All equations of the plane, which are discussed in the following paragraphs, can be obtained from the general equation of the plane, and also reduced to the general equation of the plane. Thus, when they talk about the equation of a plane, they mean the general equation of a plane, unless otherwise stated.

Equation of a plane in segments.

View plane equation , where a, b and c are non-zero real numbers, is called equation of the plane in segments.

This name is not accidental. The absolute values ​​of the numbers a, b and c are equal to the lengths of the segments that the plane cuts off on the coordinate axes Ox, Oy and Oz, respectively, counting from the origin. The sign of the numbers a, b and c indicates in which direction (positive or negative) the segments should be plotted on the coordinate axes.

For example, let’s construct a plane in the rectangular coordinate system Oxyz, defined by the equation of the plane in segments . To do this, mark a point that is 5 units away from the origin in the negative direction of the abscissa axis, 4 units in the negative direction of the ordinate axis, and 4 units in the positive direction of the applicate axis. All that remains is to connect these points with straight lines. The plane of the resulting triangle is the plane corresponding to the equation of the plane in segments of the form .

For more complete information, refer to the article equation of a plane in segments, it shows the reduction of the equation of a plane in segments to the general equation of a plane, and there you will also find detailed solutions to typical examples and problems.

Normal plane equation.

The general plane equation of the form is called normal plane equation, If equal to one, that is, , And .

You can often see that the normal equation of a plane is written as . Here are the direction cosines of the normal vector of a given plane of unit length, that is, and p is a non-negative number equal to the distance from the origin to the plane.

The normal equation of a plane in the rectangular coordinate system Oxyz defines a plane that is removed from the origin by a distance p in the positive direction of the normal vector of this plane . If p=0, then the plane passes through the origin.

Let us give an example of a normal plane equation.

Let the plane be specified in the rectangular coordinate system Oxyz by the general equation of the plane of the form . This general equation of the plane is the normal equation of the plane. Indeed, the normal vector of this plane is has length equal to unity, since .

The plane equation in normal form allows you to find the distance from a point to a plane.

We recommend that you understand this type of plane equation in more detail, look at detailed solutions to typical examples and problems, and also learn how to reduce the general plane equation to normal form. You can do this by referring to the article.

Bibliography.

• Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Kiseleva L.S., Poznyak E.G. Geometry. Textbook for 10-11 grades of secondary school.
• Bugrov Ya.S., Nikolsky S.M. Higher mathematics. Volume one: elements of linear algebra and analytical geometry.
• Ilyin V.A., Poznyak E.G. Analytic geometry.

LECTURE 6-7. Elements of analytical geometry.

Surfaces and their equations.

Example 1.

Sphere .

Example 2.

F(x,y,z)=0(*),

This - surface equation

Examples:

x 2 + y 2 – z 2 = 0 (cone)

Plane.

Equation of a plane passing through a given point perpendicular to a given vector.

Let's consider a plane in space. Let M 0 (x 0, y 0, z 0) be a given point of the plane P, and be a vector perpendicular to the plane ( normal vector plane).

(1) – vector equation of the plane.

In coordinate form:

A(x - x 0) + B(y - y 0) + C(z - z 0) = 0 (2)

We obtained the equation of a plane passing through a given point.

General equation of the plane.

Let's open the brackets in (2): Ax + By + Cz + (-Ax 0 – By 0 – Cz 0) = 0 or

Ax + By + Cz + D = 0 (3)

The resulting equation of the plane linear, i.e. 1st degree equation with respect to coordinates x, y, z. Therefore the plane is first order surface .

Statement: Any equation linear with respect to x, y, z defines a plane.

Any plane m.b. is given by equation (3), which is called general equation of the plane.

Special cases of the general equation.

a) D=0: Ax + By + Cz = 0. Because the coordinates of the point O(0, 0, 0) satisfy this equation, then the plane specified by it passes through the origin.

b) С=0: Ax + By + D = 0. In this case, the normal vector of the plane , therefore the plane defined by the equation is parallel to the OZ axis.

c) C=D=0: Ax + By = 0. The plane is parallel to the OZ axis (since C=0) and passes through the origin of coordinates (since D=0). This means that it passes through the OZ axis.

d) B=C=0: Ax + D = 0 or . Vector, i.e. And . Consequently, the plane is parallel to the OY and OZ axes, i.e. is parallel to the YOZ plane and passes through the point .

Consider the cases yourself: B=0, B=D=0, A=0, A=D=0, A=C=0, A=B=0/

Equation of a plane passing through three given points.

Because all four points belong to the plane, then these vectors are coplanar, i.e. their mixed product is equal to zero:

We obtained the equation of a plane passing through three points in vector form.

In coordinate form:

(7)

If we expand the determinant, we obtain the equation of the plane in the form:

Ax + By + Cz + D = 0.

Example. Write the equation of the plane passing through the points M 1 (1,-1,0);

M 2 (-2,3,1) and M 3 (0,0,1).

, (x - 1) 3 - (y + 1)(-2) + z 1 = 0;

3x + 2y + z – 1 = 0.

Equation of a plane in segments

Let the general equation of the plane be given: Ax + By + Cz + D = 0 and D ≠ 0, i.e. the plane does not pass through the origin. Divide both sides by –D: and denote: ; ; . Then

got plane equation in segments .

where a, b, c are the values ​​of the segments cut off by the plane on the coordinate axes.

Example 1. Write the equation of the plane passing through the points A(3, 0, 0);

B(0, 2, 0) and C(0, 0, -3).

a=3; b=2; c=-3, or 2x + 3y - 2z – 6 = 0.

Example 2. Find the values ​​of the segments cut off by the plane

4x – y – 3z – 12 = 0 on the coordinate axes.

4x – y – 3z = 12 a=3, b=-12, c=-4.

Normal plane equation.

Let a certain plane Q be given. From the origin of coordinates, draw a perpendicular OP to the plane. Let |OP|=p and vector : . Let's take the current point M(x, y, z) of the plane and calculate the scalar product of the vectors and : .

If we project the point M onto the direction , then we will get to the point P.T.O., we get the equation

(9).

Setting a line in space.

Line L in space can be defined as the intersection of two surfaces. Let the point M(x, y, z) lying on the line L belong to both the surface P1 and the surface P2. Then the coordinates of this point must satisfy the equations of both surfaces. Therefore, under equation of line L in space understand a set of two equations, each of which is the equation of the corresponding surface:

The line L contains those and only those points whose coordinates satisfy both equations in (*). Later we will look at other ways to define lines in space.

A bunch of planes.

Bunch of planes– the set of all planes passing through a given straight line – the beam axis.

To define a bundle of planes, it is enough to specify its axis. Let the equation of this line be given in general form:

.

Write a beam equation- means to compose an equation from which, under an additional condition, one can obtain the equation of any plane of the beam, except, b.m. one. Let's multiply equation II by l and add it to equation I:

A 1 x + B 1 y + C 1 z + D 1 + l(A 2 x + B 2 y + C 2 z + D 2) = 0 (1) or

(A 1 + lA 2)x + (B 1 + lB 2)y + (C 1 + lC 2)z + (D 1 + lD 2) = 0 (2).

l – parameter – a number that can take real values. For any chosen value of l, equations (1) and (2) are linear, i.e. these are the equations of a certain plane.

1. We'll show you that this plane passes through the beam axis L. Take an arbitrary point M 0 (x 0, y 0, z 0) L. Consequently, M 0 P 1 and M 0 P 2. Means:

Consequently, the plane described by equation (1) or (2) belongs to the beam.

2. The opposite can also be proven: any plane passing through the straight line L is described by equation (1) with an appropriate choice of the parameter l.

Example 1. Write down the equation of a plane passing through the line of intersection of the planes x + y + 5z – 1 = 0 and 2x + 3y – z + 2 = 0 and through the point M(3, 2, 1).

We write the beam equation: x + y + 5z – 1 + l(2x + 3y – z + 2) = 0. To find l, we take into account that M R:

Any surface in space can be considered as a locus of points that has some property common to all points.

Example 1.

Sphere – a set of points equidistant from a given point C (center). С(x 0 ,y 0 ,z 0). By definition |CM|=R or or . This equation is valid for all points of the sphere and only for them. If x 0 =0, y 0 =0, z 0 =0, then .

In a similar way, you can create an equation for any surface if a coordinate system is selected.

Example 2. x=0 – equation of the YOZ plane.

Expressing the geometric definition of a surface in terms of the coordinates of its current point and collecting all the terms in one part, we obtain an equality of the form

F(x,y,z)=0(*),

This - surface equation , if the coordinates of all points on the surface satisfy this equality, but the coordinates of points not lying on the surface do not.

Thus, each surface in the selected coordinate system has its own equation. However, not every equation of the form (*) corresponds to a surface in the sense of the definition.

Examples:

2x – y + z – 3 = 0 (plane)

x 2 + y 2 – z 2 = 0 (cone)

x 2 + y 2 +3 = 0 – the coordinates of no point satisfy.

x 2 + y 2 + z 2 =0 – the only point (0,0,0).

x 2 = 3y 2 = 0 – straight line (OZ axis).

Graphic method. Coordinate plane (x;y)

Equations with a parameter cause serious logical difficulties. Each such equation is essentially a short version of a family of equations. It is clear that it is impossible to write down every equation from an infinite family, but, nevertheless, each of them must be solved. The easiest way to do this is to use a graphical representation of the dependence of a variable on a parameter.

On the plane, the function defines a family of curves depending on the parameter. We will be interested in which plane transformation can be used to move to other curves of the family (see , , , , , , ).

Parallel transfer

Example. For each parameter value, determine the number of solutions to the equation.

Solution. Let's build a graph of the function.

Let's consider. This is a straight line parallel to the OX axis.

Answer. If, then there are no solutions;

if, then 3 solutions;

if, then 2 solutions;

if, 4 solutions.

Turn

It should be noted right away that the choice of a family of curves is not monotonous (unlike the problems themselves), or rather, it is the same: in all problems - straight lines. Moreover, the center of rotation belongs to the straight line.

Example. For what values ​​of the parameter does the equation have a unique solution?

Solution. Let's consider the function and. The graph of the second function is a semicircle with a center at a point with coordinates and radius =1 (Fig. 2).

Arc AB.

All rays passing between OA and OB intersect at one point, and OB and OM (tangent) also intersect at one point. The angular coefficients OA and OB are equal, respectively. The slope of the tangent is equal to. Easily found from the system

So, straight families have only one common point with an arc at.

Answer. .

Example. At what conditions does the equation have a solution?

Solution. Let's consider the function. Examining it for monotonicity, we find out that it increases on the interval and decreases on. Point - is the maximum point.

A function is a family of lines passing through a point. Let's look at Figure 2. The graph of the function is the arc AB. The straight lines that will be located between the straight lines OA and OB satisfy the conditions of the problem. The slope coefficient of the straight line OA is a number, and OB is .

Answer. When the equation has 1 solution;

for other values ​​of the parameter there are no solutions.

Homothety. Compression to straight

Example. Find all values ​​of the parameter for each of which the equation has exactly 8 solutions.

Solution. We have. Let's consider the function. The first of them specifies a family of semicircles with a center at a point with coordinates, the second a family of straight lines parallel to the abscissa axis.

The number of roots will correspond to the number 8 when the radius of the semicircle is larger and smaller, that is. Note that there is.

Answer. or.

Graphic method. Coordinate plane (x;a)

In general, the equations, containing a parameter, are not provided with any clear, methodically designed solution system. One has to search for certain parameter values ​​by touch, by searching, solving a large number of intermediate equations. This approach does not always ensure success in finding all parameter values ​​for which the equation has no solutions, or has one, two or more solutions. Often, some parameter values ​​are lost or extra values ​​appear. In order to do these latter, it is necessary to conduct a special study which can be quite difficult.

Let's consider a method that simplifies the work of solving equations with a parameter. The method is as follows

1. From an equation with a variable x and parameter a Let's express the parameter as a function of x: .

2. In the coordinate plane x O a build a graph of the function.

3. Consider the straight lines and select those intervals of the O axis a, on which these lines satisfy the following conditions: a) does not intersect the graph of the function, b) intersects the graph of the function at one point, c) at two points, d) at three points, and so on.

4. If the task is to find the values x, then we express x through a for each of the found value intervals a separately.

The view of a parameter as an equal variable is reflected in graphical methods. Thus, a coordinate plane appears. It would seem that such an insignificant detail as the rejection of the traditional designation of the coordinate plane by letters x And y defines one of the most effective methods for solving problems with parameters.

The described method is very clear. In addition, almost all the basic concepts of the algebra course and the principles of analysis find application in it. The entire set of knowledge related to the study of a function is involved: applying the derivative to determine extremum points, finding the limit of the function, asymptotes, etc.. d. (see , , ).

Example. At what parameter values does the equation have two roots?

Solution. Let's move on to an equivalent system

The graph shows that the equation has 2 roots.

Answer. When the equation has two roots.

Example. Find the set of all numbers for each of which the equation has only two different roots.

Solution. Let's rewrite this equation in the following form:

Now it is important not to miss that, and - roots of the original equation only under the condition. Let us pay attention to the fact that it is more convenient to construct a graph on a coordinate plane. In Figure 5, the desired graph is a union of solid lines. Here the answer is “read” by vertical lines.

Answer. At, or, or.

To obtain the general equation of a plane, let us analyze the plane passing through a given point.

Let there be three coordinate axes already known to us in space - Ox, Oy And Oz. Hold the sheet of paper so that it remains flat. The plane will be the sheet itself and its continuation in all directions.

Let P arbitrary plane in space. Every vector perpendicular to it is called normal vector to this plane. Naturally, we are talking about a non-zero vector.

If any point on the plane is known P and some normal vector to it, then by these two conditions the plane in space is completely defined(through a given point you can draw a single plane perpendicular to the given vector). The general equation of the plane will be:

So, the conditions that define the equation of the plane are. To get yourself plane equation, having the above form, take on the plane P arbitrary point M with variable coordinates x, y, z. This point belongs to the plane only if vector perpendicular to the vector(Fig. 1). For this, according to the condition of perpendicularity of vectors, it is necessary and sufficient that the scalar product of these vectors be equal to zero, that is

The vector is specified by condition. We find the coordinates of the vector using the formula :

.

Now, using the scalar product of vectors formula , we express the scalar product in coordinate form:

Since the point M(x; y; z) is chosen arbitrarily on the plane, then the last equation is satisfied by the coordinates of any point lying on the plane P. For a point N, not lying on a given plane, i.e. equality (1) is violated.

Example 1. Write an equation for a plane passing through a point and perpendicular to the vector.

Solution. Let’s use formula (1) and look at it again:

In this formula the numbers A , B And C vector coordinates, and numbers x0 , y0 And z0 - coordinates of the point.

The calculations are very simple: we substitute these numbers into the formula and get

We multiply everything that needs to be multiplied and add just numbers (which do not have letters). Result:

.

The required equation of the plane in this example turned out to be expressed by a general equation of the first degree with respect to variable coordinates x, y, z arbitrary point of the plane.

So, an equation of the form

called general plane equation .

Example 2. Construct in a rectangular Cartesian coordinate system a plane given by the equation .

Solution. To construct a plane, it is necessary and sufficient to know any three of its points that do not lie on the same straight line, for example, the points of intersection of the plane with the coordinate axes.

How to find these points? To find the point of intersection with the axis Oz, you need to substitute zeros for X and Y in the equation given in the problem statement: x = y= 0 . Therefore we get z= 6. Thus, the given plane intersects the axis Oz at the point A(0; 0; 6) .

In the same way we find the point of intersection of the plane with the axis Oy. At x = z= 0 we get y= −3, that is, the point B(0; −3; 0) .

And finally, we find the point of intersection of our plane with the axis Ox. At y = z= 0 we get x= 2, that is, a point C(2; 0; 0) . Based on the three points obtained in our solution A(0; 0; 6) , B(0; −3; 0) and C(2; 0; 0) construct the given plane.

Let's now consider special cases of the general plane equation. These are cases when certain coefficients of equation (2) become zero.

1. When D= 0 equation defines a plane passing through the origin, since the coordinates of the point 0 (0; 0; 0) satisfy this equation.

2. When A= 0 equation defines a plane parallel to the axis Ox, since the normal vector of this plane is perpendicular to the axis Ox(its projection onto the axis Ox equal to zero). Similarly, when B= 0 plane parallel to the axis Oy, and when C= 0 plane parallel to the axis Oz.

3. When A=D= 0 equation defines a plane passing through the axis Ox, since it is parallel to the axis Ox (A=D= 0). Similarly, the plane passes through the axis Oy, and the plane through the axis Oz.

4. When A=B= 0 equation defines a plane parallel to the coordinate plane xOy, since it is parallel to the axes Ox (A= 0) and Oy (B= 0). Similarly, the plane is parallel to the plane yOz, and the plane is the plane xOz.

5. When A=B=D= 0 equation (or z = 0) defines the coordinate plane xOy, since it is parallel to the plane xOy (A=B= 0) and passes through the origin ( D= 0). Likewise, Eq. y = 0 in space defines the coordinate plane xOz, and the equation x = 0 - coordinate plane yOz.

Example 3. Create an equation of the plane P, passing through the axis Oy and period.

Solution. So the plane passes through the axis Oy. Therefore, in her equation y= 0 and this equation has the form . To determine the coefficients A And C let's take advantage of the fact that the point belongs to the plane P .

Therefore, among its coordinates there are those that can be substituted into the plane equation that we have already derived (). Let's look again at the coordinates of the point:

M0 (2; −4; 3) .

Among them x = 2 , z= 3 . We substitute them into the general equation and get the equation for our particular case:

2A + 3C = 0 .

Leave 2 A on the left side of the equation, move 3 C to the right side and we get

A = −1,5C .

Substituting the found value A into the equation, we get

or .

This is the equation required in the example condition.

Solve the plane equation problem yourself, and then look at the solution

Example 4. Define a plane (or planes, if more than one) with respect to coordinate axes or coordinate planes if the plane(s) is given by the equation.

Solutions to typical problems that occur during tests are in the textbook “Problems on a plane: parallelism, perpendicularity, intersection of three planes at one point.”

Equation of a plane passing through three points

As already mentioned, a necessary and sufficient condition for constructing a plane, in addition to one point and the normal vector, are also three points that do not lie on the same line.

Let three different points , and , not lying on the same line, be given. Since the indicated three points do not lie on the same straight line, the vectors are not collinear, and therefore any point in the plane lies in the same plane with the points, and if and only if the vectors , and coplanar, i.e. then and only when mixed product of these vectors equals zero.

Using the expression for the mixed product in coordinates, we obtain the equation of the plane

(3)

After revealing the determinant, this equation becomes an equation of the form (2), i.e. general equation of the plane.

Example 5. Write an equation for a plane passing through three given points that do not lie on the same straight line:

and determine a special case of the general equation of a line, if one occurs.

Solution. According to formula (3) we have:

Normal plane equation. Distance from point to plane

The normal equation of a plane is its equation, written in the form

Every first degree equation with respect to coordinates x, y, z

Ax + By + Cz +D = 0 (3.1)

defines a plane, and vice versa: any plane can be represented by equation (3.1), which is called plane equation.

Vector n(A, B, C) orthogonal to the plane is called normal vector plane. In equation (3.1), the coefficients A, B, C are not equal to 0 at the same time.

Special cases of equation (3.1):

1. D = 0, Ax+By+Cz = 0 - the plane passes through the origin.

2. C = 0, Ax+By+D = 0 - the plane is parallel to the Oz axis.

3. C = D = 0, Ax + By = 0 - the plane passes through the Oz axis.

4. B = C = 0, Ax + D = 0 - the plane is parallel to the Oyz plane.

Equations of coordinate planes: x = 0, y = 0, z = 0.

A straight line in space can be specified:

1) as a line of intersection of two planes, i.e. system of equations:

A 1 x + B 1 y + C 1 z + D 1 = 0, A 2 x + B 2 y + C 2 z + D 2 = 0; (3.2)

2) by its two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the straight line passing through them is given by the equations:

= ; (3.3)

3) the point M 1 (x 1, y 1, z 1) belonging to it, and the vector a(m, n, p), collinear to it. Then the straight line is determined by the equations:

. (3.4)

Equations (3.4) are called canonical equations of the line.

Vector a called direction vector straight.

We obtain parametric ones by equating each of the relations (3.4) to the parameter t:

x = x 1 +mt, y = y 1 + nt, z = z 1 + rt. (3.5)

Solving system (3.2) as a system of linear equations for unknowns x And y, we arrive at the equations of the line in projections or to given equations of the straight line:

x = mz + a, y = nz + b. (3.6)

From equations (3.6) we can go to the canonical equations, finding z from each equation and equating the resulting values:

.

From general equations (3.2) we can move to canonical ones in another way, if we find any point of this line and its directing line n= [n 1 , n 2 ], where n 1 (A 1, B 1, C 1) and n 2 (A 2 , B 2 , C 2 ) - normal vectors of given planes. If one of the denominators m, n or R in equations (3.4) turns out to be equal to zero, then the numerator of the corresponding fraction must be set equal to zero, i.e. system

is equivalent to the system ; such a straight line is perpendicular to the Ox axis.

System is equivalent to the system x = x 1, y = y 1; the straight line is parallel to the Oz axis.

Example 1.15. Write an equation for the plane, knowing that point A(1,-1,3) serves as the base of a perpendicular drawn from the origin to this plane.

Solution. According to the problem conditions, the vector OA(1,-1,3) is a normal vector of the plane, then its equation can be written as
x-y+3z+D=0. Substituting the coordinates of point A(1,-1,3) belonging to the plane, we find D: 1-(-1)+3×3+D = 0, D = -11. So x-y+3z-11=0.

Example 1.16. Write an equation for a plane passing through the Oz axis and forming an angle of 60° with the plane 2x+y-z-7=0.

Solution. The plane passing through the Oz axis is given by the equation Ax+By=0, where A and B do not simultaneously vanish. Let B not
equals 0, A/Bx+y=0. Using the cosine formula for the angle between two planes

.

Solving the quadratic equation 3m 2 + 8m - 3 = 0, we find its roots
m 1 = 1/3, m 2 = -3, from where we get two planes 1/3x+y = 0 and -3x+y = 0.

Example 1.17. Compose the canonical equations of the line:
5x + y + z = 0, 2x + 3y - 2z + 5 = 0.

Solution. The canonical equations of the line have the form:

Where m, n, p- coordinates of the directing vector of the straight line, x 1 , y 1 , z 1- coordinates of any point belonging to a line. A straight line is defined as the line of intersection of two planes. To find a point belonging to a line, one of the coordinates is fixed (the easiest way is to set, for example, x=0) and the resulting system is solved as a system of linear equations with two unknowns. So, let x=0, then y + z = 0, 3y - 2z+ 5 = 0, hence y=-1, z=1. We found the coordinates of the point M(x 1, y 1, z 1) belonging to this line: M (0,-1,1). The direction vector of a straight line is easy to find, knowing the normal vectors of the original planes n 1 (5,1,1) and n 2 (2,3,-2). Then

The canonical equations of the line have the form: x/(-5) = (y + 1)/12 =
= (z - 1)/13.

Example 1.18. In the beam defined by the planes 2x-y+5z-3=0 and x+y+2z+1=0, find two perpendicular planes, one of which passes through the point M(1,0,1).

Solution. The equation of the beam defined by these planes has the form u(2x-y+5z-3) + v(x+y+2z+1)=0, where u and v do not vanish simultaneously. Let us rewrite the beam equation as follows:

(2u +v)x + (- u + v)y + (5u +2v)z - 3u + v = 0.

In order to select a plane from the beam that passes through point M, we substitute the coordinates of point M into the equation of the beam. We get:

(2u+v)×1 + (-u + v) ×0 + (5u + 2v)×1 -3u + v =0, or v = - u.

Then we find the equation of the plane containing M by substituting v = - u into the beam equation:

u(2x-y +5z - 3) - u (x + y +2z +1) = 0.

Because u ¹0 (otherwise v=0, and this contradicts the definition of a beam), then we have the equation of the plane x-2y+3z-4=0. The second plane belonging to the beam must be perpendicular to it. Let us write down the condition for the orthogonality of planes:

(2u+ v) ×1 + (v - u) ×(-2) + (5u +2v)×3 = 0, or v = - 19/5u.

This means that the equation of the second plane has the form:

u(2x -y+5z - 3) - 19/5 u(x + y +2z +1) = 0 or 9x +24y + 13z + 34 = 0.