Every first degree equation with respect to coordinates x, y, z

Ax + By + Cz +D = 0 (3.1)

defines a plane, and vice versa: any plane can be represented by equation (3.1), which is called plane equation.

Vector n(A, B, C) orthogonal to the plane is called normal vector plane. In equation (3.1), the coefficients A, B, C are not equal to 0 at the same time.

Special cases of equation (3.1):

1. D = 0, Ax+By+Cz = 0 - the plane passes through the origin.

2. C = 0, Ax+By+D = 0 - the plane is parallel to the Oz axis.

3. C = D = 0, Ax + By = 0 - the plane passes through the Oz axis.

4. B = C = 0, Ax + D = 0 - the plane is parallel to the Oyz plane.

Equations coordinate planes: x = 0, y = 0, z = 0.

A straight line in space can be specified:

1) as a line of intersection of two planes, i.e. system of equations:

A 1 x + B 1 y + C 1 z + D 1 = 0, A 2 x + B 2 y + C 2 z + D 2 = 0; (3.2)

2) by its two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the straight line passing through them is given by the equations:

= ; (3.3)

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

. (3.4)

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

Vector a called directing vector straight.

We obtain parametric ones by equating each of the relations (3.4) to the parameter t:

x = x 1 +mt, y = y 1 + nt, z = z 1 + rt. (3.5)

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

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

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

.

From general equations (3.2) we can move to canonical ones in another way, if we find any point of this line and its directing line n= [n 1 , n 2 ], where n 1 (A 1, B 1, C 1) and n 2 (A 2 , B 2 , C 2 ) - normal vectors of given planes. If one of the denominators m, n or R in equations (3.4) turns out to be equal to zero, then the numerator of the corresponding fraction must be set equal to zero, i.e. system

is equivalent to the system ; such a straight line is perpendicular to the Ox axis.

System is equivalent to the system x = x 1, y = y 1; the straight line is parallel to the Oz axis.

Example 1.15. Write an equation for the plane, knowing that point A(1,-1,3) serves as the base of a perpendicular drawn from the origin to this plane.

Solution. According to the problem conditions, the vector OA(1,-1,3) is a normal vector of the plane, then its equation can be written as
x-y+3z+D=0. Substituting the coordinates of point A(1,-1,3), belonging to the plane, let's find D: 1-(-1)+3×3+D = 0, D = -11. So x-y+3z-11=0.

Example 1.16. Write an equation for a plane passing through the Oz axis and forming an angle of 60° with the plane 2x+y-z-7=0.

Solution. The plane passing through the Oz axis is given by the equation Ax+By=0, where A and B do not simultaneously vanish. Let B not
equals 0, A/Bx+y=0. Using the cosine formula for the angle between two planes

.

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

Example 1.17. Compose canonical equations straight:
5x + y + z = 0, 2x + 3y - 2z + 5 = 0.

Solution. The canonical equations of the line have the form:

Where m, n, p- coordinates of the directing vector of the straight line, x 1 , y 1 , z 1- coordinates of any point belonging to a line. A straight line is defined as the line of intersection of two planes. To find a point belonging to a line, one of the coordinates is fixed (the easiest way is to set, for example, x=0) and the resulting system is solved as a system of linear equations with two unknowns. So, let x=0, then y + z = 0, 3y - 2z+ 5 = 0, whence y=-1, z=1. We found the coordinates of the point M(x 1, y 1, z 1) belonging to this line: M (0,-1,1). The direction vector of a straight line is easy to find, knowing the normal vectors of the original planes n 1 (5,1,1) and n 2 (2,3,-2). Then

The canonical equations of the line have the form: x/(-5) = (y + 1)/12 =
= (z - 1)/13.

Example 1.18. In the beam defined by the planes 2x-y+5z-3=0 and x+y+2z+1=0, find two perpendicular planes, one of which passes through the point M(1,0,1).

Solution. The equation of the beam defined by these planes has the form u(2x-y+5z-3) + v(x+y+2z+1)=0, where u and v do not vanish simultaneously. Let us rewrite the beam equation as follows:

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

In order to select a plane from the beam that passes through point M, we substitute the coordinates of point M into the equation of the beam. We get:

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

Then we find the equation of the plane containing M by substituting v = - u into the beam equation:

u(2x-y +5z - 3) - u (x + y +2z +1) = 0.

Because u ¹0 (otherwise v=0, and this contradicts the definition of a beam), then we have the equation of the plane x-2y+3z-4=0. The second plane belonging to the beam must be perpendicular to it. Let us write down the condition for the orthogonality of planes:

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

This means that the equation of the second plane has the form:

u(2x -y+5z - 3) - 19/5 u(x + y +2z +1) = 0 or 9x +24y + 13z + 34 = 0.

The canonical equations of a line in space are the equations that determine the line passing through given point collinear to the direction vector.

Let a point and a direction vector be given. An arbitrary point lies on a line l only if the vectors and are collinear, i.e., the condition is satisfied for them:

.

The above equations are the canonical equations of the straight line.

Numbers m , n And p are projections of the direction vector onto the coordinate axes. Since the vector is non-zero, then all numbers m , n And p cannot simultaneously be equal to zero. But one or two of them may turn out to be zero. In analytical geometry, for example, the following entry is allowed:

,

which means that the projections of the vector on the axis Oy And Oz are equal to zero. Therefore, both the vector and the straight line defined by the canonical equations are perpendicular to the axes Oy And Oz, i.e. planes yOz .

Example 1. Write equations for a line in space perpendicular to a plane and passing through the point of intersection of this plane with the axis Oz .

Solution. Let's find the point of intersection of this plane with the axis Oz. Since any point lying on the axis Oz, has coordinates , then, assuming in given equation plane x = y = 0, we get 4 z- 8 = 0 or z= 2 . Therefore, the point of intersection of this plane with the axis Oz has coordinates (0; 0; 2) . Since the desired line is perpendicular to the plane, it is parallel to its normal vector. Therefore, the directing vector of the straight line can be the normal vector given plane.

Now let’s write down the required equations for a straight line passing through a point A= (0; 0; 2) in the direction of the vector:

Equations of a line passing through two given points

A straight line can be defined by two points lying on it And In this case, the directing vector of the straight line can be the vector . Then the canonical equations of the line take the form

.

The above equations determine a line passing through two given points.

Example 2. Write an equation for a line in space passing through the points and .

Solution. Let us write down the required equations of the straight line in the form given above in the theoretical reference:

.

Since , then the desired straight line is perpendicular to the axis Oy .

Straight as the line of intersection of planes

A straight line in space can be defined as the line of intersection of two non-parallel planes and, i.e., as a set of points satisfying a system of two linear equations

The equations of the system are also called the general equations of a straight line in space.

Example 3. Compose canonical equations of a line in space given by general equations

Solution. To write the canonical equations of a line or, what is the same thing, the equations of a line passing through two given points, you need to find the coordinates of any two points on the line. They can be the points of intersection of a straight line with any two coordinate planes, for example yOz And xOz .

Point of intersection of a line and a plane yOz has an abscissa x= 0 . Therefore, assuming in this system of equations x= 0, we get a system with two variables:

Her decision y = 2 , z= 6 together with x= 0 defines a point A(0; 2; 6) the desired line. Then assuming in the given system of equations y= 0, we get the system

Her decision x = -2 , z= 0 together with y= 0 defines a point B(-2; 0; 0) intersection of a line with a plane xOz .

Now let's write down the equations of the line passing through the points A(0; 2; 6) and B (-2; 0; 0) :

,

or after dividing the denominators by -2:

,


All plane equations that are discussed in the following paragraphs can be obtained from general equation plane, and are 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, called equation of the plane in segments.

This name is not accidental. Absolute values 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 .

To get 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, there you will also find detailed solutions typical examples and tasks.

Normal plane equation.

The general plane equation of the form is called normal equation plane, 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 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 a length equal to one, 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 linear algebra and analytical geometry.
  • Ilyin V.A., Poznyak E.G. Analytic geometry.

To obtain the general equation of a plane, let us 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 P arbitrary plane in space. Every vector perpendicular to it is called normal vector to this plane. Naturally, we are talking about a non-zero vector.

If any point on the plane is known P and some normal vector to it, then by these two conditions the plane in space is completely defined(through a given point you can draw a single plane perpendicular to this vector). The general equation of the plane will be:

So, the conditions that define the equation of the plane are. To get yourself plane equation, having the above form, 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 specified by condition. We find the coordinates of the vector using the formula :

.

Now, using the scalar product of vectors formula , let's express 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 a 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. Let’s use formula (1) and look at it again:

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

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

We multiply everything that needs to be multiplied and add just numbers (which do not have letters). Result:

.

The required equation of the plane in this example turned out to be expressed by a 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 general plane equation .

Example 2. Construct in a rectangular Cartesian coordinate system a 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 the same 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 for X and Y in the equation given in the problem statement: x = y= 0 . Therefore we get z= 6. Thus, given plane crosses 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, the 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) . Based on the three points obtained in our solution A(0; 0; 6) , B(0; −3; 0) and C(2; 0; 0) construct the given plane.

Let's now consider special cases of the general plane equation. These are cases when certain coefficients of equation (2) become zero.

1. When D= 0 equation defines a plane passing through the origin, since the coordinates of the 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 onto the axis Ox equal to zero). Similarly, when B= 0 plane parallel to the axis Oy, and when C= 0 plane parallel to the axis Oz.

3. When A=D= 0 equation defines a plane passing through the axis Ox, since 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, since it is parallel to the axes Ox (A= 0) and Oy (B= 0). Similarly, the plane is parallel to the plane yOz, and the plane is 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). Likewise, Eq. y= 0 in space defines the coordinate plane xOz, and the equation x = 0 - coordinate plane yOz.

Example 3. Create an equation of the plane P, passing through the axis Oy and period.

Solution. So the plane passes through the axis Oy. Therefore, in her equation y= 0 and this equation has the form . To determine the coefficients A And C let's take advantage of the fact that the point belongs to the plane P .

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

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 .

Leave 2 A on the left side of the equation, move 3 C to the right side and we 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 plane equation problem yourself, and then look at the solution

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

Solutions to typical problems that occur in tests- in the manual "Plane problems: 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 the same line.

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

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

(3)

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

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

and determine special case general equation of the line, if such exists.

Solution. According to formula (3) we have:

Normal plane equation. Distance from point to plane

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