Equation of a plane. How to write an equation of a plane?
Mutual arrangement of planes. Tasks

Spatial geometry is not much more complicated than “flat” geometry, and our flights in space begin with this article. To master the topic, you need to have a good understanding of vectors, in addition, it is advisable 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 left the flat TV screen and is launching from the Baikonur Cosmodrome.

Let's start with drawings and symbols. Schematically, the plane can be drawn in the form of a parallelogram, which creates 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 exactly this way and in exactly this position. Real planes that we will consider in practical examples, can be positioned in any way - mentally take the drawing in your hands and rotate it in space, giving the plane any inclination, any angle.

Designations: planes are usually denoted in small Greek letters, apparently so as not to confuse them with straight line on a plane or with straight line in space. I'm used to using the letter . In the drawing it is the letter “sigma”, and not a hole at all. Although, the holey plane is certainly quite funny.

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

It is obvious that the plane is uniquely defined by three different points that do not lie on the same line. Therefore, three-letter designations of planes are quite popular - by 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 quick access menu:

• How to create an equation of a plane using a point and two vectors?
• How to create an equation of a plane using a point and a normal vector?

and we will not languish in long waits:

General plane equation

The general equation of the plane has the form , where the coefficients are not equal to zero at the same time.

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

Now let’s practice our spatial imagination a little. It’s okay if yours is bad, now we’ll develop it a little. Even playing on nerves requires training.

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.

Let's consider the simplest equations of planes:

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

Likewise:
– equation of the coordinate plane;
– 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 should we understand it? “X” is ALWAYS, for any values ​​of “Y” and “Z”, 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.

Likewise:
– equation of a plane that is parallel to the coordinate plane;
– equation of a plane that is parallel to the coordinate plane.

Let's add members: . The equation can be rewritten as follows: , that is, “zet” can be anything. What does it mean? “X” and “Y” are connected by the relation, which draws a certain straight line in the plane (you will find out equation of a line in a plane?). Since “z” can be anything, this straight line is “replicated” at any height. Thus, the equation defines a plane parallel to the coordinate axis

Likewise:
– equation of a plane that is parallel to the coordinate axis;
– equation of a plane that 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 defined by the equation passes through the coordinate axis.

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

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

Linear inequalities in space

To understand the information you need to study well linear inequalities in the plane, because many things will be similar. The paragraph will be of a brief overview nature with several 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, also includes the plane itself.

Example 5

Find the unit normal vector of the plane .

Solution: A unit vector is a vector whose length is one. Let's denote given vector through . It is absolutely clear that the vectors are collinear:

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

How to find a unit vector? In order to find the unit vector, you need every divide the vector coordinate by the vector length.

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

According to the above:

Answer:

Verification: what was required to be verified.

Readers who 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 take a break from the problem at hand: when you are given an arbitrary non-zero vector, and according to the condition it is required to find its direction cosines (see the last problems of the lesson Dot product of vectors), then you, in fact, find a unit vector collinear to this one. Actually two tasks in one bottle.

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

We’ve figured out how to fish out a normal vector, now let’s answer the opposite question:

How to create an equation of a plane using a point and a normal vector?

This rigid construction of a normal vector and a point is well known to the dartboard. Please stretch your hand forward and mentally select an arbitrary point in space, for example, a small cat in the 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:

In this material, we will look at how to find the equation of a plane if we know the coordinates of three different points that do not lie on the same straight line. To do this, we need to remember what a rectangular coordinate system is in three-dimensional space. To begin, we will introduce the basic principle of this equation and show exactly how to use it to solve specific problems.

Yandex.RTB R-A-339285-1

First, we need to remember one axiom, which sounds like this:

Definition 1

If three points do not coincide with each other and do not lie on the same line, then in three-dimensional space only one plane passes through them.

In other words, if we have three different points whose coordinates do not coincide and which cannot be connected by a straight line, then we can determine the plane passing through it.

Let's say we have a rectangular coordinate system. Let's denote it O x y z. It contains three points M with coordinates M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3), which cannot be connected straight line. Based on these conditions, we can write down the equation of the plane we need. There are two approaches to solving this problem.

1. The first approach uses the general plane equation. In letter form, it is written as A (x - x 1) + B (y - y 1) + C (z - z 1) = 0. With its help, you can define in a rectangular coordinate system a certain alpha plane that passes through the first given point M 1 (x 1, y 1, z 1). It turns out that the normal vector of the plane α will have coordinates A, B, C.

Definition of N

Knowing the coordinates of the normal vector and the coordinates of the point through which the plane passes, we can write down the general equation of this plane.

This is what we will proceed from in the future.

Thus, according to the conditions of the problem, we have the coordinates of the desired point (even three) through which the plane passes. To find the equation, you need to calculate the coordinates of its normal vector. Let's denote it n → .

Let us remember the rule: any non-zero vector of a given plane is perpendicular to the normal vector of the same plane. Then we have that n → will be perpendicular to the vectors composed of the original points M 1 M 2 → and M 1 M 3 → . Then we can denote n → as a vector product of the form M 1 M 2 → · M 1 M 3 → .

Since M 1 M 2 → = (x 2 - x 1, y 2 - y 1, z 2 - z 1) and M 1 M 3 → = x 3 - x 1, y 3 - y 1, z 3 - z 1 (proofs of these equalities are given in the article devoted to calculating the coordinates of a vector from the coordinates of points), then it turns out that:

n → = M 1 M 2 → × M 1 M 3 → = i → j → k → x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1

If we calculate the determinant, we will obtain the coordinates of the normal vector n → we need. Now we can write down the equation we need for a plane passing through three given points.

2. The second approach to finding the equation passing through M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3), is based on such a concept as coplanarity of vectors.

If we have a set of points M (x, y, z), then in a rectangular coordinate system they define a plane for given points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2 ) , M 3 (x 3 , y 3 , z 3) only in the case when the vectors M 1 M → = (x - x 1 , y - y 1 , z - z 1) , M 1 M 2 → = ( x 2 - x 1 , y 2 - y 1 , z 2 - z 1) and M 1 M 3  → = (x 3 - x 1 , y 3 - y 1 , z 3 - z 1) will be coplanar.

In the diagram it will look like this:

This will mean that the mixed product of the vectors M 1 M → , M 1 M 2 → , M 1 M 3 → will be equal to zero: M 1 M → · M 1 M 2 → · M 1 M 3 → = 0 , since this is the main condition of coplanarity: M 1 M → = (x - x 1, y - y 1, z - z 1), M 1 M 2 → = (x 2 - x 1, y 2 - y 1, z 2 - z 1 ) and M 1 M 3 → = (x 3 - x 1, y 3 - y 1, z 3 - z 1).

Let us write the resulting equation in coordinate form:

After we calculate the determinant, we can obtain the plane equation we need for three points that do not lie on the same line M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3 , y 3 , z 3) .

From the resulting equation, you can go to the equation of the plane in segments or to the normal equation of the plane, if the conditions of the problem require it.

In the next paragraph we will give examples of how the approaches we have indicated are implemented in practice.

Examples of problems on composing an equation of a plane passing through 3 points

Previously, we identified two approaches that can be used to find the desired equation. Let's look at how they are used to solve problems and when you should choose each one.

Example 1

There are three points that do not lie on the same line, with coordinates M 1 (- 3, 2, - 1), M 2 (- 1, 2, 4), M 3 (3, 3, - 1). Write an equation for the plane passing through them.

Solution

We use both methods alternately.

1. Find the coordinates of the two vectors we need M 1 M 2 → , M 1 M 3 → :

M 1 M 2 → = - 1 - - 3 , 2 - 2 , 4 - - 1 ⇔ M 1 M 2 → = (2 , 0 , 5) M 1 M 3 → = 3 - - 3 , 3 - 2 , - 1 - - 1 ⇔ M 1 M 3 → = 6 , 1 , 0

Now let's calculate their vector product. We will not describe the calculations of the determinant:

n → = M 1 M 2 → × M 1 M 3 → = i → j → k → 2 0 5 6 1 0 = - 5 i → + 30 j → + 2 k →

We have a normal vector of the plane that passes through the three required points: n → = (- 5, 30, 2) . Next, we need to take one of the points, for example, M 1 (- 3, 2, - 1), and write down the equation for the plane with vector n → = (- 5, 30, 2). We get that: - 5 (x - (- 3)) + 30 (y - 2) + 2 (z - (- 1)) = 0 ⇔ - 5 x + 30 y + 2 z - 73 = 0

This is the equation we need for a plane that passes through three points.

2. Let's take a different approach. Let us write the equation for a plane with three points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3) in the following form:

x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1 = 0

Here you can substitute data from the problem statement. Since x 1 = - 3, y 1 = 2, z 1 = - 1, x 2 = - 1, y 2 = 2, z 2 = 4, x 3 = 3, y 3 = 3, z 3 = - 1, as a result we get:

x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1 = x - (- 3) y - 2 z - (- 1) - 1 - (- 3) 2 - 2 4 - (- 1) 3 - (- 3) 3 - 2 - 1 - (- 1) = = x + 3 y - 2 z + 1 2 0 5 6 1 0 = - 5 x + 30 y + 2 z - 73

We got the equation we needed.

Answer:- 5 x + 30 y + 2 z - 73 .

But what if the given points still lie on the same line and we need to create a plane equation for them? Here it must be said right away that this condition will not be entirely correct. An infinite number of planes can pass through such points, so it is impossible to calculate a single answer. Let us consider such a problem to prove the incorrectness of such a formulation of the question.

Example 2

We have a rectangular coordinate system in three-dimensional space, in which three points are placed with coordinates M 1 (5, - 8, - 2), M 2 (1, - 2, 0), M 3 (- 1, 1, 1) . It is necessary to create an equation of the plane passing through it.

Solution

Let's use the first method and start by calculating the coordinates of two vectors M 1 M 2 → and M 1 M 3 →. Let's calculate their coordinates: M 1 M 2 → = (- 4, 6, 2), M 1 M 3 → = - 6, 9, 3.

The cross product will be equal to:

M 1 M 2 → × M 1 M 3 → = i → j → k → - 4 6 2 - 6 9 3 = 0 i ⇀ + 0 j → + 0 k → = 0 →

Since M 1 M 2 → × M 1 M 3 → = 0 →, then our vectors will be collinear (re-read the article about them if you forgot the definition of this concept). Thus, the initial points M 1 (5, - 8, - 2), M 2 (1, - 2, 0), M 3 (- 1, 1, 1) are on the same line, and our problem has infinitely many options answer.

If we use the second method, we will get:

x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1 = 0 ⇔ x - 5 y - (- 8) z - (- 2) 1 - 5 - 2 - (- 8) 0 - (- 2) - 1 - 5 1 - (- 8) 1 - (- 2) = 0 ⇔ ⇔ x - 5 y + 8 z + 2 - 4 6 2 - 6 9 3 = 0 ⇔ 0 ≡ 0

From the resulting equality it also follows that the given points M 1 (5, - 8, - 2), M 2 (1, - 2, 0), M 3 (- 1, 1, 1) are on the same line.

If you want to find at least one answer to this problem from the infinite number of its options, then you need to follow these steps:

1. Write down the equation of the line M 1 M 2, M 1 M 3 or M 2 M 3 (if necessary, look at the material about this action).

2. Take a point M 4 (x 4, y 4, z 4), which does not lie on the straight line M 1 M 2.

3. Write down the equation of the plane that passes through three different points M 1, M 2 and M 4, not lying on the same straight line.

If you notice an error in the text, please highlight it and press Ctrl+Enter

In order for a single plane to be drawn through any three points in space, it is necessary that these points do not lie on the same straight line.

Consider the points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3) in general Cartesian system coordinates

In order for an arbitrary point M(x, y, z) to lie in the same plane with points M 1, M 2, M 3, it is necessary that the vectors be coplanar.

(
) = 0

Thus,

Equation of a plane passing through three points:

Equation of a plane given two points and a vector collinear to the plane.

Let the points M 1 (x 1,y 1,z 1),M 2 (x 2,y 2,z 2) and the vector be given
.

Let's create an equation for a plane passing through the given points M 1 and M 2 and an arbitrary point M (x, y, z) parallel to the vector .

Vectors
and vector
must be coplanar, i.e.

(
) = 0

Plane equation:

Equation of a plane using one point and two vectors,

collinear to the plane.

Let two vectors be given
And
, collinear planes. Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors
must be coplanar.

Plane equation:

Equation of a plane by point and normal vector .

Theorem. If a point M is given in space 0 (X 0 , y 0 , z 0 ), then the equation of the plane passing through the point M 0 perpendicular to the normal vector (A, B, C) has the form:

A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.

Proof. For an arbitrary point M(x, y, z) belonging to the plane, we compose a vector. Because vector is the normal vector, then it is perpendicular to the plane, and, therefore, perpendicular to the vector
. Then the scalar product

= 0

Thus, we obtain the equation of the plane

The theorem is proven.

Equation of a plane in segments.

If in general equation Ax + Bu + Cz + D = 0 divide both sides by (-D)

,

replacing
, we obtain the equation of the plane in segments:

The numbers a, b, c are the intersection points of the plane with the x, y, z axes, respectively.

Equation of a plane in vector form.

Where

- radius vector of the current point M(x, y, z),

A unit vector having the direction of a perpendicular dropped onto a plane from the origin.

,  and  are the angles formed by this vector with the x, y, z axes.

p is the length of this perpendicular.

In coordinates, this equation looks like:

xcos + ycos + zcos - p = 0.

Distance from a point to a plane.

The distance from an arbitrary point M 0 (x 0, y 0, z 0) to the plane Ax+By+Cz+D=0 is:

Example. Find the equation of the plane, knowing that point P(4; -3; 12) is the base of the perpendicular dropped from the origin to this plane.

So A = 4/13; B = -3/13; C = 12/13, we use the formula:

A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.

Example. Find the equation of a plane passing through two points P(2; 0; -1) and

Q(1; -1; 3) perpendicular to the plane 3x + 2y – z + 5 = 0.

Normal vector to the plane 3x + 2y – z + 5 = 0
parallel to the desired plane.

We get:

Example. Find the equation of the plane passing through points A(2, -1, 4) and

B(3, 2, -1) perpendicular to the plane X + at + 2z – 3 = 0.

The required equation of the plane has the form: A x+B y+C z+ D = 0, normal vector to this plane (A, B, C). Vector
(1, 3, -5) belongs to the plane. The plane given to us, perpendicular to the desired one, has a normal vector (1, 1, 2). Because points A and B belong to both planes, and the planes are mutually perpendicular, then

So the normal vector (11, -7, -2). Because point A belongs to the desired plane, then its coordinates must satisfy the equation of this plane, i.e. 112 + 71 - 24 +D= 0;D= -21.

In total, we get the equation of the plane: 11 x - 7y – 2z – 21 = 0.

Example. Find the equation of the plane, knowing that point P(4, -3, 12) is the base of the perpendicular dropped from the origin to this plane.

Finding the coordinates of the normal vector
= (4, -3, 12). The required equation of the plane has the form: 4 x – 3y + 12z+ D = 0. To find the coefficient D, we substitute the coordinates of point P into the equation:

16 + 9 + 144 + D = 0

In total, we get the required equation: 4 x – 3y + 12z – 169 = 0

Example. The coordinates of the vertices of the pyramid are given: A 1 (1; 0; 3), A 2 (2; -1; 3), A 3 (2; 1; 1),

Find the length of edge A 1 A 2.

Find the angle between edges A 1 A 2 and A 1 A 4.

Find the angle between edge A 1 A 4 and face A 1 A 2 A 3.

First we find the normal vector to the face A 1 A 2 A 3 as a cross product of vectors
And
.

= (2-1; 1-0; 1-3) = (1; 1; -2);

Let's find the angle between the normal vector and the vector
.

-4 – 4 = -8.

The desired angle  between the vector and the plane will be equal to  = 90 0 - .

Find the area of ​​face A 1 A 2 A 3.

Find the volume of the pyramid.

Find the equation of the plane A 1 A 2 A 3.

Let's use the formula for the equation of a plane passing through three points.

2x + 2y + 2z – 8 = 0

x + y + z – 4 = 0;

When using the computer version “ Higher mathematics course” you can run a program that will solve the above example for any coordinates of the vertices of the pyramid.

To start the program, double-click on the icon:

In the program window that opens, enter the coordinates of the vertices of the pyramid and press Enter. In this way, all decision points can be obtained one by one.

Note: To run the program, the Maple program ( Waterloo Maple Inc.) of any version, starting with MapleV Release 4, must be installed on your computer.

Suppose we need to find the equation of a plane passing through three given points that do not lie on the same line. Denoting their radius vectors by and the current radius vector by , we can easily obtain the required equation in vector form. In fact, 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.

Moving on to the coordinates, we get the equation in coordinates:

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

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

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

To obtain the equation of the desired plane, it is necessary to require that equation (17) be satisfied by the coordinates of two other points:

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

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

The equation of the 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 find:

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 obtain:

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 the passage of this plane through points (1, 1, 1) and (2, 2, 2) are:

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

From here we get . Now substituting the value of the plane into the equation, we find:

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

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

Remark 2. The problem of drawing a plane through three given points that do not lie on the same line is easily solved in general view, if we 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, different from zero (Part 1, Chapter VI, § 6):

Having expanded this determinant into 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.

You can also verify this latter directly by substituting the coordinates of any of these points instead of . On the left side we get a determinant in which either the elements of the first row are zeros or there are two identical rows. Thus, the equation constructed represents a plane passing through the three given points.

In this lesson we will look at how to use the determinant to create plane equation. If you don’t 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 using three points

Why do we need a plane equation at all? It's simple: knowing it, we can easily calculate angles, distances and other crap in problem C2. In general, you cannot do without this equation. Therefore, we formulate the problem:

Task. Three points are given in space that do not lie on the same line. Their coordinates:

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

You need to create an equation for the plane passing through these three points. Moreover, the equation should look like:

Ax + By + Cz + D = 0

where the numbers A, B, C and D are the coefficients that, in fact, need to be found.

Well, how to get the equation of a 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 can be easily solved.

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

Therefore, the most advanced teachers began to look for simpler and more elegant solutions. And they found it! True, the technique obtained rather relates 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 or evidence.

Equation of a plane through a determinant

Enough of the lyrics, let's get down to business. To begin with, a theorem about how the determinant of a matrix 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 = (x 2, y 2, z 2); K = (x 3, y 3, z 3). Then the equation of this plane can be written through the determinant:

As an example, let's try to find a pair of planes that actually occur in problems C2. Look how quickly everything is calculated:

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

We compose a determinant and equate it to zero:

We expand 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 “combed” the equation a little so that the variables x, y and z were in the correct sequence. That's it! The plane equation 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);

We immediately substitute the coordinates of the points into the determinant:

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

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

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

This could end 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 contains just x. To really get this out of the way, let's look at 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 appear in Problem C2 are defined by three points. These points are always marked on the drawing, or even indicated directly in the text of the problem. In any case, to create an equation we will need to write down their coordinates:

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

Let's consider another point on our plane with arbitrary coordinates:

T = (x, y, z)

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 compose from these vectors square matrix and equate its determinant to zero. The coordinates of the vectors will become rows of the matrix - and we will get the very determinant that is indicated in the theorem:

This formula means that the volume of a parallelepiped 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 lines of a determinant

Determinants have several great properties that make it even easier solution to problem C2. For example, it doesn’t matter to us from which point we draw the 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 is convenient for you:

Some people are confused that in one of the lines there are variables x, y and z, which do not disappear when substituting points. But they shouldn’t disappear! Substituting the numbers into the determinant, you should get this construction:

Then the determinant is expanded according to the diagram 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. It's the last one in today's lesson. I will deliberately swap the lines to make sure that the answer will give 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);
D 1 = (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 create a standard determinant and equate it to zero:

We expand 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 = a − b = y − (2 − x − z ) = y − 2 + x + z = 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 the variables x, y, z not at the bottom, but at the top:

We again expand the resulting determinant:

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. This means that it really does not depend on the order of the rows. All that remains is to write down the answer.

So, we are convinced that the equation of the plane does not depend on the sequence of lines. We can carry out similar calculations and prove that the equation of the plane does not depend on the point whose coordinates we subtract from 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.