example 4. Plane given by the general equation:

Find point deviation
from a given plane.

Solution... Let us bring the equation of the plane to its normal form.

,

.

Substitute in the resulting normal equation the coordinates of the point M *.

.

Answer:
.

Example 5. Whether the line intersects the plane.

Solution... To segment AB crossed the plane, deviations and from the plane must have different signs:

.

Example 6. Intersection of three planes at one point.

.

The system has a unique solution, therefore, three planes have one common point.

Example 7. Finding the bisectors of a dihedral angle formed by two given planes.

Let be and - deviation of some point
from the first and second planes.

On one of the bisectral planes (corresponding to the angle at which the origin of coordinates lies), these deviations are equal in magnitude and sign, and on the other, they are equal in magnitude and opposite in sign.

This is the equation of the first bisectral plane.

This is the equation of the second bisectral plane.

Example 8. Locating two given points and relative to the dihedral angles formed by these planes.

Let be
... Determine if there are points in one, adjacent or vertical corners and .

a). If and lie on one side of and from , then they lie in one dihedral corner.

b). If and lie on one side of and different from , then they lie in adjacent corners.

v). If and lie on opposite sides of and , then they lie in the vertical corners.

Coordinate systems 3

Lines on plane 8

First order lines. Straight lines on a plane. ten

Angle between straight lines 12

General equation of the straight line 13

Incomplete Equation of First Degree 14

Equation of a straight line “in line segments” 14

Joint study of equations of two straight lines 15

Normal to straight 15

Angle between two straight lines 16

Canonical equation of the line 16

Parametric Equations of a Line 17

Normal (normalized) equation of a straight line 18

Distance from point to line 19

Equation of a beam of straight lines 20

Examples of problems on the topic "line on a plane" 22

Vector product of vectors 24

Cross Product Properties 24

Geometric properties 24

Algebraic properties 25

Expression of the cross product in terms of the coordinates of the factors 26

Mixed product of three vectors 28

The geometric meaning of the mixed work 28

Expression of the mixed product in terms of the coordinates of the vectors 29

Examples of problem solving

, Competition "Presentation for the lesson"

Class: 11

Lesson presentation

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the presentation options. If you are interested in this work, please download the full version.

Goals:

• generalization and systematization of knowledge and skills of students;
• development of skills to analyze, compare, draw conclusions.

Equipment:

• multimedia projector;
• a computer;
• worksheets with texts of tasks

PROCESS OF THE LESSON

I. Organizational moment

II. Knowledge update stage(slide 2)

We repeat how the distance from a point to a plane is determined

III. Lecture(slides 3-15)

In this lesson, we will look at various ways to find the distance from a point to a plane.

First method: step-by-step computational

Distance from point M to plane α:
- is equal to the distance to the plane α from an arbitrary point P lying on the straight line a, which passes through the point M and is parallel to the plane α;
- is equal to the distance to the plane α from an arbitrary point P lying on the plane β, which passes through the point M and is parallel to the plane α.

Let's solve the following tasks:

№1. In the cube A ... D 1 find the distance from point C 1 to plane AB 1 C.

It remains to calculate the value of the length of the segment O 1 N.

№2. In a regular hexagonal prism A ... F 1, all edges of which are equal to 1, find the distance from point A to the plane DEA 1.

The next method: volume method.

If the volume of the pyramid ABCM is equal to V, then the distance from point M to the plane α containing ∆ABS is calculated by the formula ρ (M; α) = ρ (M; ABC) =
When solving problems, we use the equality of the volumes of one figure, expressed in two different ways.

Let's solve the following problem:

№3. The edge AD of the pyramid DABC is perpendicular to the plane of the base ABC. Find the distance from A to the plane passing through the midpoints of the ribs AB, AC and AD, if.

When solving problems coordinate method the distance from point M to plane α can be calculated by the formula ρ (M; α) = , where M (x 0; y 0; z 0), and the plane is given by the equation ax + by + cz + d = 0

Let's solve the following problem:

№4. In the unit cube A ... D 1 find the distance from point A1 to plane BDC 1.

We introduce a coordinate system with the origin at point A, the y-axis will run along the AB edge, the x-axis along the AD edge, and the z-axis along the AA 1 edge. Then the coordinates of points B (0; 1; 0) D (1; 0; 0;) C 1 (1; 1; 1)
Let's compose the equation of the plane passing through the points B, D, C 1.

Then - dx - dy + dz + d = 0 x + y - z - 1 = 0. Therefore, ρ =

The next method that can be used when solving problems of this type is - method of support tasks.

The application of this method consists in the application of known support problems, which are formulated as theorems.

Let's solve the following problem:

№5. In the unit cube A ... D 1 find the distance from point D 1 to plane AB 1 C.

Consider the application vector method.

№6. In the unit cube A ... D 1 find the distance from point A 1 to the plane BDC 1.

So, we looked at the various methods that can be used to solve this type of problem. The choice of this or that method depends on the specific task and your preferences.

IV. Working in groups

Try to solve the problem in different ways.

№1. The edge of the cube A ... D 1 is equal to. Find the distance from vertex C to plane BDC 1.

№2. In a regular tetrahedron ABCD with an edge, find the distance from point A to the plane BDC

№3. In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance from A to the plane BCA 1.

№4. In a regular rectangular pyramid SABCD, all edges of which are equal to 1, find the distance from A to the SCD plane.

V. Lesson summary, homework, reflection