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 the same 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 general Cartesian coordinate system.
In order for an arbitrary point M(x, y, z) to lie in the same plane with points M 1, M 2, M 3, it is necessary that the vectors be coplanar.
(
)
= 0
Thus, 
Equation of a plane passing through three points:
Equation of a plane given two points and a vector collinear to the plane.
Let the points M 1 (x 1,y 1,z 1),M 2 (x 2,y 2,z 2) and the vector be given 
.
Let's create an equation for a 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 using one point and two vectors,
collinear to the plane.
Let two vectors be given 
And 
, collinear planes. Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors 
must be coplanar.
Plane equation:
Equation of a plane by point and normal vector .
Theorem.
If a point M is given in space 0
(X 0
, y 0
,
z 0
), then the equation of the plane passing through the point M 0
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
is 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 + Bi + Cz + D = 0 we divide both sides by (-D)
,
replacing 
, we obtain the equation of the plane in segments:
The numbers a, b, c are the intersection points of the plane with the x, y, z axes, respectively.
Equation of a plane in vector form.
Where
- radius vector of the current point M(x, y, z),
A unit vector having the direction of a perpendicular dropped onto a 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 looks like:
xcos + ycos + zcos - p = 0.
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+By+Cz+D=0 is:
Example. Find the equation of the plane, knowing that 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, we 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) 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 points A(2, -1, 4) and
B(3, 2, -1) perpendicular to the plane X + at + 2z – 3 = 0.
The required equation of the plane has the form: A x+B y+C z+ D = 0, 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. 112 + 71 - 24 +D= 0;D= -21.
In total, we get the equation of the plane: 11 x - 7y – 2z – 21 = 0.
Example. Find the equation of the plane, knowing that 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 required equation of the plane has the form: 4 x
– 3y
+ 12z+ D = 0. To find the coefficient D, we substitute the coordinates of point P into the equation:
16 + 9 + 144 + D = 0
In total, we get the required equation: 4 x – 3y + 12z – 169 = 0
Example. Given are the coordinates of the vertices of the pyramid A 1 (1; 0; 3), A 2 (2; -1; 3), A 3 (2; 1; 1),
Find the length of edge A 1 A 2.
Find the angle between edges A 1 A 2 and A 1 A 4.
Find the angle between edge A 1 A 4 and face A 1 A 2 A 3.
First we 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);
Let's 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 A 1 A 2 A 3.
Let's use the formula for the equation of a plane passing through three points.
2x + 2y + 2z – 8 = 0
x + y + z – 4 = 0;
When using the computer version “ Higher mathematics course” you can run a program that will solve the above example for any coordinates of the vertices of the pyramid.
To start the program, double-click on the icon:
In the program window that opens, enter the coordinates of the vertices of the pyramid and press Enter. In this way, all decision points can be obtained one by one.
Note: To run the program, the Maple program ( Waterloo Maple Inc.) of any version, starting with MapleV Release 4, must be installed on your computer.
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.
Properties of a straight line in Euclidean geometry.
An infinite number of straight lines can be drawn through any point.
Through any two non-coinciding points a single straight line can be drawn.
Two divergent lines in a plane either intersect at a single point or are
parallel (follows from the previous one).
In three-dimensional space, there are three options for the relative position of two lines:
- lines intersect;
- lines are parallel;
- straight lines intersect.
Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
and constant A, B are not equal to zero at the same time. This first order equation is called general
equation of a straight line. Depending on the values of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠0, B ≠ 0- a straight line passes through the origin
. A = 0, B ≠0, C ≠0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠0- the straight line coincides with the axis OU
. A = C = 0, B ≠0- the straight line coincides with the axis Oh
The equation of a straight line can be represented in in various forms depending on any given
initial conditions.
Equation of a straight line from a point and a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ax + Wu + C = 0.
Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
Let's substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the required equation: 3x - y - 1 = 0.
Equation of a line passing through two points.
Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2, z 2), Then equation of a line,
passing through these points:
If any of the denominators equal to zero, the corresponding numerator should be set equal to zero. On
plane, the equation of the straight line written above is simplified:
If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .
Fraction = k called slope straight.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Solution. Applying the formula written above, we get:
Equation of a straight line using a point and slope.
If the general equation of the line Ax + Wu + C = 0 lead to:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
Equation of a straight line from a point and a direction vector.
By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task
a straight line through a point and a directing vector of a straight line.
Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition
Aα 1 + Bα 2 = 0 called directing vector of a straight line.
Ax + Wu + C = 0.
Example. Find the equation of a straight line with a direction vector (1, -1) and passing through the point A(1, 2).
Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. According to the definition,
coefficients must satisfy the following conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.
at x = 1, y = 2 we get C/A = -3, i.e. required equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by -С, we get:
or where
Geometric meaning coefficients is that coefficient a is the coordinate of the intersection point
straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis OU.
Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.
C = 1, , a = -1, b = 1.
Normal equation of a line.
If both sides of the equation Ax + Wu + C = 0 divide by number 
which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a line.
The sign ± of the normalizing factor must be chosen so that μ*C< 0.
R- the length of the perpendicular dropped from the origin to the straight line,
A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. The general equation of the line is given 12x - 5y - 65 = 0. Required to write Various types equations
this straight line.
The equation of this line in segments:
The equation of this line with the slope: (divide by 5)
Equation of a line:
cos φ = 12/13; sin φ= -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,
parallel to the axes or passing through the origin.
The angle between straight lines on a plane.
Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, That sharp corner between these lines
will be defined as
Two lines are parallel if k 1 = k 2. Two lines are perpendicular
If k 1 = -1/ k 2 .
Theorem.
Direct Ax + Wu + C = 0 And A 1 x + B 1 y + C 1 = 0 parallel when the coefficients are proportional
A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these lines.
The equation of a line passing through a given point perpendicular to a given line.
Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
Distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the straight line Ax + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of a perpendicular dropped from a point M for a given
direct. Then the distance between points M And M 1:
(1)
Coordinates x 1 And at 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly
given straight line. If we transform the first equation of the system to the form:
A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
Equation of a plane. How to write an equation of a plane?
Mutual arrangement of planes. Tasks
Spatial geometry is not much more complicated than “flat” geometry, and our flights in space begin with this article. To master the topic, you need to have a good understanding of vectors, in addition, it is advisable to be familiar with the geometry of the plane - there will be many similarities, many analogies, so the information will be digested much better. In a series of my lessons, the 2D world opens with an article Equation of a straight line on a plane. But now Batman has left the flat TV screen and is launching from the Baikonur Cosmodrome.
Let's start with drawings and symbols. Schematically, the plane can be drawn in the form of a parallelogram, which creates the impression of space: 
The plane is infinite, but we have the opportunity to depict only a piece of it. In practice, in addition to the parallelogram, an oval or even a cloud is also drawn. For technical reasons, it is more convenient for me to depict the plane in exactly this way and in exactly this position. Real planes, which we will consider in practical examples, can be located in any way - mentally take the drawing in your hands and rotate it in space, giving the plane any slope, any angle.
Designations: planes are usually denoted in small Greek letters, apparently so as not to confuse them with straight line on a plane or with straight line in space. I'm used to using the letter . In the drawing it is the letter “sigma”, and not a hole at all. Although, the holey plane is certainly quite funny.
In some cases, it is convenient to use the same Greek letters with lower subscripts to designate planes, for example, .
It is obvious that the plane is uniquely defined by three different points that do not lie on the same line. Therefore, three-letter designations of planes are quite popular - by the points belonging to them, for example, etc. Often letters are enclosed in parentheses: 
, so as not to confuse the plane with another geometric figure.
For experienced readers I will give quick access menu:
- How to create an equation of a plane using a point and two vectors?
- How to create an equation of a plane using a point and a normal vector?
and we will not languish in long waits:
General plane equation
The general equation of the plane has the form , where the coefficients are not equal to zero at the same time.
A number of theoretical calculations and practical problems are valid both for the usual orthonormal basis and for the affine basis of space (if the oil is oil, return to the lesson Linear (non) dependence of vectors. Basis of vectors). For simplicity, we will assume that all events occur in an orthonormal basis and a Cartesian rectangular coordinate system.
Now let’s practice our spatial imagination a little. It’s okay if yours is bad, now we’ll develop it a little. Even playing on nerves requires training.
In the most general case, when the numbers are not equal to zero, the plane intersects all three coordinate axes. For example, like this:
I repeat once again that the plane continues indefinitely in all directions, and we have the opportunity to depict only part of it.
Let's consider the simplest equations of planes:
How to understand this equation? Think about it: “Z” is ALWAYS equal to zero, for any values of “X” and “Y”. This is the equation of the "native" coordinate plane. Indeed, formally the equation can be rewritten as follows: 
, from where you can clearly see that we don’t care what values “x” and “y” take, it is important that “z” is equal to zero.
Likewise:
– equation of the coordinate plane;
– equation of the coordinate plane.
Let's complicate the problem a little, consider a plane (here and further in the paragraph we assume that the numerical coefficients are not equal to zero). Let's rewrite the equation in the form: . How to understand it? “X” is ALWAYS, for any values of “Y” and “Z”, equal to a certain number. This plane is parallel to the coordinate plane. For example, a plane is parallel to a plane and passes through a point.
Likewise:
– equation of a plane that is parallel to the coordinate plane;
– equation of a plane that is parallel to the coordinate plane.
Let's add members: . The equation can be rewritten as follows: , that is, “zet” can be anything. What does it mean? “X” and “Y” are connected by the relation, which draws a certain straight line in the plane (you will find out equation of a line in a plane?). Since “z” can be anything, this straight line is “replicated” at any height. Thus, the equation defines a plane parallel to the coordinate axis
Likewise:
– equation of a plane that is parallel to the coordinate axis;
– equation of a plane that is parallel to the coordinate axis.
If the free terms are zero, then the planes will directly pass through the corresponding axes. For example, the classic “direct proportionality”: . Draw a straight line in the plane and mentally multiply it up and down (since “Z” is any). Conclusion: the plane defined by the equation passes through the coordinate axis.
We complete the review: the equation of the plane 
passes through the origin. Well, here it is quite obvious that the point satisfies this equation.
And finally, the case shown in the drawing: – the plane is friendly with all coordinate axes, while it always “cuts off” a triangle, which can be located in any of the eight octants.
Linear inequalities in space
To understand the information you need to study well linear inequalities in the plane, because many things will be similar. The paragraph will be of a brief overview nature with several examples, since the material is quite rare in practice.
If the equation defines a plane, then the inequalities
ask half-spaces. If the inequality is not strict (the last two in the list), then the solution of the inequality, in addition to the half-space, also includes the plane itself.
Example 5
Find the unit normal vector of the plane 
.
Solution: A unit vector is a vector whose length is one. Let us denote this vector by . It is absolutely clear that the vectors are collinear: 
First, we remove the normal vector from the equation of the plane: .
How to find a unit vector? In order to find the unit vector, you need every divide the vector coordinate by the vector length.
Let's rewrite the normal vector in the form and find its length:
According to the above:
Answer: 
Verification: what was required to be verified.
Readers who carefully studied the last paragraph of the lesson probably noticed that the coordinates of the unit vector are exactly the direction cosines of the vector:
Let's take a break from the problem at hand: when you are given an arbitrary non-zero vector, and according to the condition it is required to find its direction cosines (see the last problems of the lesson Dot product of vectors), then you, in fact, find a unit vector collinear to this one. Actually two tasks in one bottle.
The need to find the unit normal vector arises in some problems of mathematical analysis.
We’ve figured out how to fish out a normal vector, now let’s answer the opposite question:
How to create an equation of a plane using a point and a normal vector?
This rigid construction of a normal vector and a point is well known to the dartboard. Please stretch your hand forward and mentally select an arbitrary point in space, for example, a small cat in the sideboard. Obviously, through this point you can draw a single plane perpendicular to your hand.
The equation of a plane passing through a point perpendicular to the vector is expressed by the formula:
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 . Substitute them into the equation general view and we 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 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
