Botany

Equation of a plane, types of equation of a plane. Plane equations: general, through three points, normal Plane in space

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, given by the equation 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 work equals 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 it be given general equation plane Ax + By + Cz + D = 0 and D ≠ 0, i.e. the plane does not pass through the origin. Let's 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 view:

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 values of the parameter 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. B 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 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 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.

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 Exam 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 all! 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 a square matrix from these vectors 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 by the fact that one of the lines contains 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.

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 equation of a plane 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 equation of a plane 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.

Consider the rectangular coordinate system Oxyz in space.

Surface equation An equation is called F(x,y,z)=0, which is satisfied by the coordinates of each point lying on the surface, and not satisfied by the coordinates of points not lying on the surface.

For example, a sphere is the locus of points equidistant from a certain point called the center of the sphere. So all points satisfying the equation
lie on a sphere with center at point O(0.0.0) and radius R (Fig. 1).

The coordinates of any point not lying on a given sphere do not satisfy this equation.

Line in space can be considered as the line of intersection of two surfaces. So in Figure 1, the intersection of the sphere with the Oxy plane is a circle with a center at point O and radius R.

The simplest surface is plane, the simplest line in space is straight.

2. Plane in space.

2.1. Equation of a plane using a point and a normal vector.

In the Oxyz coordinate system, consider the plane (Fig.2). Its position is determined by specifying the vector perpendicular to this plane, and a fixed point
lying in this plane. Vector
perpendicular to the plane
called normal vector(normal vector). Consider an arbitrary point M(x,y,z) of the plane . Vector
flat
will be perpendicular to the normal vector Using the condition of vector orthogonality
we get the equation: where

The equation ( 2.2.1 )

called the plane equation with respect to a point and a normal vector.

If we open the brackets in equation (2.1.1) and rearrange the terms, we obtain the equation orAx + By + Cz + D = 0, where

D=
.

2.2. General equation of the plane.

The equation Ax + By + Cz +D = 0 ( 2.2.1 )

is called the general equation of the plane, where
- normal vector.

Let's consider special cases of this equation.

1).D = 0. The equation looks like: Ax + By + Cz = 0. Such a plane passes through the origin. Its normal vector

2). C = 0:Ax + By + D = 0
the plane is parallel to the oz axis (Fig. 3).

3). B = 0: Ax + Cz + D = 0
the plane is parallel to the axis oy (Fig. 4).

4). A = 0: By + Cz + D = 0

the plane is parallel to the ox axis (Fig. 5).

5). C = D = 0: Ax + By = 0
the plane passes through the oz axis (Fig. 6).

6).B = D = 0: Ax + Cz = 0
the plane passes through the axis oy (Fig. 7).

7). A = D = 0: By + Cz = 0
the plane passes through the ox axis (Fig. 8).

8).A = B = 0: Cz + D = 0

||oz
the plane is parallel to the Oxy plane (Fig. 9).

9). B = C = 0: Ax + D = 0

||ox
plane

P parallel to the Oyz plane (Fig. 10).

10).A = C = 0: By + D = 0

||oy
the plane is parallel to the Oxz plane (Fig. 11).

Example 1. Write an equation for a plane passing through a point
perpendicular to the vector
Find the points of intersection of this plane with the coordinate axes.

Solution. According to formula (2.1.1) we have

2x – y + 3z + 3 = 0.

In order to find the intersection of this plane with the ox axis, we substitute y = 0, z = 0 into the resulting equation. We have 2x + 3 = 0; x = – 1.5.

The point of intersection of the desired plane with the ox axis has the coordinates:

Let's find the intersection of the plane with the axis oy. To do this, let's take x = 0; z = 0. We have

– y + 3 = 0 y = 3. So,

To find the point of intersection with the oz axis, take x = 0; y = 0
3z + 3 = 0
z = – 1. So,

Answer: 2x – y + 3z + 3 = 0,
,
,
.

Example 2. Explore planes defined by the equations:

a).

3x – y + 2z = 0

b). 2x + z – 1 = 0

Solution. V). – y + 5 = 0 A). This plane

passes through the origin (D = 0) and has a normal vector
b). In Eq.
coefficientB = 0. Therefore,

The plane is parallel to the axis.

V). In the equation – y + 5 = 0, the coefficients are A = 0, C = 0. This means

The plane is parallel to the oxz plane.

G). The equation x = 0 defines the oyz plane, since at B = 0, C = 0 the plane is parallel to the oyz plane, and from the condition D = 0 it follows that the plane passes through the origin. Example 3.
Write an equation for a plane passing through the point A(2,3,1) and perpendicular to the vector

Solution. where B(1,0, –1), C(–2,2,0).

Let's find the vector
Vector

is the normal vector of the desired plane passing through the point A(2,3,1). According to formula (2.1.1) we have:
– 3x + 2y + z + 6 – 6 – 1 = 0 – 3x + 2y + z – 1 = 0

Answer: 3x – 2y – z + 1 = 0.

2.3. 3x – 2y – z + 1 = 0.

Equation of a plane passing through three points.
Three points that do not lie on the same line define a single plane (see Fig. 12). Let the points not lie on the same line. To create an equation of a plane, you need to know one point of the plane and the normal vector. The points lying on the plane are known:
You can take any one. To find a normal vector, we use the definition of the vector product of vectors. Let
Then, therefore,
Knowing the coordinates of the point and normal vector

Let's find the equation of the plane using formula (2.1.1).
In another way, the equation of a plane passing through three given points can be obtained using the condition of coplanarity of three vectors. Indeed, vectors

where M(x,y,z) is an arbitrary point of the desired plane, coplanar (see Fig. 13). Therefore, their mixed product is 0:

(2.3.1)

Example 1. Applying the mixed product formula in coordinate form, we obtain:

Solution. Write an equation for a plane passing through the points

According to formula (2.3.1) we have

The resulting plane is parallel to the oy axis. Its normal vector

Answer: x + z – 4 = 0.

2.4. The angle between two straight lines.

Two planes, intersecting, form four dihedral angles, equal in pairs (see Fig. 14). One of the dihedral angles equal to angle between the normal vectors of these planes.

Let the planes be given:

Their normal vectors have coordinates:

From vector algebra it is known that
or

(2.4.1)

Example: Find the angle between the planes:

Solution: Let's find the coordinates of the normal vectors: Using formula (2.4.1) we have:

One of the dihedral angles obtained by intersecting these planes is equal to
You can also find the second angle:

Answer:

2.5. Condition for parallelism of two planes.

Let two planes be given:

And

If these planes are parallel, then their normal vectors

collinear (see Fig. 15).

If the vectors are collinear, then their corresponding coordinates are proportional:

(2.5.1 )

The converse statement is also true: if the normal vectors of the planes are collinear, then the planes are parallel.

Example 1. Which of the following planes are parallel:

Solution: A). Let's write down the coordinates of the normal vectors.

Let's check their collinearity:

It follows that

b). Let's write down the coordinates

Let's check collinearity:

Vectors
not collinear, planes
not parallel.

Example 2. Write an equation for a plane passing through a point

M(2, 3, –2) parallel to the plane

Solution: The desired plane is parallel to the given plane. Therefore the normal vector of the plane can be taken as the normal vector of the desired plane.
Applying equation (2.1.1), we obtain:

Answer:
.

Solution: Let's write down the coordinates of normal vectors:

Since the planes are parallel, then the vectors
collinear. By condition (2.5.1)
Hence b = – 2; a = 3.

Answer: a = 3; b = –2.

2.6. The condition of perpendicularity of two planes.

If the plane
are perpendicular, then their normal vectors
are also perpendicular (see Fig. 16). It follows that their scalar product is equal to zero, i.e.
or in coordinates:

This is the condition of perpendicularity of two planes. The converse statement is also true, that is, if condition (2.6.1) is satisfied, then the vectors
hence,