Lecture 9

Analytical geometry in space.

General equation of the plane.

Definition. plane a surface is called, all points of which satisfy the general equation:

Ax + By + Cz + D = 0,

where A, B, C are the coordinates of the vector -vector normals to the plane.

The following special cases are possible:

A \u003d 0 - the plane is parallel to the Ox axis

B \u003d 0 - the plane is parallel to the Oy axis

C \u003d 0 - the plane is parallel to the Oz axis

D = 0 - the plane passes through the origin

A \u003d B \u003d 0 - the plane is parallel to the xOy plane

A \u003d C \u003d 0 - the plane is parallel to the xOz plane

B = C = 0 - the plane is parallel to the plane yOz

A \u003d D \u003d 0 - the plane passes through the Ox axis

B \u003d D \u003d 0 - the plane passes through the Oy axis

C \u003d D \u003d 0 - the plane passes through the axis Oz

A \u003d B \u003d D \u003d 0 - the plane coincides with the xOy plane

A = C = D = 0 - the plane coincides with the xOz plane

B = C = D = 0 - the plane coincides with the plane yOz

Equation of a plane passing through three points.

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 one 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 the Cartesian coordinate system.

In order for an arbitrary point M(x, y, z) to lie in the same plane with the points M 1, M 2, M 3, it is necessary that the vectors
were coplanar i.e. their mixed product:

(
) = 0

In this way,

Equation of a plane passing through three points:

Equation of a plane passing through two points parallel to a vector.

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

Let us compose the equation of the 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 passing through a point parallel to two vectors.

Let two vectors be given
and
, collinear planes and point M 1 (x 1, y 1, z 1). Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors
must be coplanar.

Plane equation:

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

Theorem. If a point M 0 is given in space (x 0, y 0, z 0), then the equation of the plane passing through the point M 0 is 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 - 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 has been proven.

Equation of a plane in segments.

If in the general equation Ax + Wu + Cz + D = 0 divide both parts by -D

,

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

The numbers a, b, c are the segments cut off by the plane at the intersection of the x, y, z axes, respectively, of the Cartesian rectangular coordinate system.

Plane equation in vector form.

where

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

A unit vector that has the direction of the perpendicular dropped to the 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 has the form:

xcos + ycos + zcos - p = 0.

Parametric plane equation

Let a point M 0 (x 0, y 0, z 0) and two non-collinear vectors be given in space

(p 1 , p 2 , p 3) and (q 1 , q 2 , q 3). Let M(x, y, z) be the current point of the plane. Since the vectors and are noncollinear, then they form a basis on the plane, in which we expand the vector
=t+ s, where t,s are parameters. Let us arbitrarily place a Cartesian rectangular coordinate system on the plane so that the Ox and Oy axes lie in the plane. From the center O we draw the radius vectors to the points M 0 and M and . Then
=-and

=+t+ s .

This is a parametric equation of the plane in vector form, and in scalar form

x=x 0 + p 1 t + q 1 s

y=y 0 + p 2 t + q 2 s

z=z 0 + p 3 t + q 3 s

The 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 + Vy + Cz + D \u003d 0 is:

Example. Find the equation of the plane, knowing that the 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, 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) are 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 the points A(2, -1, 4) and

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

The desired plane equation has the form: A x+ B y+C z+ D = 0, the 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.

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

Example . Find the equation of the plane, knowing that the 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 desired equation of the plane has the form: 4 x – 3y + 12z+ D = 0. To find the coefficient D, we substitute the coordinates of the point Р into the equation:

16 + 9 + 144 + D = 0

So, we get the desired equation: 4 x – 3y + 12z – 169 = 0

Example . Given the coordinates of the vertices of the pyramid

A 1 (1; 0; 3), A 2 (2; -1; 3), A 3 (2; 1; 1), A 4 (1; 2; 5).

Find the length of the edge A 1 A 2 .

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

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

First, 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);

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 А 1 А 2 А 3 .

We use the formula for the equation of a plane passing through three points.

2x + 2y + 2z - 8 = 0

Surface equation in space

Definition. Any equation relating the x, y, z coordinates of any point on a surface is an equation of that surface.

General equation of the plane

Definition. A plane is a surface, all points of which satisfy the general equation:

Ax + By + Cz + D = 0,

where A, B, C are the coordinates of the vector

the normal vector to the plane. The following special cases are possible:

A \u003d 0 - the plane is parallel to the Ox axis

B \u003d 0 - the plane is parallel to the Oy axis

C \u003d 0 - the plane is parallel to the Oz axis

D = 0 - the plane passes through the origin

A \u003d B \u003d 0 - the plane is parallel to the xOy plane

A \u003d C \u003d 0 - the plane is parallel to the xOz plane

B \u003d C \u003d 0 - the plane is parallel to the plane yOz

A \u003d D \u003d 0 - the plane passes through the Ox axis

B \u003d D \u003d 0 - the plane passes through the Oy axis

C \u003d D \u003d 0 - the plane passes through the Oz axis

A \u003d B \u003d D \u003d 0 - the plane coincides with the xOy plane

A \u003d C \u003d D \u003d 0 - the plane coincides with the xOz plane

B \u003d C \u003d D \u003d 0 - the plane coincides with the plane yOz

Equation of a plane passing through three points

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 one straight line. Consider the points М1(x1, y1, z1), M2(x2, y2, z2), M3(x3, y3, z3) in the general Cartesian coordinate system. In order for an arbitrary point M(x, y, z) to lie in the same plane as the points M1, M2, M3, the vectors must be coplanar.

In this way,

Equation of a plane passing through three points:

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

Let points M1(x1, y1, z1), M2(x2, y2, z2) and a vector be given.

Let us compose the equation of the plane passing through the given points M1 and M2 and an arbitrary point M(x, y, z) parallel to the vector.

Vectors and the vector must be coplanar, i.e.

Plane equation:

Equation of a plane with respect to one point and two vectors collinear to the plane

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

Plane equation by point and normal vector

Theorem. If a point M0 (x0, y0, z0) is given in space, then the equation of the plane passing through the point M0 perpendicular to the normal vector (A, B, C) has the form:

A(x - x0) + B(y - y0) + C(z - z0) = 0.

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

Thus, we obtain the equation of the plane

The theorem has been proven.

The general equation of a straight line is called complete, if all its coefficients are not equal to 0. Otherwise, the equation is called incomplete.

D=0 Ax+Vu+Сz=0- plane, passing through the origin of coordinates.

The remaining cases are determined by the position of the normal vector n=( A; B; C).

A=0 Ву+Сz+D=0 is the equation of the plane, parallel axis Ox.(Because the normal vector n=( 0;B;C) is perpendicular to the Ox axis).

B=0 Ah+Сz+D=0 - plane Equation, parallel to the y axis.(Because the normal vector n=( A; 0; C) is perpendicular to the Oy axis).

C=0 Ah+Wu+D=0 - plane Equation, parallel axis Oz. (Because the normal vector n=( A; B; 0) is perpendicular to the Oz axis).

A=B=0 Сz+D=0 – z=-D/C – the equation of a plane parallel to the Oxy plane (because this plane is parallel to the Ox and Oy axes).

A=C=0 Wu+D=0 - y=-D/B- the equation of a plane parallel to the Oxz plane (because this plane is parallel to the Ox and Oz axes).

B=C=0 Ah+D=0 – x=-D/A- the equation of a plane parallel to the Oyz plane (because this plane is parallel to the Oy and Oz axes).

A=D=0 By+Cz=0 - equation of the plane passing through the x-axis.

B=D=0 Ax+Cz=0 - equation of the plane passing through the Oy axis.

A=B=D=0 Cz=0 (z=0) – Oxy coordinate plane.(because this plane is parallel to Oxy and passes through the origin).

A=C=D=0 By=0 (y=0) – coordinate plane Охz.(because this plane is parallel to Oxz and passes through the origin).

B=C=D=0 ax=0 (x=0) – coordinate plane Оуz.(because this plane is parallel to Oyz and passes through the origin).

Equation of a plane passing through three given points.

We derive the equation of a plane passing through 3 different 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), not lying on one straight line. Then the vectors M 1 M 2 \u003d (x 2 -x 1; y 2 ​​-y 1; z 2 -z 1) and M 1 M 3 \u003d (x 3 -x 1; y 3 -y 1; z 3 -z 1) are not collinear. Therefore, the point M(x, y, z) lies in the same plane with the points M 1 , M 2 and M 3 if and only if the vectors M 1 M 2 , M 1 M 3 and M 1 M\u003d (x-x 1; y-y 1; z-z 1) - coplanar, i.e.  when their mixed product is 0

(M 1 MM 1 M 2 M 1 M 3 =0) , i.e.

(4) Equation of a plane passing through 3 given points.

(Expanding the determinant along the 1st line and simplifying, we get the general equation of the plane: Ax + Vy + Cz + D \u003d 0).

That. three points uniquely define a plane.

The equation of the plane in segments on the axes.

The plane Π intersects the coordinate axes at the points M 1 (a; 0; 0), M 2 (0; b; 0), M 3 (0; 0; c).

M (x; y; z) is a variable point of the plane.

M 1 M=(x-a; y; z)

M 1 M 2 =(0-а;b;0) define the given plane

M 1 M 3 =(-a;0;c)

Those. M 1 MM 1 M 2 M 1 M 3 =0

Let's expand on the 1st line: (х-а)bc-y(-ac)+zab=xbc-abc+yac+zab=0

Divide the equality by abc≠0. We get:

(5) the equation of the plane in segments on the axes.

Equation (5) can be obtained from the general equation of the plane, assuming that D≠0, divide by D

Denoting –D/A=a, -D/B=b, -C/D=c – we get Equation 4.

Angle between two planes. Conditions of parallelism and perpendicularity of planes.

The angle φ between two planes α 1 and α 2 is measured by a flat angle between 2 rays perpendicular to the line along which these planes intersect. Any two intersecting planes form two angles that sum up to . It is enough to define one of these angles.

Let the planes be given by the general equations:

 1 : A 1 x+B 1 y+C 1 z+D 1 =0

 2 : A 2 x+ B 2 y+ C 2 z+ D 2 =0

Consider PDSC (O, i,j,k) in the space R 3 . Let  be some plane and vector  N perpendicular to a. We fix an arbitrary point M 0 on the plane  and take the current point M of the space.. Denote ` r =
and` r 0 =
. Then
=`r – `r 0 , and the point М if and only if the vectors ` N and
orthogonal. The latter is possible when

N .
= 0, i.e. N . (`r-`r 0) = 0, (9)

this equation is called vector equation planes. Vector ` N called normal plane vector.

If a ` N =(BUT, AT, FROM), M 0 ( X 0 , at 0 , z 0) , M( X, at, z) , then equation (9) takes the form

BUT( X – X 0) + B( at – at 0) + C( z – z 0) = 0, (10).

This equation is called the equation of a plane passing through a given point perpendicular to a given vector.

To It is known that through three points one plane can be drawn. Let M 1 ( X 1 , at 1 , z 1), M 3 ( X 2 , at 2 , z 2), M 3 ( X 3 , at 3 , z 3). Let's find the equation of this plane. According to the vector equation (9), in order to write this equation, it is necessary to know the point of the plane and the normal vector. We have a point (for example, M 1). And as a normal vector, any vector perpendicular to this plane will do. It is known that the cross product of two vectors is perpendicular to the plane in which these vectors lie. Therefore, the cross product of vectors
and
can be taken as a normal vector of the plane :

` N =


Then the equation of the plane  in vector form has the form

. (

) =
.
.
= 0.

(note that we have obtained the condition for the complanarity of vectors
,
,
).

Through the coordinates of the points M 1, M 2, M 3 and M, this equation can be written as

, (11)

and is called the equation of the plane, passing through three given points M 1 ( X 1 , at 1 , z 1), M 2 ( X 2 , at 2 , z 2), M 3 ( X 3 , at 3 , z 3).

Consider equation (9) again, transform it:

Oh + Wu + cz +(–Oh 0 – Wu 0 – cz 0) = 0 ,

Oh + Wu + cz+D = 0, where D = (– Oh 0 – Wu 0 – cz 0) .

The equation

Oh + Wu + cz+D = 0, (12)

called general equation planes. Here the vectorN = ( A, B, C) is the normal vector of the plane (i.e., the vector perpendicular to the plane). The theorem is true:

Theorem 4.2.

In the space R 3 any plane can be described linear with respect to the variables x y, z equation and vice versa. any equation of the first degree defines some plane.

Let us study the location of the plane relative to the coordinate system according to its general equation Oh + Wu + cz+D = 0 .

If the coefficient D = 0, then the coordinates of the point O(0, 0, 0) satisfy the equation Oh + Wu + cz= 0, so this point lies on the plane, i.e. plane with equation Oh + Wu + cz= 0 passes through the origin.

If in the general equation of the plane missing one from the variables (the corresponding coefficient is equal to zero), then the plane is parallel to the coordinate axis of the same name. For example, the equation Oh + cz + D= 0 defines a plane parallel to the y-axis. Indeed, the normal vector has coordinates ` N= (A, 0, C) and it is easy to check that ` Nj. But if a plane and a vector are perpendicular to the same vector, then they are parallel. Plane with equation Wu + cz= 0, in this case, passes through the OX axis (that is, this axis lies on the plane)

The absence of two variables in the plane equation means that the plane is parallel to the corresponding coordinate plane, for example, an equation of the form Oh + D= 0 defines a plane parallel to the YOZ plane. The normal vector has coordinates ` N= (A, 0, 0), it is collinear to the vector  i, and, therefore, the plane is perpendicular to the vector  i, or parallel to the UOZ plane.

Equations of coordinate planes look like: HOW: z= 0, pl. XOZ: y= 0, pl. YOZ: x = 0.

Indeed, the HOW plane passes through the origin (D = 0) and the vector  k=(0, 0, 1) is its normal vector. Similarly, the XOZ and YOZ planes pass through the origin (D = 0) and the vectors  j=(0, 1, 0) and  i = (1,0,0) are their normals, respectively.

If D0, then we transform the general equation as follows

Oh + Wu+C z = –D,
,
.

O denoting here
,
,
, we get the equation
, (13)

which is called the plane equation in segments on the axes. Here a, b, c are the values ​​of the segments cut off by the plane on the coordinate axes (Fig.). This equation is convenient to use to construct a plane in a coordinate system. It is easy to verify that the points ( a, 0, 0), (0. b, 0), (0, 0, With) lie on a plane. The lines passing through these points are called traces planes on coordinate planes.

For example, let's build a plane

2X – 3at + 4z –12 = 0.

Let us bring this equation to the form (13), we obtain

D To build a plane in the coordinate system, mark the point (6, 0, 0) on the OX axis, the point (0, -4, 0) on the OY axis, (0, 0, 3) on the OZ axis, connect them with straight line segments ( plane traces). The resulting triangle is a part of the desired plane, enclosed between the coordinate axes.

So that find the equation of the plane enough to know

Either the normal vector of this plane and any of its points (equation (10));

Or three points lying on a plane (equation (11)).

Mutual arrangement of planes in space it is convenient to study using the vectors corresponding to them. If  is a plane with a normal vector N, then

.

The derivation of the formula is similar to how it was done for a straight line on a plane. Carry it out on your own.

It can be specified in different ways (one point and a vector, two points and a vector, three points, etc.). It is with this in mind that the equation of the plane can have different forms. Also, under certain conditions, the planes can be parallel, perpendicular, intersecting, etc. We will talk about this in this article. We will learn how to write the general equation of the plane and not only.

Normal form of the equation

Let's say there is a space R 3 that has a rectangular coordinate system XYZ. We set the vector α, which will be released from the initial point O. Through the end of the vector α we draw the plane P, which will be perpendicular to it.

Denote by P an arbitrary point Q=(x, y, z). We will sign the radius vector of the point Q with the letter p. The length of the vector α is p=IαI and Ʋ=(cosα,cosβ,cosγ).

This is a unit vector that points sideways, just like the vector α. α, β and γ are the angles that form between the vector Ʋ and the positive directions of the space axes x, y, z, respectively. The projection of some point QϵП onto the vector Ʋ is a constant value equal to р: (р,Ʋ) = р(р≥0).

This equation makes sense when p=0. The only thing is that the plane P in this case will intersect the point O (α=0), which is the origin, and the unit vector Ʋ released from the point O will be perpendicular to P, regardless of its direction, which means that the vector Ʋ is determined from sign-accurate. The previous equation is the equation of our P plane, expressed in vector form. But in coordinates it will look like this:

P here is greater than or equal to 0. We have found the equation of a plane in space in its normal form.

General Equation

If we multiply the equation in coordinates by any number that is not equal to zero, we get an equation equivalent to the given one, which determines that same plane. It will look like this:

Here A, B, C are numbers that are simultaneously different from zero. This equation is referred to as the general plane equation.

Plane equations. Special cases

The equation in general form can be modified in the presence of additional conditions. Let's consider some of them.

Assume that the coefficient A is 0. This means that the given plane is parallel to the given axis Ox. In this case, the form of the equation will change: Ву+Cz+D=0.

Similarly, the form of the equation will change under the following conditions:

• Firstly, if B = 0, then the equation will change to Ax + Cz + D = 0, which will indicate parallelism to the Oy axis.
• Secondly, if С=0, then the equation is transformed into Ах+Ву+D=0, which will indicate parallelism to the given axis Oz.
• Thirdly, if D=0, the equation will look like Ax+By+Cz=0, which will mean that the plane intersects O (the origin).
• Fourth, if A=B=0, then the equation will change to Cz+D=0, which will prove parallel to Oxy.
• Fifth, if B=C=0, then the equation becomes Ax+D=0, which means that the plane to Oyz is parallel.
• Sixth, if A=C=0, then the equation will take the form Ву+D=0, that is, it will report parallelism to Oxz.

Type of equation in segments

In the case when the numbers A, B, C, D are non-zero, the form of equation (0) can be as follows:

x/a + y/b + z/c = 1,

in which a \u003d -D / A, b \u003d -D / B, c \u003d -D / C.

We get as a result It is worth noting that this plane will intersect the Ox axis at a point with coordinates (a,0,0), Oy - (0,b,0), and Oz - (0,0,c).

Taking into account the equation x/a + y/b + z/c = 1, it is easy to visually represent the placement of the plane relative to a given coordinate system.

Normal vector coordinates

The normal vector n to the plane P has coordinates that are the coefficients of the general equation of the given plane, that is, n (A, B, C).

In order to determine the coordinates of the normal n, it is sufficient to know the general equation of a given plane.

When using the equation in segments, which has the form x/a + y/b + z/c = 1, as well as when using the general equation, one can write the coordinates of any normal vector of a given plane: (1/a + 1/b + 1/ With).

It should be noted that the normal vector helps to solve various problems. The most common are tasks that consist in proving the perpendicularity or parallelism of planes, problems in finding angles between planes or angles between planes and lines.

View of the equation of the plane according to the coordinates of the point and the normal vector

A non-zero vector n perpendicular to a given plane is called normal (normal) for a given plane.

Suppose that in the coordinate space (rectangular coordinate system) Oxyz are given:

• point Mₒ with coordinates (xₒ,yₒ,zₒ);
• zero vector n=A*i+B*j+C*k.

It is necessary to compose an equation for a plane that will pass through the point Mₒ perpendicular to the normal n.

In space, we choose any arbitrary point and denote it by M (x y, z). Let the radius vector of any point M (x, y, z) be r=x*i+y*j+z*k, and the radius vector of the point Mₒ (xₒ,yₒ,zₒ) - rₒ=xₒ*i+yₒ *j+zₒ*k. The point M will belong to the given plane if the vector MₒM is perpendicular to the vector n. We write the orthogonality condition using the scalar product:

[MₒM, n] = 0.

Since MₒM \u003d r-rₒ, the vector equation of the plane will look like this:

This equation can take another form. To do this, the properties of the scalar product are used, and the left side of the equation is transformed. = - . If denoted as c, then the following equation will be obtained: - c \u003d 0 or \u003d c, which expresses the constancy of the projections onto the normal vector of the radius vectors of the given points that belong to the plane.

Now you can get the coordinate form of writing the vector equation of our plane = 0. Since r-rₒ = (x-xₒ)*i + (y-yₒ)*j + (z-zₒ)*k, and n = A*i+B *j+C*k, we have:

It turns out that we have an equation for a plane passing through a point perpendicular to the normal n:

A*(x-xₒ)+B*(y-yₒ)C*(z-zₒ)=0.

View of the plane equation according to the coordinates of two points and a vector collinear to the plane

We define two arbitrary points M′ (x′,y′,z′) and M″ (x″,y″,z″), as well as the vector a (a′,a″,a‴).

Now we can compose an equation for a given plane, which will pass through the available points M′ and M″, as well as any point M with coordinates (x, y, z) parallel to the given vector a.

In this case, the vectors M′M=(x-x′;y-y′;z-z′) and M″M=(x″-x′;y″-y′;z″-z′) must be coplanar with the vector a=(a′,a″,a‴), which means that (M′M, M″M, a)=0.

So, our equation of a plane in space will look like this:

Type of the equation of a plane intersecting three points

Suppose we have three points: (x′, y′, z′), (x″,y″,z″), (x‴,y‴,z‴), which do not belong to the same straight line. It is necessary to write the equation of the plane passing through the given three points. The theory of geometry claims that this kind of plane really exists, only it is the only one and inimitable. Since this plane intersects the point (x′, y′, z′), the form of its equation will be as follows:

Here A, B, C are different from zero at the same time. Also, the given plane intersects two more points: (x″,y″,z″) and (x‴,y‴,z‴). In this regard, the following conditions must be met:

Now we can compose a homogeneous system with unknowns u, v, w:

In our case, x, y or z is an arbitrary point that satisfies equation (1). Taking into account the equation (1) and the system of equations (2) and (3), the system of equations indicated in the figure above satisfies the vector N (A, B, C), which is non-trivial. That is why the determinant of this system is equal to zero.

Equation (1), which we have obtained, is the equation of the plane. It passes exactly through 3 points, and this is easy to check. To do this, we need to expand our determinant over the elements in the first row. It follows from the existing properties of the determinant that our plane simultaneously intersects three initially given points (x′, y′, z′), (x″,y″,z″), (x‴,y‴,z‴). That is, we have solved the task set before us.

Dihedral angle between planes

A dihedral angle is a spatial geometric figure formed by two half-planes that emanate from one straight line. In other words, this is the part of space that is limited by these half-planes.

Let's say we have two planes with the following equations:

We know that the vectors N=(A,B,C) and N¹=(A¹,B¹,C¹) are perpendicular according to the given planes. In this regard, the angle φ between the vectors N and N¹ is equal to the angle (dihedral), which is between these planes. The scalar product has the form:

NN¹=|N||N¹|cos φ,

precisely because

cosφ= NN¹/|N||N¹|=(AA¹+BB¹+CC¹)/((√(A²+B²+C²))*(√(A¹)²+(B¹)²+(C¹)²)).

It suffices to take into account that 0≤φ≤π.

In fact, two planes that intersect form two (dihedral) angles: φ 1 and φ 2 . Their sum is equal to π (φ 1 + φ 2 = π). As for their cosines, their absolute values ​​are equal, but they differ in signs, that is, cos φ 1 =-cos φ 2. If in equation (0) we replace A, B and C with the numbers -A, -B and -C, respectively, then the equation that we get will determine the same plane, the only angle φ in the equation cos φ= NN 1 /| N||N 1 | will be replaced by π-φ.

Perpendicular plane equation

Planes are called perpendicular if the angle between them is 90 degrees. Using the material outlined above, we can find the equation of a plane perpendicular to another. Let's say we have two planes: Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D=0. We can state that they will be perpendicular if cosφ=0. This means that NN¹=AA¹+BB¹+CC¹=0.

Parallel plane equation

Parallel are two planes that do not contain common points.

The condition (their equations are the same as in the previous paragraph) is that the vectors N and N¹, which are perpendicular to them, are collinear. This means that the following conditions of proportionality are satisfied:

A/A¹=B/B¹=C/C¹.

If the proportionality conditions are extended - A/A¹=B/B¹=C/C¹=DD¹,

this indicates that these planes coincide. This means that the equations Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D¹=0 describe one plane.

Distance to plane from point

Let's say we have a plane P, which is given by equation (0). It is necessary to find the distance to it from the point with coordinates (xₒ,yₒ,zₒ)=Qₒ. To do this, you need to bring the equation of the plane P into normal form:

(ρ,v)=p (p≥0).

In this case, ρ(x,y,z) is the radius vector of our point Q located on P, p is the length of the perpendicular to P that was released from the zero point, v is the unit vector that is located in the a direction.

The difference ρ-ρº of the radius vector of some point Q \u003d (x, y, z) belonging to P, as well as the radius vector of a given point Q 0 \u003d (xₒ, yₒ, zₒ) is such a vector, the absolute value of the projection of which on v is equal to the distance d, which must be found from Q 0 \u003d (xₒ, yₒ, zₒ) to P:

D=|(ρ-ρ 0 ,v)|, but

(ρ-ρ 0 ,v)= (ρ,v)-(ρ 0 ,v) =р-(ρ 0 ,v).

So it turns out

d=|(ρ 0 ,v)-p|.

Thus, we will find the absolute value of the resulting expression, that is, the desired d.

Using the language of parameters, we get the obvious:

d=|Axₒ+Vuₒ+Czₒ|/√(A²+B²+C²).

If the given point Q 0 is on the other side of the plane P, as well as the origin, then between the vector ρ-ρ 0 and v is therefore:

d=-(ρ-ρ 0 ,v)=(ρ 0 ,v)-p>0.

In the case when the point Q 0, together with the origin, is located on the same side of P, then the angle created is acute, that is:

d \u003d (ρ-ρ 0, v) \u003d p - (ρ 0, v)>0.

As a result, it turns out that in the first case (ρ 0 ,v)> р, in the second (ρ 0 ,v)<р.

Tangent plane and its equation

The tangent plane to the surface at the point of contact Mº is the plane containing all possible tangents to the curves drawn through this point on the surface.

With this form of the surface equation F (x, y, z) \u003d 0, the equation of the tangent plane at the tangent point Mº (xº, yº, zº) will look like this:

F x (xº, yº, zº)(x- xº)+ F x (xº, yº, zº)(y-yº)+ F x (xº, yº, zº)(z-zº)=0.

If you specify the surface in explicit form z=f (x, y), then the tangent plane will be described by the equation:

z-zº = f(xº, yº)(x- xº)+f(xº, yº)(y-yº).

Intersection of two planes

In the coordinate system (rectangular) Oxyz is located, two planes П′ and П″ are given, which intersect and do not coincide. Since any plane located in a rectangular coordinate system is determined by the general equation, we will assume that P′ and P″ are given by the equations A′x+B′y+C′z+D′=0 and A″x+B″y+ С″z+D″=0. In this case, we have the normal n′ (A′, B′, C′) of the P′ plane and the normal n″ (A″, B″, C″) of the P″ plane. Since our planes are not parallel and do not coincide, these vectors are not collinear. Using the language of mathematics, we can write this condition as follows: n′≠ n″ ↔ (A′, B′, C′) ≠ (λ*A″,λ*B″,λ*C″), λϵR. Let the line that lies at the intersection of P′ and P″ be denoted by the letter a, in this case a = P′ ∩ P″.

a is a straight line consisting of the set of all points of (common) planes П′ and П″. This means that the coordinates of any point belonging to the line a must simultaneously satisfy the equations A′x+B′y+C′z+D′=0 and A″x+B″y+C″z+D″=0. This means that the coordinates of the point will be a particular solution of the following system of equations:

As a result, it turns out that the (general) solution of this system of equations will determine the coordinates of each of the points of the straight line, which will act as the intersection point of П′ and П″, and determine the straight line a in the coordinate system Oxyz (rectangular) in space.