Corner φ general equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0, is calculated by the formula:
Corner φ between two straight lines canonical equations(x-x 1) / m 1 \u003d (y-y 1) / n 1 and (x-x 2) / m 2 \u003d (y-y 2) / n 2, is calculated by the formula:
Distance from point to line
Each plane in space can be represented as linear equation called general equation plane
Special cases.
o If in equation (8), then the plane passes through the origin.
o With (,) the plane is parallel to the axis(axis, axis), respectively.
o When (,) the plane is parallel to the plane(plane, plane).
Solution: use (7)
Answer: the general equation of the plane.
Example.
plane in rectangular system coordinates Oxyz is given by the general equation of the plane 
. Write down the coordinates of all normal vectors in this plane.
We know that the coefficients of the variables x, y, and z in the general equation of the plane are the corresponding coordinates of the normal vector of that plane. Therefore, the normal vector of the given plane 
has coordinates. The set of all normal vectors can be given as.
Write the equation of a plane if in a rectangular coordinate system Oxyz in space it passes through a point 
, A 
is the normal vector of this plane.
We present two solutions to this problem.
From the condition we have . We substitute these data into the general equation of the plane passing through the point:
Write the general equation for a plane parallel to the coordinate plane Oyz and passing through the point 
.
A plane that is parallel to the coordinate plane Oyz can be given by a general incomplete equation of the plane of the form . Since the point 
belongs to the plane by condition, then the coordinates of this point must satisfy the equation of the plane, that is, equality must be true. From here we find. Thus, the desired equation has the form.
Solution. The vector product, by definition 10.26, is orthogonal to the vectors p and q. Therefore, it is orthogonal to the desired plane and the vector can be taken as its normal vector. Find the coordinates of the vector n:
that is 
. Using formula (11.1), we obtain
Opening the brackets in this equation, we arrive at the final answer.
Answer: 
.
Let's rewrite the normal vector in the form and find its length:
According to the above:
Answer: 
Parallel planes have the same normal vector. 1) From the equation we find the normal vector of the plane:.
2) We compose the equation of the plane according to the point and the normal vector: 
Answer:
Vector equation of a plane in space
Parametric equation of a plane in space
Equation of a plane passing through a given point perpendicular to a given vector
Let in three-dimensional space a rectangular Cartesian coordinate system is specified. Let's formulate the following problem:
Write an equation for a plane passing through a given point M(x 0, y 0, z 0) perpendicular to the given vector n = ( A, B, C} .
Solution. Let P(x, y, z) is an arbitrary point in space. Dot P belongs to the plane if and only if the vector MP = {x − x 0, y − y 0, z − z 0) orthogonal to vector n = {A, B, C) (Fig. 1).
Having written the orthogonality condition for these vectors (n, MP) = 0 in coordinate form, we get:
|
A(x − x 0) + B(y − y 0) + C(z − z 0) = 0 |
Equation of a plane by three points
In vector form
In coordinates
Mutual arrangement of planes in space
are general equations of two planes. Then:
1) if 
, then the planes coincide;
2) if 
, then the planes are parallel;
3) if or , then the planes intersect and the system of equations
(6)
are the equations of the line of intersection of the given planes.
|
Solution: We compose the canonical equations of the straight line by the formula: Answer: |
We take the resulting equations and mentally “pin off”, for example, the left piece: . Now we equate this piece to any number(remember that there was already a zero), for example, to one: . Since , then the other two "pieces" must also be equal to one. Essentially, you need to solve the system: |
Write parametric equations for the following lines: 
Solution: The lines are given by canonical equations and at the first stage one should find some point belonging to the line and its direction vector.
a) From the equations 
remove the point and the direction vector: . You can choose another point (how to do this is described above), but it is better to take the most obvious one. By the way, to avoid mistakes, always substitute its coordinates into the equations.
Let us compose the parametric equations of this straight line: 
The convenience of parametric equations is that with their help it is very easy to find other points of the line. For example, let's find a point whose coordinates, say, correspond to the value of the parameter : 
Thus: b) Consider the canonical equations 
. The choice of a point here is simple, but insidious: (be careful not to mix up the coordinates!!!). How to pull out a guide vector? You can argue what this straight line is parallel to, or you can use a simple formal trick: there are “y” and “z” in proportion, so we write the direction vector , and put zero in the remaining space: .
We compose the parametric equations of the straight line: 
c) Let's rewrite the equations in the form , that is, "Z" can be anything. And if any, then let, for example, . Thus, the point belongs to this line. To find the direction vector, we use the following formal technique: in the initial equations there are "x" and "y", and in the direction vector at these places we write zeros: . In the remaining place we put unit: . Instead of one, any number, except zero, will do.
We write the parametric equations of the straight line: 
Let two lines l and m on the plane in the Cartesian coordinate system be given general equations: l: A 1 x + B 1 y + C 1 = 0, m: A 2 x + B 2 y + C 2 = 0
The vectors of normals to these lines: = (A 1 , B 1) - to the line l,
= (A 2 , B 2) to the line m.
Let j be the angle between lines l and m.
Since angles with mutually perpendicular sides are either equal or add up to p, then 
, i.e. cos j = .
So, we have proved the following theorem.
Theorem. Let j be the angle between two straight lines in the plane, and let these straight lines be given in the Cartesian coordinate system by the general equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0. Then cos j = 
.
Exercises.
1) Derive a formula for calculating the angle between lines if:
(1) both lines are given parametrically; (2) both lines are given by canonical equations; (3) one straight line is given parametrically, the other straight line – by the general equation; (4) both lines are given by the slope equation.
2) Let j be the angle between two straight lines in the plane, and let these straight lines be given to the Cartesian coordinate system by the equations y = k 1 x + b 1 and y =k 2 x + b 2 .
Then tan j = .
3) Explore mutual arrangement two straight lines given by general equations in the Cartesian coordinate system, and fill in the table:
The distance from a point to a line in a plane.
Let the line l on the plane in the Cartesian coordinate system be given by the general equation Ax + By + C = 0. Find the distance from the point M(x 0 , y 0) to the line l.
The distance from the point M to the line l is the length of the perpendicular HM (H н l, HM ^ l).
The vector and the normal vector to the line l are collinear, so that | | = | | | | and | | = .
Let the coordinates of the point H be (x,y).
Since the point H belongs to the line l, then Ax + By + C = 0 (*).
The coordinates of the vectors and: = (x 0 - x, y 0 - y), = (A, B).
| | = 
= 
= 
(C = -Ax - By , see (*))
Theorem. Let the line l be given in the Cartesian coordinate system by the general equation Ax + By + C = 0. Then the distance from the point M(x 0 , y 0) to this line is calculated by the formula: r (M; l) = .
Exercises.
1) Derive a formula for calculating the distance from a point to a line if: (1) the line is given parametrically; (2) the line is given by the canonical equations; (3) the straight line is given by the slope equation.
2) Write the equation of a circle tangent to the line 3x - y = 0 centered at Q(-2,4).
3) Write the equations of the lines dividing the angles formed by the intersection of the lines 2x + y - 1 = 0 and x + y + 1 = 0 in half.
§ 27. Analytical definition of a plane in space
Definition. The normal vector to the plane we will call a non-zero vector, any representative of which is perpendicular to the given plane.
Comment. It is clear that if at least one representative of the vector is perpendicular to the plane, then all other representatives of the vector are perpendicular to this plane.
Let a Cartesian coordinate system be given in space.
Let the plane a be given, = (A, B, C) – the normal vector to this plane, the point M (x 0 , y 0 , z 0) belongs to the plane a.
For any point N(x, y, z) of the plane a, the vectors and are orthogonal, that is, their scalar product is equal to zero: = 0. Let's write the last equality in coordinates: A(x - x 0) + B(y - y 0) + C(z - z0) = 0.
Let -Ax 0 - By 0 - Cz 0 = D, then Ax + By + Cz + D = 0.
Take a point K (x, y) such that Ax + By + Cz + D \u003d 0. Since D \u003d -Ax 0 - By 0 - Cz 0, then A(x - x 0) + B(y - y 0) + C(z - z 0) = 0. Since the coordinates of the directed segment = (x - x 0 , y - y 0 , z - z 0), the last equality means that ^ , and, therefore, K н a.
So we have proved the following theorem:
Theorem. Any plane in space in the Cartesian coordinate system can be defined by an equation of the form Ax + By + Cz + D = 0 (A 2 + B 2 + C 2 ≠ 0), where (A, B, C) are the coordinates of the normal vector to this plane.
The reverse is also true.
Theorem. Any equation of the form Ax + By + Cz + D \u003d 0 (A 2 + B 2 + C 2 ≠ 0) in the Cartesian coordinate system defines a certain plane, while (A, B, C) are the coordinates of the normal vector to this plane.
Proof.
Take a point M (x 0 , y 0 , z 0) such that Ax 0 + By 0 + Cz 0 + D = 0 and vector = (A, B, C) ( ≠ q).
A plane (and only one) passes through the point M perpendicular to the vector. According to the previous theorem, this plane is given by the equation Ax + By + Cz + D = 0.
Definition. An equation of the form Ax + By + Cz + D = 0 (A 2 + B 2 + C 2 ≠ 0) is called the general equation of the plane.
Example.
Let's write the equation of the plane passing through the points M (0.2.4), N (1,-1.0) and K (-1.0.5).
1. Find the coordinates of the normal vector to the plane (MNK). Since the vector product ´ is orthogonal to non-collinear vectors and , the vector is collinear to ´ .
= (1, -3, -4), = (-1, -2, 1);
´ = 
,
´ = (-11, 3, -5).
So, as a normal vector, take the vector = (-11, 3, -5).
2. Let us now use the results of the first theorem:
the equation of this plane A(x - x 0) + B(y - y 0) + C(z - z 0) = 0, where (A, B, C) are the coordinates of the normal vector, (x 0 , y 0 , z 0) – coordinates of a point lying in the plane (for example, point M).
11(x - 0) + 3(y - 2) - 5(z - 4) = 0
11x + 3y - 5z + 14 = 0
Answer: -11x + 3y - 5z + 14 = 0.
Exercises.
1) Write the equation of the plane if
(1) the plane passes through the point M (-2,3,0) parallel to the plane 3x + y + z = 0;
(2) the plane contains the (Ox) axis and is perpendicular to the x + 2y – 5z + 7 = 0 plane.
2) Write the equation for a plane passing through three given points.
§ 28. Analytical specification of a half-space*
Comment*. Let some plane be fixed. Under half-space we will understand the set of points lying on one side of a given plane, that is, two points lie in the same half-space if the segment connecting them does not intersect the given plane. This plane is called boundary of this half-space. The union of a given plane and a half-space will be called closed half-space.
Let a Cartesian coordinate system be fixed in space.
Theorem. Let the plane a be given by the general equation Ax + By + Cz + D = 0. Then one of the two half-spaces into which the plane a divides the space is given by the inequality Ax + By + Cz + D > 0, and the second half-space is given by the inequality Ax + By + Cz + D< 0.
Proof.
Let us plot the normal vector = (A, B, С) to the plane a from the point M (x 0 , y 0 , z 0) lying on this plane: = , M н a, MN ^ a. The plane divides the space into two half-spaces: b 1 and b 2 . It is clear that the point N belongs to one of these half-spaces. Without loss of generality, we assume that N н b 1 .
Let us prove that the half-space b 1 is defined by the inequality Ax + By + Cz + D > 0.
1) Take a point K(x,y,z) in the half-space b 1 . The angle Ð NMK is the angle between the vectors and is acute, therefore the scalar product of these vectors is positive: > 0. Let's write this inequality in coordinates: A(x - x 0) + B(y - y 0) + C(z - z 0) > 0, i.e. Ax + By + Cy - Ax 0 - By 0 - C z 0 > 0.
Since M н b 1 , then Ax 0 + By 0 + C z 0 + D = 0, therefore -Ax 0 - By 0 - C z 0 = D. Therefore, the last inequality can be written as follows: Ax + By + Cz + D > 0.
2) Take a point L(x,y) such that Ax + By + Cz + D > 0.
Let us rewrite the inequality, replacing D with (-Ax 0 - By 0 - C z 0) (since M н b 1, then Ax 0 + By 0 + C z 0 + D = 0): A(x - x 0) + B(y - y 0) + C(z - z 0) > 0.
The vector with coordinates (x - x 0 ,y - y 0 , z - z 0) is a vector , so the expression A(x - x 0) + B(y - y 0) + C(z - z 0) can be understood , as the scalar product of the vectors and . Since the scalar product of the vectors and is positive, the angle between them is acute and the point L н b 1 .
Similarly, one can prove that the half-space b 2 is given by the inequality Ax + By + Cz + D< 0.
Remarks.
1) It is clear that the above proof does not depend on the choice of the point M in the plane a.
2) It is clear that the same half-space can be defined by different inequalities.
The reverse is also true.
Theorem. Any linear inequality of the form Ax + By + Cz + D > 0 (or Ax + By + Cz + D< 0) (A 2 + B 2 + C 2 ≠ 0) задает в пространстве в декартовой системе координат полупространство с границей Ax + By + Cz + D = 0.
Proof.
The equation Ax + By + Cz + D = 0 (A 2 + B 2 + C 2 ≠ 0) in space defines some plane a (see § ...). As was proved in the previous theorem, one of the two half-spaces into which the plane divides the space is given by the inequality Ax Ax + By + Cz + D > 0.
Remarks.
1) It is clear that a closed half-space can be defined by a non-strict linear inequality, and any non-strict linear inequality in the Cartesian coordinate system defines a closed half-space.
2) Any convex polyhedron can be defined as the intersection of closed half-spaces (the boundaries of which are planes containing the faces of the polyhedron), that is, analytically, by a system of linear non-strict inequalities.
Exercises.
1) Prove the two theorems presented for an arbitrary affine coordinate system.
2) Is the converse true that any system of nonstrict linear inequalities defines a convex polygon?
Exercise.1) Explore the relative position of two planes given by general equations in the Cartesian coordinate system and fill in the table.
I will be brief. The angle between two lines is equal to the angle between their direction vectors. Thus, if you manage to find the coordinates of the direction vectors a \u003d (x 1; y 1; z 1) and b \u003d (x 2; y 2; z 2), you can find the angle. More precisely, the cosine of the angle according to the formula:
Let's see how this formula works on specific examples:
Task. Points E and F are marked in the cube ABCDA 1 B 1 C 1 D 1 - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AE and BF.
Since the edge of the cube is not specified, we set AB = 1. We introduce a standard coordinate system: the origin is at point A, and the x, y, z axes are directed along AB, AD, and AA 1, respectively. The unit segment is equal to AB = 1. Now let's find the coordinates of the direction vectors for our lines.
Find the coordinates of the vector AE. To do this, we need points A = (0; 0; 0) and E = (0.5; 0; 1). Since the point E is the middle of the segment A 1 B 1 , its coordinates are equal to the arithmetic mean of the coordinates of the ends. Note that the origin of the vector AE coincides with the origin, so AE = (0.5; 0; 1).
Now let's deal with the BF vector. Similarly, we analyze the points B = (1; 0; 0) and F = (1; 0.5; 1), because F - the middle of the segment B 1 C 1 . We have:
BF = (1 - 1; 0.5 - 0; 1 - 0) = (0; 0.5; 1).
So, the direction vectors are ready. The cosine of the angle between the lines is the cosine of the angle between the direction vectors, so we have:
Task. In a regular trihedral prism ABCA 1 B 1 C 1 , all edges of which are equal to 1, points D and E are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AD and BE.
We introduce a standard coordinate system: the origin is at point A, the x-axis is directed along AB, z - along AA 1 . We direct the y axis so that the OXY plane coincides with the ABC plane. The unit segment is equal to AB = 1. Find the coordinates of the direction vectors for the desired lines.
First, let's find the coordinates of the AD vector. Consider the points: A = (0; 0; 0) and D = (0.5; 0; 1), because D - the middle of the segment A 1 B 1 . Since the beginning of the vector AD coincides with the origin, we get AD = (0.5; 0; 1).
Now let's find the coordinates of the vector BE. Point B = (1; 0; 0) is easy to calculate. With point E - the middle of the segment C 1 B 1 - a little more complicated. We have:
It remains to find the cosine of the angle:
Task. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 , all edges of which are equal to 1, the points K and L are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AK and BL.
We introduce a standard coordinate system for a prism: we place the origin of coordinates at the center of the lower base, direct the x-axis along FC, the y-axis through the midpoints of segments AB and DE, and the z-axis vertically upwards. The unit segment is again equal to AB = 1. Let us write out the coordinates of the points of interest to us:
Points K and L are the midpoints of the segments A 1 B 1 and B 1 C 1, respectively, so their coordinates are found through the arithmetic mean. Knowing the points, we find the coordinates of the direction vectors AK and BL:
Now let's find the cosine of the angle:
Task. In a regular quadrangular pyramid SABCD, all edges of which are equal to 1, points E and F are marked - the midpoints of the sides SB and SC, respectively. Find the angle between lines AE and BF.
We introduce a standard coordinate system: the origin is at point A, the x and y axes are directed along AB and AD, respectively, and the z axis is directed vertically upwards. The unit segment is equal to AB = 1.
Points E and F are the midpoints of the segments SB and SC, respectively, so their coordinates are found as the arithmetic mean of the ends. We write down the coordinates of the points of interest to us:
A = (0; 0; 0); B = (1; 0; 0)
Knowing the points, we find the coordinates of the direction vectors AE and BF:
The coordinates of the vector AE coincide with the coordinates of point E, since point A is the origin. It remains to find the cosine of the angle:
Definition
A geometric figure consisting of all points of a plane enclosed between two rays emanating from one point is called flat corner.
Definition
Angle between two intersecting direct called the value of the smallest plane angle at the intersection of these lines. If two lines are parallel, then the angle between them is assumed to be zero.
The angle between two intersecting lines (if measured in radians) can take values from zero to $\dfrac(\pi)(2)$.
Definition
Angle between two intersecting lines is called the quantity equal to the angle between two intersecting lines parallel to the intersecting lines. The angle between lines $a$ and $b$ is denoted by $\angle (a, b)$.
The correctness of the introduced definition follows from the following theorem.
Plane angle theorem with parallel sides
The values of two convex plane angles with correspondingly parallel and equally directed sides are equal.
Proof
If the angles are straight, then they are both equal to $\pi$. If they are not developed, then we plot equal segments $ON=O_1ON_1$ and $OM=O_1M_1$ on the corresponding sides of the angles $\angle AOB$ and $\angle A_1O_1B_1$.
The quadrilateral $O_1N_1NO$ is a parallelogram because its opposite sides $ON$ and $O_1N_1$ are equal and parallel. Similarly, the quadrilateral $O_1M_1MO$ is a parallelogram. Hence $NN_1 = OO_1 = MM_1$ and $NN_1 \parallel OO_1 \parallel MM_1$, hence $NN_1=MM_1$ and $NN_1 \parallel MM_1$ by transitivity. The quadrilateral $N_1M_1MN$ is a parallelogram because its opposite sides are equal and parallel. Hence, the segments $NM$ and $N_1M_1$ are also equal. Triangles $ONM$ and $O_1N_1M_1$ are equal according to the third triangle equality criterion, hence the corresponding angles $\angle NOM$ and $\angle N_1O_1M_1$ are also equal.
It will be useful for every student who is preparing for the exam in mathematics to repeat the topic “Finding the angle between lines”. As statistics show, when passing a certification test, tasks in this section of stereometry cause difficulties for a large number students. At the same time, tasks that require finding the angle between straight lines are found in the USE, both basic and profile level. This means that everyone should be able to solve them.
Basic moments
There are 4 types of mutual arrangement of lines in space. They can coincide, intersect, be parallel or intersecting. The angle between them can be acute or straight.
To find the angle between the lines in the Unified State Examination or, for example, in the solution, schoolchildren in Moscow and other cities can use several methods for solving problems in this section of stereometry. You can complete the task by classical constructions. To do this, it is worth learning the basic axioms and theorems of stereometry. The student needs to be able to logically build reasoning and create drawings in order to bring the task to a planimetric problem.
You can also use the vector-coordinate method, using simple formulas, rules and algorithms. The main thing in this case is to correctly perform all the calculations. Hone your problem solving skills in stereometry and other topics school course will help you educational project"Shkolkovo".