Plane equation. How to write an equation for a plane?
Mutual arrangement planes. Tasks

Spatial geometry is not much more complicated than "flat" geometry, and our flights in space begin with this article. In order to understand the topic, one must have a good understanding of vectors, in addition, it is desirable to be familiar with the geometry of the plane - there will be many similarities, many analogies, so the information will be digested much better. In a series of my lessons, the 2D world opens with an article Equation of a straight line on a plane. But now Batman has stepped off the flat screen TV and is launching from the Baikonur Cosmodrome.

Let's start with drawings and symbols. Schematically, the plane can be drawn as a parallelogram, which gives the impression of space:

The plane is infinite, but we have the opportunity to depict only a piece of it. In practice, in addition to the parallelogram, an oval or even a cloud is also drawn. For technical reasons, it is more convenient for me to depict the plane in this way and in this position. Real planes, which we will consider in practical examples, can be arranged as you like - mentally take the drawing in your hands and twist it in space, giving the plane any slope, any angle.

Notation: it is customary to designate planes in small Greek letters, apparently so as not to confuse them with straight on the plane or with straight in space. I'm used to using the letter . In the drawing, it is the letter "sigma", and not a hole at all. Although, a holey plane, it is certainly very funny.

In some cases, it is convenient to use the same Greek letters with subscripts, for example, .

Obviously, the plane is uniquely determined by three different points not lying on the same straight line. Therefore, three-letter designations of planes are quite popular - according to the points belonging to them, for example, etc. Often letters are enclosed in parentheses: , so as not to confuse the plane with another geometric figure.

For experienced readers, I will give shortcut menu:

• How to write an equation for a plane using a point and two vectors?
• How to write an equation for a plane using a point and a normal vector?

and we will not languish in long waits:

General equation of the plane

The general equation of the plane has the form , where the coefficients are simultaneously non-zero.

A number of theoretical calculations and practical problems are valid both for the usual orthonormal basis and for the affine basis of space (if oil is oil, return to the lesson Linear (non) dependence of vectors. Vector basis). For simplicity, we will assume that all events occur in an orthonormal basis and a Cartesian rectangular coordinate system.

And now let's train a little spatial imagination. It's okay if you have it bad, now we'll develop it a little. Even playing on nerves requires practice.

In the most general case, when the numbers are not equal to zero, the plane intersects all three coordinate axes. For example, like this:

I repeat once again that the plane continues indefinitely in all directions, and we have the opportunity to depict only part of it.

Consider the simplest equations of planes:

How to understand given equation? Think about it: “Z” ALWAYS, for any values ​​of “X” and “Y” is equal to zero. This is the equation of the "native" coordinate plane. Indeed, formally the equation can be rewritten as follows: , from where it is clearly visible that we don’t care, what values ​​“x” and “y” take, it is important that “z” is equal to zero.

Similarly:
is the equation of the coordinate plane ;
is the equation of the coordinate plane.

Let's complicate the problem a little, consider a plane (here and further in the paragraph we assume that the numerical coefficients are not equal to zero). Let's rewrite the equation in the form: . How to understand it? "X" is ALWAYS, for any value of "y" and "z" is equal to a certain number. This plane is parallel to the coordinate plane. For example, a plane is parallel to a plane and passes through a point.

Similarly:
- the equation of the plane, which is parallel to the coordinate plane;
- the equation of a plane that is parallel to the coordinate plane.

Add members: . The equation can be rewritten like this: , that is, "Z" can be anything. What does it mean? "X" and "Y" are connected by a ratio that draws a certain straight line in the plane (you will recognize equation of a straight line in a plane?). Since Z can be anything, this line is "replicated" at any height. Thus, the equation defines a plane parallel to the coordinate axis

Similarly:
- the equation of the plane, which is parallel to the coordinate axis;
- the equation of the plane, which is parallel to the coordinate axis.

If the free terms are zero, then the planes will directly pass through the corresponding axes. For example, the classic "direct proportionality":. Draw a straight line in the plane and mentally multiply it up and down (since “z” is any). Conclusion: the plane given by the equation passes through the coordinate axis.

We conclude the review: the equation of the plane passes through the origin. Well, here it is quite obvious that the point satisfies the given equation.

And, finally, the case that is shown in the drawing: - the plane is friends with all coordinate axes, while it always “cuts off” a triangle that can be located in any of the eight octants.

Linear inequalities in space

In order to understand the information, it is necessary to study well linear inequalities in the plane because many things will be similar. The paragraph will be of a brief overview with a few examples, since the material is quite rare in practice.

If the equation defines a plane, then the inequalities
ask half-spaces. If the inequality is not strict (the last two in the list), then the solution of the inequality, in addition to the half-space, includes the plane itself.

Example 5

Find the unit normal vector of the plane .

Decision: A unit vector is a vector whose length is one. Let's denote this vector by . It is quite clear that the vectors are collinear:

First, we remove the normal vector from the equation of the plane: .

How to find the unit vector? To find the unit vector, you need every vector coordinate divided by vector length.

Let's rewrite the normal vector in the form and find its length:

According to the above:

Answer:

Check: , which was required to check.

Readers who have carefully studied the last paragraph of the lesson, probably noticed that the coordinates of the unit vector are exactly the direction cosines of the vector:

Let's digress from the disassembled problem: when you are given an arbitrary nonzero vector , and by the condition it is required to find its direction cosines (see the last tasks of the lesson Dot product of vectors), then you, in fact, also find a unit vector collinear to the given one. In fact, two tasks in one bottle.

The need to find a unit normal vector arises in some problems of mathematical analysis.

We figured out the fishing of the normal vector, now we will answer the opposite question:

How to write an equation for a plane using a point and a normal vector?

This rigid construction of a normal vector and a point is well known by a darts target. Please stretch your hand forward and mentally select an arbitrary point in space, for example, a small cat in a sideboard. Obviously, through this point, you can draw a single plane perpendicular to your hand.

The equation of a plane passing through a point perpendicular to the vector is expressed by the formula:

This article gives an idea of ​​how to write an equation for a plane passing through a given point in three-dimensional space perpendicular to a given line. Let us analyze the above algorithm using the example of solving typical problems.

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Finding the equation of a plane passing through a given point in space perpendicular to a given line

Let a three-dimensional space and a rectangular coordinate system O x y z be given in it. The point M 1 (x 1, y 1, z 1), the straight line a and the plane α passing through the point M 1 perpendicular to the straight line a are also given. It is necessary to write down the equation of the plane α.

Before proceeding to solve this problem, let's recall the geometry theorem from the program for grades 10 - 11, which reads:

Definition 1

A single plane passes through a given point in three-dimensional space and is perpendicular to a given line.

Now consider how to find the equation of this single plane passing through the starting point and perpendicular to the given line.

It is possible to write the general equation of a plane if the coordinates of a point belonging to this plane are known, as well as the coordinates of the normal vector of the plane.

By the condition of the problem, we are given the coordinates x 1, y 1, z 1 of the point M 1 through which the plane α passes. If we determine the coordinates of the normal vector of the plane α, then we will be able to write the desired equation.

The normal vector of the plane α, since it is non-zero and lies on the line a, perpendicular to the plane α, will be any directing vector of the line a. So, the problem of finding the coordinates of the normal vector of the plane α is transformed into the problem of determining the coordinates of the directing vector of the straight line a .

The determination of the coordinates of the directing vector of the straight line a can be carried out different methods: depends on the option of specifying the straight line a in the initial conditions. For example, if the line a in the condition of the problem is given by canonical equations of the form

x - x 1 a x = y - y 1 a y = z - z 1 a z

or parametric equations of the form:

x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ

then the directing vector of the straight line will have coordinates a x, a y and a z. In the case when the straight line a is represented by two points M 2 (x 2, y 2, z 2) and M 3 (x 3, y 3, z 3), then the coordinates of the direction vector will be determined as (x3 - x2, y3 - y2 , z3 – z2).

Definition 2

Algorithm for finding the equation of a plane passing through a given point perpendicular to a given line:

Determine the coordinates of the directing vector of the straight line a: a → = (a x, a y, a z) ;

We define the coordinates of the normal vector of the plane α as the coordinates of the directing vector of the straight line a:

n → = (A , B , C) , where A = a x , B = a y , C = a z;

We write the equation of the plane passing through the point M 1 (x 1, y 1, z 1) and having a normal vector n→=(A, B, C) in the form A (x - x 1) + B (y - y 1) + C (z - z 1) = 0. This will be the required equation of a plane that passes through a given point in space and is perpendicular to a given line.

The resulting general equation of the plane: A (x - x 1) + B (y - y 1) + C (z - z 1) \u003d 0 makes it possible to obtain the equation of the plane in segments or the normal equation of the plane.

Let's solve some examples using the algorithm obtained above.

Example 1

A point M 1 (3, - 4, 5) is given, through which the plane passes, and this plane is perpendicular to the coordinate line O z.

Decision

the direction vector of the coordinate line O z will be the coordinate vector k ⇀ = (0 , 0 , 1) . Therefore, the normal vector of the plane has coordinates (0 , 0 , 1) . Let's write the equation of a plane passing through a given point M 1 (3, - 4, 5) whose normal vector has coordinates (0, 0, 1) :

A (x - x 1) + B (y - y 1) + C (z - z 1) = 0 ⇔ ⇔ 0 (x - 3) + 0 (y - (- 4)) + 1 (z - 5) = 0 ⇔ z - 5 = 0

Answer: z - 5 = 0 .

Consider another way to solve this problem:

Example 2

A plane that is perpendicular to the line O z will be given by an incomplete general equation of the plane of the form С z + D = 0 , C ≠ 0 . Let's define the values ​​of C and D: those for which the plane passes through a given point. Substitute the coordinates of this point in the equation C z + D = 0 , we get: C · 5 + D = 0 . Those. numbers, C and D are related by - D C = 5 . Taking C \u003d 1, we get D \u003d - 5.

Substitute these values ​​into the equation C z + D = 0 and obtain the required equation for a plane perpendicular to the line O z and passing through the point M 1 (3, - 4, 5) .

It will look like: z - 5 = 0.

Answer: z - 5 = 0 .

Example 3

Write an equation for a plane passing through the origin and perpendicular to the line x - 3 = y + 1 - 7 = z + 5 2

Decision

Based on the conditions of the problem, it can be argued that the guiding vector of a given straight line can be taken as a normal vector n → of a given plane. Thus: n → = (- 3 , - 7 , 2) . Let's write the equation of a plane passing through the point O (0, 0, 0) and having a normal vector n → \u003d (- 3, - 7, 2) :

3 (x - 0) - 7 (y - 0) + 2 (z - 0) = 0 ⇔ - 3 x - 7 y + 2 z = 0

We have obtained the required equation for the plane passing through the origin perpendicular to the given line.

Answer:- 3x - 7y + 2z = 0

Example 4

Given a rectangular coordinate system O x y z in three-dimensional space, it contains two points A (2 , - 1 , - 2) and B (3 , - 2 , 4) . The plane α passes through the point A perpendicular to the line AB. It is necessary to compose the equation of the plane α in segments.

Decision

The plane α is perpendicular to the line A B, then the vector A B → will be the normal vector of the plane α. The coordinates of this vector are determined as the difference between the corresponding coordinates of points B (3, - 2, 4) and A (2, - 1, - 2):

A B → = (3 - 2 , - 2 - (- 1) , 4 - (- 2)) ⇔ A B → = (1 , - 1 , 6)

The general equation of the plane will be written in the following form:

1 x - 2 - 1 y - (- 1 + 6 (z - (- 2)) = 0 ⇔ x - y + 6 z + 9 = 0

Now we compose the desired equation of the plane in the segments:

x - y + 6 z + 9 = 0 ⇔ x - y + 6 z = - 9 ⇔ x - 9 + y 9 + z - 3 2 = 1

Answer:x - 9 + y 9 + z - 3 2 = 1

It should also be noted that there are problems whose requirement is to write an equation for a plane passing through a given point and perpendicular to two given planes. In general, the solution to this problem is to write an equation for a plane passing through a given point perpendicular to a given line, since two intersecting planes define a straight line.

Example 5

A rectangular coordinate system O x y z is given, in it is a point M 1 (2, 0, - 5) . The equations of two planes 3 x + 2 y + 1 = 0 and x + 2 z - 1 = 0 are also given, which intersect along the straight line a . It is necessary to compose an equation for a plane passing through the point M 1 perpendicular to the line a.

Decision

Let's determine the coordinates of the directing vector of the straight line a . It is perpendicular to both the normal vector n 1 → (3 , 2 , 0) of the plane n → (1 , 0 , 2) and the normal vector 3 x + 2 y + 1 = 0 of the plane x + 2 z - 1 = 0 .

Then the directing vector α → straight line a we take the vector product of vectors n 1 → and n 2 → :

a → = n 1 → × n 2 → = i → j → k → 3 2 0 1 0 2 = 4 i → - 6 j → - 2 k → ⇒ a → = (4 , - 6 , - 2 )

Thus, the vector n → = (4, - 6, - 2) will be the normal vector of the plane perpendicular to the line a. We write the desired equation of the plane:

4 (x - 2) - 6 (y - 0) - 2 (z - (- 5)) = 0 ⇔ 4 x - 6 y - 2 z - 18 = 0 ⇔ ⇔ 2 x - 3 y - z - 9 = 0

Answer: 2 x - 3 y - z - 9 = 0

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Let it be necessary to find the equation of a plane passing through three given points that do not lie on one straight line. Denoting their radius vectors by and the current radius vector by , we can easily obtain the desired equation in vector form. Indeed, the vectors , must be coplanar (they all lie in the desired plane). Therefore, the vector-scalar product of these vectors must be equal to zero:

This is the equation of a plane passing through three given points, in vector form.

Turning to the coordinates, we get the equation in coordinates:

If the three given points lie on the same straight line, then the vectors would be collinear. Therefore, the corresponding elements of the last two rows of the determinant in equation (18) would be proportional and the determinant would be identically equal to zero. Therefore, equation (18) would become an identity for any values ​​of x, y, and z. Geometrically, this means that a plane passes through each point of space, in which three given points also lie.

Remark 1. The same problem can be solved without using vectors.

Denoting the coordinates of the three given points, respectively, through we write the equation of any plane passing through the first point:

To obtain the equation of the desired plane, one must require that equation (17) be satisfied by the coordinates of the other two points:

From equations (19), it is necessary to determine the ratios of two coefficients to the third and enter the found values ​​into equation (17).

Example 1. Write an equation for a plane passing through points.

The equation for a plane passing through the first of these points will be:

The conditions for the plane (17) to pass through two other points and the first point are:

Adding the second equation to the first, we get:

Substituting into the second equation, we get:

Substituting into equation (17) instead of A, B, C, respectively, 1, 5, -4 (numbers proportional to them), we get:

Example 2. Write an equation for a plane passing through the points (0, 0, 0), (1, 1, 1), (2, 2, 2).

The equation of any plane passing through the point (0, 0, 0) will be]

The conditions for passing this plane through the points (1, 1, 1) and (2, 2, 2) are:

Reducing the second equation by 2, we see that to determine the two unknowns, the relation has one equation with

From here we get . Substituting now into the plane equation instead of its value, we find:

This is the equation of the required plane; it depends on arbitrary

quantities B, C (namely, from the ratio, i.e., there are an infinite number of planes passing through three given points (three given points lie on one straight line).

Remark 2. The problem of drawing a plane through three given points that do not lie on one straight line is easily solved in general view if you use determinants. Indeed, since in equations (17) and (19) the coefficients A, B, C cannot be simultaneously equal to zero, then, considering these equations as a homogeneous system with three unknowns A, B, C, we write a necessary and sufficient condition for the existence of a solution of this system, other than zero (part 1, ch. VI, § 6):

Expanding this determinant by the elements of the first row, we obtain an equation of the first degree with respect to the current coordinates , which will be satisfied, in particular, by the coordinates of the three given points.

This latter can also be verified directly if we substitute the coordinates of any of these points instead of into the equation written using the determinant. On the left side, a determinant is obtained, in which either the elements of the first row are zero, or there are two identical rows. Thus, the formulated equation represents a plane passing through three given points.

13. Angle between planes, distance from a point to a plane.

Let the planes α and β intersect along the line c.
The angle between the planes is the angle between the perpendiculars to the line of their intersection, drawn in these planes.

In other words, in the plane α we draw a line a perpendicular to c. In the plane β - line b, also perpendicular to c. The angle between the planes α and β is equal to the angle between the lines a and b.

Note that when two planes intersect, four corners are actually formed. See them in the picture? As the angle between the planes we take spicy injection.

If the angle between the planes is 90 degrees, then the planes perpendicular,

This is the definition of perpendicularity of planes. When solving problems in stereometry, we also use sign of perpendicularity of planes:

If the plane α passes through the perpendicular to the plane β, then the planes α and β are perpendicular.

point to plane distance

Consider a point T given by its coordinates:

T \u003d (x 0, y 0, z 0)

Also consider the plane α, given by the equation:

Ax + By + Cz + D = 0

Then the distance L from the point T to the plane α can be calculated by the formula:

In other words, we substitute the coordinates of the point into the equation of the plane, and then divide this equation by the length of the normal vector n to the plane:

The resulting number is the distance. Let's see how this theorem works in practice.

We have already derived the parametric equations of a straight line in a plane, let's get the parametric equations of a straight line, which is given in a rectangular coordinate system in three-dimensional space.

Let a rectangular coordinate system be fixed in three-dimensional space Oxyz. Let's define a straight line a(see the section on how to define a straight line in space) by specifying the directing vector of a straight line and the coordinates of some point on the line . We will start from these data when compiling parametric equations of a straight line in space.

Let be an arbitrary point in three-dimensional space. If we subtract from the coordinates of the point M corresponding point coordinates M 1, then we will get the coordinates of the vector (see the article finding the coordinates of the vector by the coordinates of the points of its end and beginning), that is, .

Obviously, the set of points defines a line a if and only if the vectors and are collinear.

Let us write down the necessary and sufficient condition for the vectors to be collinear and : , where is some real number. The resulting equation is called vector-parametric equation of a straight line in rectangular coordinate system Oxyz in three-dimensional space. The vector-parametric equation of a straight line in coordinate form has the form and represents parametric equations of the straight line a. The name "parametric" is not accidental, since the coordinates of all points of the line are specified using the parameter .

Let us give an example of parametric equations of a straight line in a rectangular coordinate system Oxyz in space: . Here

15. Angle between a straight line and a plane. The point of intersection of a line with a plane.

Any equation of the first degree 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 planes. 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.

Coordinate plane equations: x = 0, y = 0, z = 0.

A straight line in space can be given:

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

. (3.4)

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

Vector a called guide vector straight.

We obtain the parametric equations of the straight line by equating each of the relations (3.4) with the parameter t:

x \u003d x 1 + mt, y \u003d y 1 + nt, z \u003d z 1 + pt. (3.5)

Solving system (3.2) as a system linear equations relatively unknown x and y, we arrive at the equations of the straight line in projections or to reduced straight line equations:

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

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

.

One can pass from general equations (3.2) to canonical equations in another way, if one finds any point of this line and its direction vector 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 the given planes. If one of the denominators m,n or R in equations (3.4) will be zero, then the numerator of the corresponding fraction must be set equal to zero, i.e. system

is tantamount to a system ; such a line is perpendicular to the x-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 the equation of the plane, knowing that the point A (1, -1,3) serves as the base of the perpendicular drawn from the origin to this plane.

Decision. By the condition of the problem, 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 the point A(1,-1,3) belonging to the plane, we find D: 1-(-1)+3×3+D = 0 Þ 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 degrees with the 2x+y-z-7=0 plane.

Decision. The plane passing through the Oz axis is given by the equation Ax+By=0, where A and B do not vanish at the same time. Let B not
is 0, A/Bx+y=0. According to the formula for the cosine of the angle between two planes

.

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

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

Decision. The canonical equations of the straight line have the form:

where m, n, p- coordinates of the directing vector of the straight line, x1, y1, z1- coordinates of any point belonging to the line. The straight line is defined as the line of intersection of two planes. To find a point belonging to a straight line, one of the coordinates is fixed (the easiest way is to put, 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, whence 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 directing 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 are: 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).

Decision. The equation of the beam defined by these planes is u(2x-y+5z-3) + v(x+y+2z+1)=0, where u and v do not vanish at the same time. We rewrite the beam equation as follows:

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

In order to select a plane passing through the point M from the beam, we substitute the coordinates of the point M into the beam equation. 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 u¹0 (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. We write the condition for the orthogonality of planes:

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

Hence, 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

In this lesson, we will look at how to use the determinant to compose plane equation. If you do not know what a determinant is, go to the first part of the lesson - " Matrices and determinants». Otherwise, you risk not understanding anything in today's material.

Equation of a plane by three points

Why do we need the equation of the plane at all? It's simple: knowing it, we can easily calculate angles, distances and other crap in problem C2. In general, this equation is indispensable. Therefore, we formulate the problem:

Task. There are three points in space that do not lie on the same straight line. Their coordinates:

M = (x 1 , y 1 , z 1);
N \u003d (x 2, y 2, z 2);
K \u003d (x 3, y 3, z 3);

It is required to write the equation of the plane passing through these three points. And the equation should look like:

Ax + By + Cz + D = 0

where the numbers A , B , C and D are the coefficients that, in fact, you want to find.

Well, how to get the equation of the plane, if only the coordinates of the points are known? The easiest way is to substitute the coordinates into the equation Ax + By + Cz + D = 0. You get a system of three equations that is easily solved.

Many students find this solution extremely tedious and unreliable. Last year's exam in mathematics showed that the probability of making a computational error is really high.

Therefore, the most advanced teachers began to look for simpler and elegant solutions. And they did find it! True, the reception obtained is more likely to higher mathematics. Personally, I had to rummage through the entire Federal list of textbooks to make sure that we have the right to use this technique without any justification and evidence.

Equation of the plane through the determinant

Enough ranting, let's get down to business. To begin with, a theorem on how the matrix determinant and the equation of the plane are related.

Theorem. Let the coordinates of three points through which the plane must be drawn be given: M = (x 1 , y 1 , z 1); N \u003d (x 2, y 2, z 2); K \u003d (x 3, y 3, z 3). Then the equation of this plane can be written in terms of the determinant:

For example, let's try to find a pair of planes that actually occur in C2 problems. Take a look at how fast everything counts:

A 1 = (0, 0, 1);
B = (1, 0, 0);
C 1 = (1, 1, 1);

We compose the determinant and equate it to zero:

Opening the determinant:

a = 1 1 (z − 1) + 0 0 x + (−1) 1 y = z − 1 − y;
b = (−1) 1 x + 0 1 (z − 1) + 1 0 y = −x;
d = a − b = z − 1 − y − (−x) = z − 1 − y + x = x − y + z − 1;
d = 0 ⇒ x − y + z − 1 = 0;

As you can see, when calculating the number d, I "brushed" the equation a little so that the variables x , y and z went into correct sequence. That's all! The equation of the plane is ready!

Task. Write an equation for a plane passing through the points:

A = (0, 0, 0);
B 1 = (1, 0, 1);
D 1 = (0, 1, 1);

Immediately substitute the coordinates of the points in the determinant:

Expanding the determinant again:

a = 1 1 z + 0 1 x + 1 0 y = z;
b = 1 1 x + 0 0 z + 1 1 y = x + y;
d \u003d a - b \u003d z - (x + y) \u003d z - x - y;
d = 0 ⇒ z − x − y = 0 ⇒ x + y − z = 0;

So, the plane equation is obtained again! Again, at the last step, I had to change the signs in it in order to get a more “beautiful” formula. It is not necessary to do this in this solution, but it is still recommended - in order to simplify the further solution of the problem.

As you can see, it is now much easier to write the equation of the plane. We substitute the points into the matrix, calculate the determinant - and that's it, the equation is ready.

This could be the end of the lesson. However, many students constantly forget what is inside the determinant. For example, which line contains x 2 or x 3 , and which line just x . To finally deal with this, let's trace where each number comes from.

Where does the formula with the determinant come from?

So, let's figure out where such a harsh equation with a determinant comes from. This will help you remember it and apply it successfully.

All planes that occur in Problem C2 are defined by three points. These points are always marked on the drawing, or even indicated directly in the problem text. In any case, to compile the equation, we need to write out their coordinates:

M = (x 1 , y 1 , z 1);
N \u003d (x 2, y 2, z 2);
K \u003d (x 3, y 3, z 3).

Consider one more point on our plane with arbitrary coordinates:

T = (x, y, z)

We take any point from the first three (for example, point M ) and draw vectors from it to each of the three remaining points. We get three vectors:

MN = (x 2 - x 1, y 2 - y 1, z 2 - z 1);
MK = (x 3 - x 1, y 3 - y 1, z 3 - z 1);
MT = (x - x 1 , y - y 1 , z - z 1).

Now let's make a square matrix from these vectors and equate its determinant to zero. The coordinates of the vectors will become the rows of the matrix - and we will get the same determinant that is indicated in the theorem:

This formula means that the volume of the box built on the vectors MN , MK and MT is equal to zero. Therefore, all three vectors lie in the same plane. In particular, an arbitrary point T = (x, y, z) is exactly what we were looking for.

Replacing points and rows of the determinant

Determinants have some wonderful properties that make it even easier to solution of problem C2. For example, it does not matter to us from which point to draw vectors. Therefore, the following determinants give the same plane equation as the one above:

You can also swap the lines of the determinant. The equation will remain unchanged. For example, many people like to write a line with the coordinates of the point T = (x; y; z) at the very top. Please, if it's convenient for you:

It confuses some that one of the lines contains variables x , y and z , which do not disappear when substituting points. But they shouldn't disappear! By substituting the numbers into the determinant, you should get the following construction:

Then the determinant is expanded according to the scheme given at the beginning of the lesson, and the standard equation of the plane is obtained:

Ax + By + Cz + D = 0

Take a look at an example. He is the last one in today's lesson. I will deliberately swap the lines to make sure that the answer will be the same equation of the plane.

Task. Write an equation for a plane passing through the points:

B 1 = (1, 0, 1);
C = (1, 1, 0);
D1 = (0, 1, 1).

So, we consider 4 points:

B 1 = (1, 0, 1);
C = (1, 1, 0);
D 1 = (0, 1, 1);
T = (x, y, z).

First, let's make a standard determinant and equate it to zero:

Opening the determinant:

a = 0 1 (z − 1) + 1 0 (x − 1) + (−1) (−1) y = 0 + 0 + y;
b = (−1) 1 (x − 1) + 1 (−1) (z − 1) + 0 0 y = 1 − x + 1 − z = 2 − x − z;
d \u003d a - b \u003d y - (2 - x - z) \u003d y - 2 + x + z \u003d x + y + z - 2;
d = 0 ⇒ x + y + z − 2 = 0;

That's it, we got the answer: x + y + z − 2 = 0 .

Now let's rearrange a couple of lines in the determinant and see what happens. For example, let's write a line with variables x, y, z not at the bottom, but at the top:

Let's expand the resulting determinant again:

a = (x − 1) 1 (−1) + (z − 1) (−1) 1 + y 0 0 = 1 − x + 1 − z = 2 − x − z;
b = (z − 1) 1 0 + y (−1) (−1) + (x − 1) 1 0 = y;
d = a − b = 2 − x − z − y;
d = 0 ⇒ 2 − x − y − z = 0 ⇒ x + y + z − 2 = 0;

We got exactly the same plane equation: x + y + z − 2 = 0. So, it really does not depend on the order of the rows. It remains to write down the answer.

So, we have seen that the equation of the plane does not depend on the sequence of lines. It is possible to carry out similar calculations and prove that the equation of the plane does not depend on the point whose coordinates we subtract from the other points.

In the problem considered above, we used the point B 1 = (1, 0, 1), but it was quite possible to take C = (1, 1, 0) or D 1 = (0, 1, 1). In general, any point with known coordinates lying on the desired plane.