General and normal equation of the plane. General equation of a plane in space. Linear inequalities in space

To get the general equation of the plane, we analyze the plane passing through a given point.

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

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

If any point of the plane is known P and some vector of the normal to it, then by these two conditions the plane in space is completely determined(through a given point, there is only one plane perpendicular to a given vector). The general equation of the plane will look like:

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

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

.

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

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

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

Solution. We use formula (1), look at it again:

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

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

We multiply everything that needs to be multiplied and add up just numbers (which are without letters). Result:

.

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

So, an equation of the form

called the general equation of the plane .

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

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

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

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

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

Consider now special cases of the general equation of the plane. These are cases when certain coefficients of equation (2) vanish.

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

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

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

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

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

Example 3 Compose the equation of the plane P passing through the axis Oy and point .

Solution. So the plane passes through the axis Oy. So in her equation y= 0 and this equation has the form . To determine the coefficients A And C we use the fact that the point belongs to the plane P .

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

M0 (2; −4; 3) .

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

2A + 3C = 0 .

We leave 2 A on the left side of the equation, we transfer 3 C to the right side and get

A = −1,5C .

Substituting the found value A into the equation , we get

or .

This is the equation required in the example condition.

Solve the problem on the equations of the plane yourself, and then look at the solution

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

Solutions to typical problems that are control work- in the manual "Tasks on a plane: parallelism, perpendicularity, intersection of three planes at one point" .

Equation of a plane passing through three points

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

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

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

(3)

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

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

and to determine a particular case of the general equation of the line, if any.

Solution. According to formula (3) we have:

Normal equation of the plane. Distance from point to plane

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

1. General equation of the plane

Definition. A plane is a surface, all points of which satisfy the general equation: Ax + By + Cz + D \u003d 0, where A, B, C are the coordinates of the vector

N = Ai + Bj + Ck is the vector of the normal to the plane. The following special cases are possible:

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

B = 0 - the plane is parallel to the Oy axis C = 0 - the plane is parallel to the Oz axis

D = 0 - the plane passes through the origin

A = B = 0 - the plane is parallel to the xOy plane A = C = 0 - the plane is parallel to the xOz plane B = C = 0 - the plane is parallel to the yOz plane A = D = 0 - the plane passes through the Ox axis

B = D = 0 - the plane passes through the Oy axis C = D = 0 - the plane passes through the Oz axis

A = B = D = 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 yOz plane

2. 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.

3. 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 points М1 (x1 , y1 , z1 ), M2 (x2 , y2 , z2 ), M3 (x3 , y3 , z3 ) in the general Cartesian system

coordinates.
In order for an arbitrary point M (x, y, z)	lay in the same plane with the points
M 1 , M 2 , M 3 vectors M 1 M 2 , M 1 M 3 , M 1 M must be coplanar, i.e.
M1 M = ( x − x1 ; y − y1 ; z − z1 )
(M 1 M 2 , M 1 M 3 , M 1 M ) = 0. Thus, M 1 M 2	= ( x 2 − x 1 ; y 2		− y 1 ; z2 − z1)
M1 M3	= ( x 3 - x 1 ; y 3 - y 1 ; z 3 - z 1)
		x − x1	y − y1	z−z1
Equation of a plane passing through three points:		x2 − x1	y2 − y1	z2 − z1
		x 3 − x 1	y 3 − y 1	z 3 − z 1

4. Equation of a plane with respect to two points and a vector collinear to the plane

Let points М1(x1, y1, z1), M2(x2, y2, z2) and vector a = (a 1 , a 2 , a 3 ) 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 a .
Vectors M1 M = ( x − x1 ; y − y1 ; z − z1 )	and vector a = (a , a					must be
M 1M 2 = ( x 2 − x 1 ; y 2 ​​− y 1 ; z 2 − z 1)
			x − x1				y − y1	z−z1
coplanar, i.e. (M 1 M , M 1 M 2 , a ) = 0. Plane equation:		x2 − x1					y2 − y1	z2 − z1

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

Let two vectors a = (a 1 , a 2 , a 3 ) and b = (b 1 ,b 2 ,b 3 ) be given, collinear planes. Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors a, b, MM 1 must be coplanar.

6. Equation of a plane with respect to a point and a normal vector

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

7. Equation of a plane in segments

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

x -

y-

z − 1 = 0 , replacing −

C , we get the plane equation

in segments:

one . The numbers a, b, c are the intersection points of the plane, respectively

with x, y, z axes.

8. Equation of the plane in vector form

r n = p , where r = xi + yj + zk is the radius vector of the current point M (x , y , z ) ,

n = i cosα + j cos β + k cosγ - unit vector having direction, perpendicular,

dropped onto the plane from the origin. α , β and γ 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:

x cosα + y cos β + z cosγ − p = 0

9. Distance from point to plane

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

d = Ax0 + By0 + Cz0 + D

A2+B2+C2

Example. Find the equation of the plane passing through the points A(2,-1,4) and B(3,2,-1) perpendicular to the plane x + y + 2z − 3 = 0 .

The desired equation of the plane has the form: Ax + By + Cz + D = 0 , the vector of the normal to this plane n 1 (A,B,C). The vector AB (1,3,-5) belongs to the plane. The plane given to us,

perpendicular to the desired one has a normal vector n 2 (1,1,2). Because points A and B belong to both planes, and the planes are mutually perpendicular, then

n = AB × n

− 5

− j

− 5

11 i − 7 j − 2 k .

− 5

So the normal vector is n 1 (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 obtain the equation of the plane: 11x − 7 y − 2z − 21 = 0

10. Line equation in space

Both on the plane and in space, any line can be defined as a set of points whose coordinates in some coordinate system chosen in space satisfy the equation:

F (x , y , z ) = 0 . This equation is called the equation of a line in space.

In addition, a line in space can be defined in another way. It can be considered as a line of intersection of two surfaces, each of which is given by some equation.

Let F (x, y, z) \u003d 0 and Ф (x, y, z) \u003d 0 are the equations of surfaces intersecting along the line L.

F(x, y, z) = 0

Then a pair of equations Ф (x, y, z) = 0 will be called the equation of a line in space.

11. Equation of a straight line in space with respect to a point and a directing vector r 0 = M 0 M .

Because vectors M 0 M and S are collinear, then the relation M 0 M = St is true, where t is some parameter. In total, we can write: r = r 0 + St .

Because this equation is satisfied by the coordinates of any point on the line, then the resulting equation is parametric equation straight.

x = x0 + mt

This vector equation can be represented in coordinate form: y = y 0 + nt

z = z0 + pt

Transforming this system and equating the values ​​of the parameter t, we obtain the canonical

equations of a straight line in space:	x − x0		y − y0		z − z0

Definition. The direction cosines of the straight line are the direction cosines of the vector S, which can be calculated by the formulas:

	cosα =			; cos β =		; cosγ =
					N 2 + p 2		m 2 + n 2 + p 2

From here we get: m : n : p = cosα : cos β : cosγ .

The numbers m, n, p are called the slope of the line. Because S is a non-zero vector, then m, n, and p cannot be zero at the same time, but one or two of these numbers can be zero. In this case, in the equation of a straight line, the corresponding numerators should be equated to zero.

12. Equation of a straight line in space passing through two points

If on a line in space we mark two arbitrary points M 1 (x 1 , y 1 , z 1 ) and

M 2 (x 2 , y 2 , z 2 ) , then the coordinates of these points must satisfy the straight line equation obtained above:

x2 − x1		y2 − y1		z2 − z1

In this article, we will consider the normal equation of the plane. Let us give examples of constructing the normal equation of the plane according to the angle of inclination of the normal vector of the plane from the axes Ox, Oy, Oz and by distance r from the origin to the plane. Let us present a method for reducing the general equation of a straight line to normal form. Consider numerical examples.

Let a Cartesian rectangular coordinate system be given in space. Then normal equation of the plane Ω represented by the following formula:

xcosα+ycosβ+zcosγ−r=0,

(1)

where r− distance from the origin to the plane Ω , but α,β,γ are the angles between the unit vector n, orthogonal plane Ω and coordinate axes Ox, Oy, Oz, respectively (Fig.1). (If r>0, then the vector n directed towards the plane Ω , if the plane passes through the origin, then the direction of the vector n chosen arbitrarily).

We derive formula (1). Let a Cartesian rectangular coordinate system and a plane be given in space Ω (Fig.1). Draw a line through the origin Q, perpendicular to the plane Ω , and the point of intersection will be denoted by R. On this line, we select the unit vector n, with the direction coinciding with the vector . (If the dots O And R match, then the direction n can be taken arbitrarily).

We express the equation of the plane Ω through the following parameters: the length of the segment and the angles of inclination α, β, γ between vector n and axes Ox, Oy, Oz, respectively.

Since the vector n is a unit vector, then its projections onto Ox, Oy, Oz will have the following coordinates:

Dot product of vectors n and has the following form:

Given that n={cosα, cosβ, cosγ}, , we will get:

xcosα+ycosβ+zcosγ−r=0.

(7)

We have obtained the normal equation of the plane Ω . Equation (7) (or (1)) is also called normalized plane equation. Vector n called plane normal vector.

As noted above, the number r in equation (1) shows the distance of the plane from the origin. Therefore, having the normal equation of the plane, it is easy to determine the distance of the plane from the origin. To check whether a given equation of a plane is an equation in normal form, you need to check the length of the normal vector of this plane and the sign of the number r, i.e. if | n|=1 and r>0, then this equation is a normal (normalized) equation of the plane.

Example 1. Given the following plane equation:

Let's determine the length of the vector n:

Since equations (1) and (8) must determine the same straight line (Proposition 2 of the article "General equation of the plane"), then there is such a number t, what

Simplify the expression and find t:

t 2 A 2 +t 2 B 2 +t 2 C 2 =t 2 (A 2 +B 2 +C 2)=1,

.

(11)

The denominator in (11) is different from zero, because at least one of the coefficients A, B, C is not equal to zero (otherwise (8) would not represent the equation of a straight line).

Find out what sign t. Let us pay attention to the fourth equality in (9). Because r is the distance from the origin to the plane, then r≥0. Then the product tD must have negative sign. Those. sign t in (11) must be opposite to the sign D.

Substituting into (1) instead of cosα, cosβ, cosγ and −r values ​​from (9), we get tAx+tBy+tCz+tD=0. Those. to bring the general equation of the plane to normal form, you need to multiply the given equation by the factor (11). The factor (11) is called normalizing factor.

Example 2. The general equation of the plane is given

Because D>0, then sign t negative:

Note that the number is the distance from the origin to the straight line (12).

is the general equation of a plane in space

Normal plane vector

A normal vector of a plane is a nonzero vector orthogonal to each vector lying in the plane.

Equation of a plane passing through a point with a given normal vector

is the equation of the plane passing through the point M0 with a given normal vector

Plane direction vectors

Two non-collinear vectors parallel to the plane are called direction vectors of the plane

Parametric plane equations

– parametric equation of the plane in vector form

is the parametric equation of the plane in coordinates

Equation of a plane through a given point and two direction vectors

-fixed point

just a dot lol

are coplanar, so their mixed product is 0.

Equation of a plane passing through three given points

– plane equation through three points

Equation of a plane in segments

- plane equation in segments

Proof

To prove it, we use the fact that our plane passes through A, B, C, and the normal vector

Let us substitute the coordinates of the point and the vector n into the equation of the plane with the normal vector

Divide everything by and get

So it goes.

Normal plane equation

is the angle between ox and the normal vector to the plane, coming out of O.

is the angle between oy and the normal vector to the plane, outgoing from O.

is the angle between oz and the normal vector to the plane, outgoing from O.

is the distance from the origin of coordinates to the plane.

Evidence or some such bullshit

The sign is opposite D.

Similarly for other cosines. End.

Distance from point to plane

Point S, plane

is the oriented distance from the point S to the plane

If , then S and O lie on opposite sides of the plane

If , then S and O lie on the same side

Multiply by n

Mutual arrangement of two lines in space

Angle between planes

At the intersection, two pairs of vertical dihedral angles are formed, the smallest is called the angle between the planes

Straight line in space

A line in space can be given as

Intersection of two planes:

Parametric equations of a straight line

- parametric equation of a straight line in vector form

is the parametric equation of a straight line in coordinates

Canonical Equation

is the canonical equation of a straight line.

Equation of a straight line passing through two given points

– canonical equation of a straight line in vector form;

Mutual arrangement of two lines in space

Mutual arrangement of a straight line and a plane in space

Angle between line and plane

Distance from a point to a line in space

a is the direction vector of our straight line.

is an arbitrary point belonging to a given line

- the point to which we are looking for the distance.

Distance between two intersecting lines

Distance between two parallel lines

M1 - point belonging to the first line

M2 is a point belonging to the second line

Curves and surfaces of the second order

An ellipse is a set of points in a plane, the sum of the distances from which to two given points (foci) is a constant value.

Canonical equation of an ellipse

Let's replace it with

Divide by

Ellipse Properties

Intersection with coordinate axes

Symmetry about

1. Origins

An ellipse is a curve lying in a limited part of a plane

An ellipse can be obtained from a circle by stretching or squeezing it

Parametric equation of an ellipse:

- directors

Hyperbola

A hyperbola is a set of points in a plane for which the modulus of the difference in distances to 2 given points (foci) is a constant value (2a)

We do everything the same as with the ellipse, we get

Replace with

Divide by

Properties of a hyperbola

;

- directors

Asymptote

An asymptote is a straight line to which the curve approaches indefinitely, receding to infinity.

Parabola

parabot properties

Relationship between ellipse, hyperbola and parabola.

The relationship between these curves has an algebraic explanation: they are all given by equations of the second degree. In any coordinate system, the equations of these curves have the form: ax 2 +bxy+cy 2 +dx+ey+f=0, where a, b, c, d, e, f are numbers

Transforming Rectangular Cartesian Coordinate Systems

Parallel translation of the coordinate system

–O’ in the old coordinate system

– coordinates of the point in the old coordinate system

-coordinates of the point in new system coordinates

Point coordinates in the new coordinate system.

Rotate in a Cartesian Coordinate System

– new coordinate system

Transition matrix from the old basis to the new one

- (under the first column I’ , under the second j’ ) the transition matrix from the basis I,j to basis I’ ,j’

General case

1 option

1. Coordinate system rotation

Option 2

1. Coordinate system rotation

   Parallel translation of the origin

General equation of second order lines and its reduction to canonical form

is the general form of the second-order curve equations

Classification of curves of the second order

Ellipsoid

Cross sections of an ellipsoid

- ellipse

- ellipse

Ellipsoids of revolution

Ellipsoids of revolution are either oblate or prolate spheroids, depending on what we are rotating around.

One-band hyperboloid

Sections of a one-strip hyperboloid

– hyperbola with real axis oy

is a hyperbola with a real x-axis

It turns out an ellipse for any h. So it goes.

Single-strip hyperboloids of revolution

A one-sheeted hyperboloid of revolution can be obtained by rotating a hyperbola around its imaginary axis.

Two-sheeted hyperboloid

Sections of a two-sheeted hyperboloid

- hyperbole with action. axisoz

is a hyperbola with real axis oz

Cone

- a pair of intersecting lines

- a pair of intersecting lines

Elliptical paraboloid

- parabola

- parabola

Rotations

If , then the elliptic paraboloid is a surface of revolution formed by the rotation of the parabola about its axis of symmetry.

Hyperbolic paraboloid

Parabola

- parabola

h>0 hyperbola with real axis parallel to x

h<0 гипербола с действительной осью паралльной оу и мнимой ох

Under the cylinder we mean the surface that will be obtained when a straight line moves in space, which does not change its direction, if the straight line moves relative to oz, then the equation of the cylinder is the equation of a section by the plane xoy.

Elliptical cylinder

hyperbolic cylinder

parabolic cylinder

Rectilinear generators of surfaces of the second order

Lines lying completely on the surface are called rectilinear generators of the surface.

Surfaces of revolution

Fuck you lol

Display

by displaying Let's call the rule according to which each element of set A is associated with one or more elements of set B. If each is assigned a single element of the set B, then the mapping is called unambiguous, otherwise ambiguous.

Transformation set is called a one-to-one mapping of a set onto itself

Injection

Injection or one-to-one mapping of set A to set B

(different elements of a correspond to different elements of B) for example y=x^2

surjection

Surjection or mapping of a set A onto a set B

For every B, there is at least one A (for example, a sine)

Each element of set B corresponds to only one element of set A. (for example, y=x)

Consider a plane Q in space. Its position is completely determined by specifying a vector N perpendicular to this plane and some fixed point lying in the plane Q. The vector N perpendicular to the plane Q is called the normal vector of this plane. If we denote by A, B and C the projections of the normal vector N, then

Let us derive the equation of the plane Q passing through the given point and having the given normal vector . To do this, consider a vector connecting a point with an arbitrary point of the plane Q (Fig. 81).

For any position of the point M on the plane Q, the MXM vector is perpendicular to the normal vector N of the plane Q. Therefore, the scalar product Let's write the scalar product in terms of projections. Since , and vector , then

and hence

We have shown that the coordinates of any point of the Q plane satisfy equation (4). It is easy to see that the coordinates of points that do not lie on the plane Q do not satisfy this equation (in the latter case, ). Therefore, we have obtained the desired equation of the plane Q. Equation (4) is called the equation of the plane passing through the given point. It is of the first degree relative to the current coordinates

So, we have shown that any plane corresponds to an equation of the first degree with respect to the current coordinates.

Example 1. Write the equation of a plane passing through a point perpendicular to the vector.

Solution. Here . Based on formula (4), we obtain

or, after simplification,

By giving the coefficients A, B and C of equation (4) different values, we can obtain the equation of any plane passing through the point . The set of planes passing through a given point is called a bunch of planes. Equation (4), in which the coefficients A, B and C can take on any values, is called the equation of a bundle of planes.

Example 2. Write an equation for a plane passing through three points, (Fig. 82).

Solution. Let us write the equation for a bunch of planes passing through a point