General plane equation - description, examples, problem solving. Equation of a plane. How to write an equation of a plane? Mutual arrangement of planes. Problems Write the equation of a perpendicular to a given plane

If all numbers A, B, C and D are different from zero, then the general equation of the plane is called complete. Otherwise, the general equation of the plane is called incomplete.

Let us consider all possible general incomplete equations of the plane in the rectangular coordinate system Oxyz in three-dimensional space.

Let D = 0, then we have a general incomplete plane equation of the form . This plane in the rectangular coordinate system Oxyz passes through the origin. Indeed, when substituting the coordinates of a point into the resulting incomplete equation of the plane, we arrive at the identity .

For , or , or we have general incomplete equations of the planes , or , or , respectively. These equations define planes parallel to the coordinate planes Oxy, Oxz and Oyz, respectively (see the article for the condition of parallel planes) and passing through the points and correspondingly. At. Since the point belongs to the plane by condition, then the coordinates of this point must satisfy the equation of the plane, that is, the equality must be true. From here we find. Thus, the required equation has the form .

Let us present the second way to solve this problem.

Since the plane, the general equation of which we need to compose, is parallel to the plane Oyz, then as its normal vector we can take the normal vector of the plane Oyz. The normal vector of the coordinate plane Oyz is the coordinate vector. Now we know the normal vector of the plane and the point of the plane, therefore, we can write its general equation (we solved a similar problem in the previous paragraph of this article):
, then its coordinates must satisfy the equation of the plane. Therefore, the equality is true where we find it from. Now we can write the desired general equation of the plane, it has the form .

Answer:

Bibliography.

• 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.

Let us consider the plane Q in space. Its position is completely determined by specifying the vector N perpendicular to this plane and some fixed point lying in the Q plane. The vector N perpendicular to the Q plane 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 a given point and having a given normal vector . To do this, consider a vector connecting a point with an arbitrary point on the Q plane (Fig. 81).

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

and therefore

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

So, we have shown that every 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 bundle of planes. Equation (4), in which the coefficients A, B and C can take any values, is called the equation of a bunch of planes.

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

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

ANGLE BETWEEN PLANES

Consider two planes α 1 and α 2, defined respectively by the equations:

Under angle between two planes we will understand one of the dihedral angles formed by these planes. It is obvious that the angle between the normal vectors and planes α 1 and α 2 is equal to one of the indicated adjacent dihedral angles or . That's why . Because And , That

.

Example. Determine the angle between planes x+2y-3z+4=0 and 2 x+3y+z+8=0.

Condition for parallelism of two planes.

Two planes α 1 and α 2 are parallel if and only if their normal vectors are parallel, and therefore .

So, two planes are parallel to each other if and only if the coefficients of the corresponding coordinates are proportional:

or

Condition of perpendicularity of planes.

It is clear that two planes are perpendicular if and only if their normal vectors are perpendicular, and therefore, or .

Thus, .

Examples.

STRAIGHT IN SPACE.

VECTOR EQUATION FOR A LINE.

PARAMETRIC DIRECT EQUATIONS

The position of a line in space is completely determined by specifying any of its fixed points M 1 and a vector parallel to this line.

A vector parallel to a line is called guides vector of this line.

So let the straight line l passes through a point M 1 (x 1 , y 1 , z 1), lying on a line parallel to the vector .

Consider an arbitrary point M(x,y,z) on a straight line. From the figure it is clear that .

Vectors and are collinear, so there is such a number t, what , where is the multiplier t can take any numeric value depending on the position of the point M on a straight line. Factor t called a parameter. Having designated the radius vectors of points M 1 and M respectively, through and , we obtain . This equation is called vector equation of a straight line. It shows that for each parameter value t corresponds to the radius vector of some point M, lying on a straight line.

Let's write this equation in coordinate form. Notice, that , and from here

The resulting equations are called parametric equations of a straight line.

When changing a parameter t coordinates change x, y And z and period M moves in a straight line.

CANONICAL EQUATIONS OF THE DIRECT

Let M 1 (x 1 , y 1 , z 1) – a point lying on a straight line l, And is its direction vector. Let us again take an arbitrary point on the line M(x,y,z) and consider the vector .

It is clear that the vectors are also collinear, so their corresponding coordinates must be proportional, therefore,

– canonical equations of a straight line.

Note 1. Note that the canonical equations of the line could be obtained from the parametric ones by eliminating the parameter t. Indeed, from the parametric equations we obtain or .

Example. Write down the equation of the line in parametric form.

Let's denote , from here x = 2 + 3t, y = –1 + 2t, z = 1 –t.

Note 2. Let the straight line be perpendicular to one of the coordinate axes, for example the axis Ox. Then the direction vector of the line is perpendicular Ox, hence, m=0. Consequently, the parametric equations of the line will take the form

Excluding the parameter from the equations t, we obtain the equations of the line in the form

However, in this case too, we agree to formally write the canonical equations of the line in the form . Thus, if the denominator of one of the fractions is zero, this means that the straight line is perpendicular to the corresponding coordinate axis.

Similar to the canonical equations corresponds to a straight line perpendicular to the axes Ox And Oy or parallel to the axis Oz.

Examples.

GENERAL EQUATIONS OF A STRAIGHT LINE AS LINES OF INTERSECTION OF TWO PLANES

Through every straight line in space there are countless planes. Any two of them, intersecting, define it in space. Consequently, the equations of any two such planes, considered together, represent the equations of this line.

In general, any two non-parallel planes given by the general equations

determine the straight line of their intersection. These equations are called general equations straight.

Examples.

Construct a line given by the equations

To construct a straight line, it is enough to find any two of its points. The easiest way is to select the points of intersection of a straight line with coordinate planes. For example, the point of intersection with the plane xOy we obtain from the equations of the straight line, assuming z= 0:

Having solved this system, we find the point M 1 (1;2;0).

Similarly, assuming y= 0, we get the point of intersection of the line with the plane xOz:

From the general equations of a straight line one can move on to its canonical or parametric equations. To do this you need to find some point M 1 on a straight line and the direction vector of a straight line.

Point coordinates M 1 we obtain from this system of equations, giving one of the coordinates an arbitrary value. To find the direction vector, note that this vector must be perpendicular to both normal vectors And . Therefore, beyond the direction vector of the straight line l you can take the vector product of normal vectors:

.

Example. Give general equations of the line to the canonical form.

Let's find a point lying on a line. To do this, we choose arbitrarily one of the coordinates, for example, y= 0 and solve the system of equations:

The normal vectors of the planes defining the line have coordinates Therefore, the direction vector will be straight

. Hence, l: .

ANGLE BETWEEN STRAIGHTS

Angle between straight lines in space we will call any of the adjacent angles formed by two straight lines drawn through an arbitrary point parallel to the data.

Let two lines be given in space:

Obviously, the angle φ between straight lines can be taken as the angle between their direction vectors and . Since , then using the formula for the cosine of the angle between vectors we get

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 the given 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 , 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 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, 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, 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 during tests are in the textbook “Problems 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 the same line.

Let three different points , and , not lying on the same line, be given. 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 for the 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 a special case of the general equation of a line, if one occurs.

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