3.1. Canonical equations of a straight line.
Let a straight line be given in the Oxyz coordinate system that passes through the point
(see Fig. 18). Denote by 
a vector parallel to the given line. Vector 
called direction vector of the straight line. Let's take a straight point 
and consider vector Vectors 
are collinear, hence their respective coordinates are proportional:
(3.3.1
)
These equations are called canonical equations straight.
Example: Write the equations of a straight line passing through the point M(1, 2, –1) parallel to the vector 
Decision: Vector 
is the direction vector of the desired line. Applying formulas (3.1.1), we obtain:
These are the canonical equations of the line.
Comment: The vanishing of one of the denominators means the vanishing of the corresponding numerator, that is, y - 2 = 0; y = 2. This line lies in the plane y = 2, parallel to the plane Oxz.
3.2. Parametric equations of a straight line.
Let the straight line be given by the canonical equations
Denote 
then 
The value t is called a parameter and can take any value: 
.
Let's express x, y and z in terms of t:
(3.2.1
)
The resulting equations are called parametric equations of the straight line.
Example 1: Compose parametric equations of a straight line passing through the point M (1, 2, –1) parallel to the vector 
Decision: The canonical equations of this line are obtained in the example of paragraph 3.1:
To find the parametric equations of the straight line, we apply the derivation of formulas (3.2.1):
So, 
- parametric equations of the given straight line.
Answer:
Example 2 Compose parametric equations of a straight line passing through the point M (–1, 0, 1) parallel to the vector 
where A (2, 1, –1), B (–1, 3, 2).
Decision: Vector 
is the direction vector of the desired line.
Let's find the vector 
.
= (–3; 2; 3). According to formulas (3.2.1), we write the equations of the straight line:
are the required parametric equations of the straight line.
3.3. Equations of a straight line passing through two given points.
A single straight line passes through two given points in space (see Fig. 20). Let points be given Vector 
can be taken as the direction vector of this straight line. Then the direct find equations 
them according to the formulas (3.1.1): 
).
(3.3.1)
Example 1 Compose canonical and parametric equations of a straight line passing through points
Decision: We apply the formula (3.3.1)
We have obtained the canonical equations of the straight line. To obtain parametric equations, we apply the derivation of formulas (3.2.1). Get
are the parametric equations of the straight line.
Example 2 Compose canonical and parametric equations of a straight line passing through points
Decision: According to formulas (3.3.1) we get:
These are canonical equations.
We turn to parametric equations:
- parametric equations.
The resulting straight line is parallel to the oz axis (see Fig. 21).
Let two planes be given in space
If these planes do not coincide and are not parallel, then they intersect in a straight line:
This system of two linear equations defines a straight line as the line of intersection of two planes. From equations (3.4.1) one can pass to canonical equations (3.1.1) or parametric equations (3.2.1). To do this, you need to find a point 
lying on the line, and the direction vector 
Point coordinates 
we obtain from system (3.4.1) by giving one of the coordinates an arbitrary value (for example, z = 0). Behind the guide vector 
you can take the cross product of vectors, that is
Example 1 Compose canonical equations of a straight line 
Decision: Let z = 0. We solve the system 
Adding these equations, we get: 3x + 6 = 0 
x = -2. Substitute the found value x = -2 into the first equation of the system and get: -2 + y + 1 = 0 
y=1.
So point 
lies on the desired line.
To find the direction vector of the straight line, we write the normal vectors of the planes: and find their vector product:
The straight line equations are found by the formulas (3.1.1):
Answer:
.
Another way: The canonical and parametric equations of the line (3.4.1) can be easily obtained by finding two different points on the line from system (3.4.1), and then applying formulas (3.3.1) and deriving formulas (3.2.1).
Example 2 Compose canonical and parametric equations of a straight line
Decision: Let y = 0. Then the system will take the form: 
Adding the equations, we get: 2x + 4 = 0; x = -2. Substitute x = -2 into the second equation of the system and get: -2 -z +1 = 0 
z = -1. So we found a point 
To find the second point, we set x = 0. We will have: 
I.e 
We have obtained the canonical equations of the straight line.
We compose the parametric equations of the straight line:
Answer:
; 
.
3.5. Mutual arrangement of two straight lines in space.
Let straight 
given by the equations:
:
;
: 
.
The angle between these lines is understood as the angle between their direction vectors (see Fig. 22). This corner 
we find by the formula from vector algebra: 
or
(3.5.1)
If straight 
perpendicular ( 
),then 
Hence,
This is the condition of perpendicularity of two straight lines in space.
If straight 
are parallel ( 
), then their direction vectors are collinear ( 
), i.e
(3.5.3
)
This is the condition for two lines to be parallel in space.
Example 1 Find the angle between lines:
a). 
and 
b). 
and 
Decision: a). Let's write the directing vector straight 
Let's find the direction vector 
planes included in the system Then we find their vector product: 
(see example 1 of point 3.4).
According to the formula (3.5.1) we get: 
Hence, 
b). Let's write the direction vectors of these lines: Vectors 
are collinear since their respective coordinates are proportional:
So straight 
are parallel ( 
), i.e 
Answer: a).
b). 
Example 2 Prove perpendicularity of lines:
and 
Decision: Let's write the direction vector of the first straight line 
Let's find the direction vector 
second line. To do this, we find the normal vectors 
planes included in the system: Calculate their vector product: 
(See example 1 of paragraph 3.4).
We apply the condition of perpendicularity of lines (3.5.2):
The condition is met; hence the lines are perpendicular ( 
).
One of the types of equations of a straight line in space is the canonical equation. We will consider this concept in detail, since it is necessary to know it to solve many practical problems.
In the first paragraph, we formulate the basic equations of a straight line located in three-dimensional space and give several examples. Next, we will show methods for calculating the coordinates of the direction vector for given canonical equations and solving the inverse problem. In the third part, we will describe how the equation of a straight line passing through 2 given points in three-dimensional space is compiled, and in the last paragraph we will point out the connections of canonical equations with others. All reasoning will be illustrated by examples of problem solving.
We have already talked about what the canonical equations of a straight line are in general in the article devoted to the equations of a straight line on a plane. We will analyze the case with three-dimensional space by analogy.
Let's say we have a rectangular coordinate system O x y z with a straight line. As we remember, you can set a straight line in different ways. We use the simplest of them - we set the point through which the line will pass, and we indicate the direction vector. If we denote the line with the letter a, and the point M, then we can write that M 1 (x 1, y 1, z 1) lies on the line a and the direction vector of this line will be a → = (a x, a y, a z) . For the set of points M (x, y, z) to define the line a, the vectors M 1 M → and a → must be collinear,
If we know the coordinates of the vectors M 1 M → and a → , then we can write in the coordinate form the necessary and sufficient condition for their collinearity. From the initial conditions, we already know the coordinates a → . In order to get the coordinates M 1 M → , we need to calculate the difference between M (x , y , z) and M 1 (x 1 , y 1 , z 1) . Let's write:
M 1 M → = x - x 1 , y - y 1 , z - z 1
After that, we can formulate the condition we need as follows: M 1 M → = x - x 1, y - y 1, z - z 1 and a → = (a x, a y, a z) : M 1 M → = λ a → ⇔ x - x 1 = λ a x y - y 1 = λ a y z - z 1 = λ a z
Here the value of the variable λ can be any real number or zero. If λ = 0 , then M (x , y , z) and M 1 (x 1 , y 1 , z 1) will coincide, which does not contradict our reasoning.
For values a x ≠ 0 , a y ≠ 0 , a z ≠ 0 we can solve with respect to parameter λ all equations of the system x - x 1 = λ a x y - y 1 = λ a y z - z 1 = λ a z
After that, it will be possible to put an equal sign between the right parts:
x - x 1 = λ a x y - y 1 = λ a y z - z 1 = λ a z ⇔ λ = x - x 1 a x λ = y - y 1 a y λ = z - z 1 a z ⇔ x - x 1 a x = y - y 1 a y = z - z 1 a z
As a result, we got the equations x - x 1 a x \u003d y - y 1 a y \u003d z - z 1 a z, with which you can determine the desired line in three-dimensional space. These are the canonical equations we need.
Such a notation is used even if one or two parameters a x , a y , a z are zero, since it will also be true in these cases. All three parameters cannot be equal to 0 because the direction vector a → = (a x , a y , a z) cannot be zero.
If one or two parameters a are equal to 0, then the equation x - x 1 a x = y - y 1 a y = z - z 1 a z is conditional. It should be considered equal to the following entry:
x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ , λ ∈ R .
We will analyze special cases of canonical equations in the third paragraph of the article.
Several important conclusions can be drawn from the definition of the canonical equation of a straight line in space. Let's consider them.
1) if the original line will pass through two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the canonical equations will take the following form:
x - x 1 a x = y - y 1 a y = z - z 1 a z or x - x 2 a x = y - y 2 a y = z - z 2 a z .
2) since a → = (a x , a y , a z) is the direction vector of the original line, then all vectors μ a → = μ a x , μ a y , μ a z , μ ∈ R , μ ≠ 0 . Then the line can be defined using the equation x - x 1 a x = y - y 1 a y = z - z 1 a z or x - x 1 μ a x = y - y 1 μ a y = z - z 1 μ a z .
Here are some examples of such equations with given values:
Example 1 Example 2
How to write the canonical equation of a straight line in space
We found out that the canonical equations of the form x - x 1 a x = y - y 1 a y = z - z 1 a z will correspond to a straight line passing through the point M 1 (x 1 , y 1 , z 1) , and the vector a → = ( a x , a y , a z) will be her guide. So, if we know the equation of a straight line, then we can calculate the coordinates of its directing vector, and given the coordinates of the vector and some point located on the straight line, we can write down its canonical equations.
Let's take a look at a couple of specific issues.
Example 3
We have a line defined in 3D space using the equation x + 1 4 = y 2 = z - 3 - 5 . Write down the coordinates of all direction vectors for it.
Decision
To get the direction vector coordinates, we just need to take the values of the denominators from the equation. We get that one of the direction vectors will be a → = (4 , 2 , - 5) , and the set of all such vectors can be formulated as μ a → = 4 μ , 2 μ , - 5 μ . Here the parameter μ is any real number (except for zero).
Answer: 4 μ , 2 μ , - 5 μ , μ ∈ R , μ ≠ 0
Example 4
Write down the canonical equations if the line in space passes through M 1 (0 , - 3 , 2) and has a direction vector with coordinates - 1 , 0 , 5 .
Decision
We have data that x 1 = 0 , y 1 = - 3 , z 1 = 2 , a x = - 1 , a y = 0 , a z = 5 . This is quite enough to go straight to writing the canonical equations.
Let's do it:
x - x 1 a x = y - y 1 a y = z - z 1 a z ⇔ x - 0 - 1 = y - (- 3) 0 = z - 2 5 ⇔ ⇔ x - 1 = y + 3 0 = z - 2 5
Answer: x - 1 = y + 3 0 = z - 2 5
These tasks are the simplest, because they have all or almost all of the initial data for writing an equation or vector coordinates. In practice, you can often find those in which you first need to find the desired coordinates, and then write down the canonical equations. We analyzed examples of such problems in articles devoted to finding the equations of a straight line passing through a point in space parallel to a given one, as well as a straight line passing through a certain point in space perpendicular to the plane.
Earlier we have already said that one or two values of the parameters a x , a y , a z in the equations can have zero values. In this case, the record x - x 1 a x \u003d y - y 1 a y \u003d z - z 1 a z \u003d λ becomes formal, since we get one or two fractions with zero denominators. It can be rewritten in the following form (for λ ∈ R):
x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ
Let's consider these cases in more detail. Suppose that a x = 0 , a y ≠ 0 , a z ≠ 0 , a x ≠ 0 , a y = 0 , a z ≠ 0 , or a x ≠ 0 , a y ≠ 0 , a z = 0 . In this case, we can write the necessary equations as follows:
- In the first case:
x - x 1 0 = y - y 1 a y = z - z 1 a z = λ ⇔ x - x 1 = 0 y = y 1 + a y λ z = z 1 + a z λ ⇔ x - x 1 = 0 y - y 1 a y = z - z 1 a z = λ
In the second case:
x - x 1 a x = y - y 1 0 = z - z 1 a z = λ ⇔ x = x 1 + a x λ y - y 1 = 0 z = z 1 + a z λ ⇔ y - y 1 = 0 x - x 1 a x = z - z 1 a z = λ
In the third case:
x - x 1 a x = y - y 1 a y = z - z 1 0 = λ ⇔ x = x 1 + a x λ y = y 1 + a y λ z - z 1 = 0 ⇔ z - z 1 = 0 x - x 1 a x = y - y 1 a y = λ
It turns out that with this value of the parameters, the required lines are in the planes x - x 1 = 0, y - y 1 = 0 or z - z 1 = 0, which are parallel to the coordinate planes (if x 1 = 0, y 1 = 0 or z1 = 0). Examples of such lines are shown in the illustration.
Therefore, we can write the canonical equations in a slightly different way.
- In the first case: x - x 1 0 = y - y 1 0 = z - z 1 a z = λ ⇔ x - x 1 = 0 y - y 1 = 0 z = z 1 + a z λ , λ ∈ R
- In the second: x - x 1 0 = y - y 1 a y = z - z 1 0 = λ ⇔ x - x 1 = 0 y = y 1 + a y λ , λ ∈ R z - z 1 = 0
- In the third: x - x 1 a x = y - y 1 0 = z - z 1 0 = λ ⇔ x = x 1 + a x λ , λ ∈ R y = y 1 = 0 z - z 1 = 0
In all three cases, the original lines will coincide with the coordinate axes or will be parallel to them: x 1 = 0 y 1 = 0, x 1 = 0 z 1 = 0, y 1 = 0 z 1 = 0. Their direction vectors have coordinates 0 , 0 , a z , 0 , a y , 0 , a x , 0 , 0 . If we denote the direction vectors of the coordinate lines as i → , j → , k → , then the direction vectors of the given lines will be collinear with respect to them. The figure shows these cases:
Let's use examples to show how these rules are applied.
Example 5
Find canonical equations that can be used to determine the coordinate lines O z , O x , O y in space.
Decision
Coordinate vectors i → = (1 , 0 , 0) , j → = 0 , 1 , 0 , k → = (0 , 0 , 1) will be guides for the original lines. We also know that our lines will necessarily pass through the point O (0 , 0 , 0) since it is the origin. Now we have all the data to write down the necessary canonical equations.
For straight line O x: x 1 = y 0 = z 0
For a straight line O y: x 0 = y 1 = z 0
For straight line O z: x 0 = y 0 = z 1
Answer: x 1 \u003d y 0 \u003d z 0, x 0 \u003d y 1 \u003d z 0, x 0 \u003d y 0 \u003d z 1.
Example 6
There is a straight line in space that passes through the point M 1 (3 , - 1 , 12) . It is also known that it is located parallel to the y-axis. Write down the canonical equations of this line.
Decision
Taking into account the condition of parallelism, we can say that the vector j → = 0 , 1 , 0 will be the guides for the required straight line. Therefore, the desired equations will have the form:
x - 3 0 = y - (- 1) 1 = z - 12 0 ⇔ x - 3 0 = y + 1 1 = z - 12 0
Answer: x - 3 0 = y + 1 1 = z - 12 0
Let's say that we have two mismatched points M 1 (x 1 , y 1 , z 1) and M 2 (x 2 , y 2 , z 2) through which a straight line passes. How then can we formulate the canonical equation for it?
To begin with, let's take the vector M 1 M 2 → (or M 2 M 1 →) as the direction vector of this line. Since we have the coordinates of the desired points, we immediately calculate the coordinates of the vector:
M 1 M 2 → = x 2 - x 1, y 2 - y 1, z 2 - z 1
x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1 = z - z 2 z 2 - z 1
The resulting equalities are the canonical equations of a straight line passing through two given points. Take a look at the illustration:
Let us give an example of solving the problem.
Example 7
in space there are two points with coordinates M 1 (- 2 , 4 , 1) and M 2 (- 3 , 2 , - 5) through which the line passes. Write down the canonical equations for it.
Decision
According to the conditions, x 1 = - 2, y 1 = - 4, z 1 = 1, x 2 = - 3, y 2 = 2, z 2 = - 5. We need to substitute these values into the canonical equation:
x - (- 2) - 3 - (- 2) = y - (- 4) 2 - (- 4) = z - 1 - 5 - 1 ⇔ x + 2 - 1 = y + 4 6 = z - 1 - 6
If we take equations of the form x - x 2 x 2 - x 1 \u003d y - y 2 y 2 - y 1 \u003d z - z 2 z 2 - z 1, then we get: x - (- 3) - 3 - ( - 2) = y - 2 2 - (- 4) = z - (- 5) - 5 - 1 ⇔ x + 3 - 1 = y - 2 6 = z + 5 - 6
Answer: x + 3 - 1 = y - 2 6 = z + 5 - 6 or x + 3 - 1 = y - 2 6 = z + 5 - 6.
Transformation of the canonical equations of a straight line in space into other types of equations
Sometimes it is not very convenient to use canonical equations of the form x - x 1 a x \u003d y - y 1 a y \u003d z - z 1 a z. To solve some problems, it is better to use the notation x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ . In some cases, it is more preferable to determine the desired line using the equations of two intersecting planes A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0. Therefore, in this paragraph, we will analyze how it is possible to move from canonical equations to other types, if this is required by us according to the conditions of the problem.
It is not difficult to understand the rules for passing to parametric equations. First, we equate each part of the equation to the parameter λ and solve these equations with respect to other variables. As a result, we get:
x - x 1 a x = y - y 1 a y = z - z 1 a z ⇔ x - x 1 a x = y - y 1 a y = z - z 1 a z ⇔ ⇔ x - x 1 a x = λ y - y 1 a y = λ z - z 1 a z = λ ⇔ x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ
The value of the parameter λ can be any real number, because x , y , z can take any real values.
Example 8
In a rectangular coordinate system in three-dimensional space, a straight line is given, which is defined by the equation x - 2 3 = y - 2 = z + 7 0 . Write the canonical equation in parametric form.
Decision
First, we equate each part of the fraction to λ.
x - 2 3 = y - 2 = z + 7 0 ⇔ x - 2 3 = λ y - 2 = λ z + 7 0 = λ
Now we resolve the first part with respect to x , the second with respect to y , the third with respect to z . We will be able to:
x - 2 3 = λ y - 2 = λ z + 7 0 = λ ⇔ x = 2 + 3 λ y = - 2 λ z = - 7 + 0 λ ⇔ x = 2 + 3 λ y = - 2 λz = - 7
Answer: x = 2 + 3 λ y = - 2 λ z = - 7
Our next step will be the transformation of the canonical equations into the equation of two intersecting planes (for the same straight line).
The equality x - x 1 a x \u003d y - y 1 a y \u003d z - z 1 a z must first be represented as a system of equations:
x - x 1 a x = y - y 1 a y x - x 1 a x = z - z 1 a x y - y 1 a y = z - z 1 a z
Since we understand p q = r s as p s = q r, we can write:
x - x 1 a x = y - y 1 a y x - x 1 a x = z - z 1 a z y - y 1 a y = z - z 1 a z ⇔ a y (x - x 1) = a x (y - y 1) a z (x - x 1) = a x (z - z 1) a z (y - y 1) = a y (z - z 1) ⇔ ⇔ a y x - a x y + a x y 1 - a y x 1 = 0 a z x - a x z + a x z 1 - a z x 1 = 0 a z y - a y z + a y z 1 - a z y 1 = 0
As a result, we got that:
x - x 1 a x = y - y 1 a y = z - z 1 a z ⇔ a y x - a x y + a x y 1 - a y x 1 = 0 a z x - a x z + a x z 1 - a z x 1 = 0 a z y - a y z + a y z 1 - a z y 1 = 0
We noted above that all three parameters a cannot be zero at the same time. This means that the rank of the main matrix of the system will be equal to 2, since a y - a x 0 a z 0 - a x 0 a z - a y = 0 and one of the second-order determinants is not equal to 0:
a y - a x a z 0 = a x a z , a y 0 a z - a x = a x a y , - a x 0 0 - a x = a x 2 a y - a x 0 a z = a y a z , a y 0 0 - a y = - a y 2 , - a x 0 a z - a y = a x a y a z 0 0 a z = a z 2 , a z - a x 0 - a y = - a y a z , 0 - a x a z - a y = a x a z
This gives us the opportunity to exclude one equation from our calculations. Thus, the canonical equations of the line can be transformed into a system of two linear equations, which will contain 3 unknowns. They will be the equations we need for two intersecting planes.
The reasoning looks rather complicated, but in practice everything is done quite quickly. Let's demonstrate this with an example.
Example 9
The straight line is given by the canonical equation x - 1 2 = y 0 = z + 2 0 . Write an equation of intersecting planes for it.
Decision
Let's start with pairwise equalization of fractions.
x - 1 2 = y 0 = z + 2 0 ⇔ x - 1 2 = y 0 x - 1 2 = z + 2 0 y 0 = z + 2 0 ⇔ ⇔ 0 (x - 1) = 2 y 0 (x - 1) = 2 (z + 2) 0 y = 0 (z + 2) ⇔ y = 0 z + 2 = 0 0 = 0
Now we exclude the last equation from the calculations, because it will be true for any x, y and z. In this case x - 1 2 = y 0 = z + 2 0 ⇔ y = 0 z + 2 = 0 .
These are the equations of two intersecting planes, which, when intersected, form a straight line given by the equation x - 1 2 = y 0 = z + 2 0
Answer: y=0 z+2=0
Example 10
The straight line is given by the equations x + 1 2 = y - 2 1 = z - 5 - 3, find the equation of two planes intersecting along this straight line.
Decision
Equate the fractions in pairs.
x + 1 2 = y - 2 1 = z - 5 - 3 ⇔ x + 1 2 = y - 2 1 x + 1 2 = z - 5 - 3 y - 2 1 = z - 5 - 3 ⇔ ⇔ 1 ( x + 1) = 2 (y - 2) - 3 (x + 1) = 2 (z - 5) - 3 (y - 2) = 1 (z - 5) ⇔ x - 2 y + 5 = 0 3 x + 2 z - 7 = 0 3 y + 7 - 11 = 0
We get that the determinant of the main matrix of the resulting system will be equal to 0:
1 - 2 0 3 0 2 0 3 1 = 1 0 1 + (- 2) 2 0 + 0 3 3 - 0 0 0 - 1 2 3 - (- 2) 3 1 = 0
The minor of the second order will not be zero: 1 - 2 3 0 = 1 0 - (- 2) 3 = 6 . Then we can take it as the basis minor.
As a result, we can calculate the rank of the main matrix of the system x - 2 y + 5 = 0 3 x + 2 z - 7 = 0 3 y + z - 11 = 0 . It will be 2. We exclude the third equation from the calculation and get:
x - 2 y + 5 = 0 3 x + 2 z - 7 = 0 3 y + z - 11 = 0 ⇔ x - 2 y + 5 = 0 3 x + 2 z - 7 = 0
Answer: x - 2 y + 5 = 0 3 x + 2 z - 7 = 0
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How to write equations of a straight line in space?
Equations of a straight line in space
Like a "flat" line, there are several ways in which we can define a line in space. Let's start with the canons - the point and the directing vector of the straight line:
If some point in space belonging to the line and the directing vector of this line are known, then the canonical equations of this line are expressed by the formulas:
The above entry assumes that the coordinates of the direction vector are not equal to zero. What to do if one or two coordinates are zero, we will consider a little later.
As in the article Plane equation, for simplicity, we will assume that in all tasks of the lesson, actions are carried out in an orthonormal basis of space.
Example 1
Compose the canonical equations of a straight line by a point and a direction vector
Decision: We compose the canonical equations of the straight line by the formula: 
Answer: 
And it’s a no-brainer… although, no, a no-brainer doesn’t understand anything at all.
What should be noted in this very simple example? First, the resulting equations DO NOT need to be reduced by one: 
. It is possible to reduce, more precisely, but this unusually hurts the eye and creates inconvenience in the course of solving problems.
And secondly, two things are inevitable in analytical geometry - this is a test and a test:
Just in case, we look at the denominators of the equations and compare - is it right the coordinates of the direction vector are written there. No, don’t think, we don’t have a lesson in the Brakes kindergarten. This advice is very important, because it allows you to completely eliminate the error due to inattention. No one is insured, but what if they rewrote it incorrectly? Awarded the Darwin Prize for Geometry.
The correct equalities are obtained, which means that the coordinates of the point satisfy our equations, and the point itself really belongs to this line.
Verification is very easy (and fast!) Orally.
In a number of tasks it is required to find some other point belonging to a given line. How to do it?
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: 
Let's check whether the found point satisfies the equations :
The correct equalities are obtained, which means that the point really lies on the given line.
Let's draw in a rectangular coordinate system. At the same time, let's remember how to correctly set aside points in space: 
Building a point:
- from the origin in the negative direction of the axis, we set aside the segment of the first coordinate (green dotted line);
- the second coordinate is zero, so we don’t “twitch” from the axis either to the left or to the right;
- in accordance with the third coordinate, we measure three units up (purple dotted line).
We build a point: we measure two units “towards ourselves” (yellow dotted line), one unit to the right (blue dotted line) and two units down (brown dotted line). The brown dotted line and the point itself are superimposed on the coordinate axis, note that they are in the lower half-space and BEFORE the axis.
The straight line itself passes over the axis and, if my eye does not fail me, over the axis. Does not fail, was convinced analytically. If the straight line passed BEHIND the axis, then it would be necessary to erase with an eraser a piece of the line above and below the crossing point.
A straight line has infinitely many direction vectors, for example:
(red arrow)
It turned out exactly the original vector, but this is pure chance, I chose the point like that. All direction vectors of the line are collinear, and their respective coordinates are proportional (for more details, see Linear (non) dependence of vectors. Vector basis). So, vectors 
will also be direction vectors of this line.
Additional information on the construction of three-dimensional drawings on checkered paper can be found at the beginning of the manual Graphs and properties of functions. In a notebook, multi-colored dotted paths to points (see drawing) are usually thinly drawn with a simple pencil with the same dotted line.
Let's deal with special cases when one or two coordinates of the direction vector are zero. Along the way, we continue the training of spatial vision, which began at the beginning of the lesson Plane equation. And again I will tell you a fairy tale about a naked king - I will draw an empty coordinate system and I will convince you that there are space lines there =) 
It is easier to list all six cases:
1) For a point and a direction vector, the canonical equations of a straight line break down into three individual equations: .
Or shorter:
Example 2: compose the equations of a straight line by a point and a direction vector:
What is this straight line? The directing vector of the straight line is collinear to the orth, which means that this straight line will be parallel to the axis. Canonical equations should be understood as follows:
a) - "Y" and "Z" constant, are equal specific numbers;
b) the variable "x" can take any value: (in practice, this equation, as a rule, is not written down).
In particular, the equations define the axis itself. Indeed, "x" takes any value, and "y" and "z" are always equal to zero.
The equations under consideration can be interpreted in another way: let's look, for example, at the analytical notation of the x-axis: . After all, these are equations of two planes! The equation defines the coordinate plane, and the equation defines the coordinate plane. You think correctly - these coordinate planes intersect along the axis. The way when a straight line in space is given by the intersection of two planes, we will consider at the very end of the lesson.
Two similar cases:
2) The canonical equations of a straight line passing through a point parallel to the vector are expressed by the formulas.
Such lines will be parallel to the coordinate axis. In particular, the equations define the coordinate axis itself.
3) The canonical equations of a straight line passing through a point parallel to the vector are expressed by the formulas.
These lines are parallel to the coordinate axis, and the equations define the applicate axis itself.
Let's drive the second three into the stall:
4) For a point and a directing vector, the canonical equations of a straight line decompose into a proportion and plane equation .
Example 3: compose the equations of a straight line by a point and a direction vector.
Properties of a straight line in Euclidean geometry.
There are infinitely many lines that can be drawn through any point.
Through any two non-coinciding points, there is only one straight line.
Two non-coincident lines in the 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;
- straight lines are parallel;
- straight lines intersect.
Straight line- algebraic curve of the first order: in the Cartesian coordinate system, a straight line
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any line in the plane can be given by a first order equation
Ah + Wu + C = 0,
and constant A, B not equal to zero at the same time. This first order equation is called general
straight line equation. Depending on the values of the constants A, B and With The following special cases are possible:
. C = 0, A ≠ 0, B ≠ 0- the 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 line coincides with the axis OU
. A = C = 0, B ≠ 0- the line coincides with the axis Oh
The equation of a straight line can be represented in various forms depending on any given
initial conditions.
Equation of a straight line by 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
Ah + Wu + C = 0.
Example. Find the equation of a straight line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Decision. Let's compose at A \u003d 3 and B \u003d -1 the equation of the straight line: 3x - y + C \u003d 0. To find the coefficient C
we substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the desired equation: 3x - y - 1 \u003d 0.
Equation of a straight 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 straight line equation,
passing through these points:
If any of the denominators is equal to zero, the corresponding numerator should be set equal to zero. On the
plane, the equation of a straight line written above is simplified:
if x 1 ≠ x 2 and x = x 1, if x 1 = x 2 .
Fraction = k called slope factor straight.
Example. Find the equation of a straight line passing through the points A(1, 2) and B(3, 4).
Decision. Applying the above formula, we get:
Equation of a straight line by a point and a slope.
If the general equation of a straight line Ah + Wu + C = 0 bring to the form:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
The equation of a straight line on a point and a directing 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 direction 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 direction vector of the straight line.
Ah + Wu + C = 0.
Example. Find the equation of a straight line with direction vector (1, -1) and passing through point A(1, 2).
Decision. We will look for the equation of the desired straight line in the form: Ax + By + C = 0. According to the definition,
coefficients must satisfy the conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of a 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. desired equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in the general equation of the straight line Ah + Wu + C = 0 C≠0, then, dividing by -C, we get:
or , where
The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point
straight with axle Oh, a b- the 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 straight line in segments.
C \u003d 1, , a \u003d -1, b \u003d 1.
Normal equation of a straight line.
If both sides of the equation Ah + Wu + C = 0 divide by number 
, which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a straight 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 line,
a φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. Given the general equation of a straight line 12x - 5y - 65 = 0. Required to write various types of equations
this straight line.
The equation of this straight line in segments:
The equation of this line with slope: (divide by 5)
Equation of a straight 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.
Angle between lines on a plane.
Definition. If two lines are given y \u003d k 1 x + b 1, y \u003d k 2 x + b 2, then the acute angle between these lines
will be defined as
Two lines are parallel if k 1 = k 2. Two lines are perpendicular
if k 1 \u003d -1 / k 2 .
Theorem.
Direct Ah + Wu + C = 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients are proportional
A 1 \u003d λA, B 1 \u003d λB. If also С 1 \u003d 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 is perpendicular to a given line.
Definition. A line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
The distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the line Ah + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of the perpendicular dropped from the point M for a given
direct. Then the distance between the points M and M 1:
(1)
Coordinates x 1 and 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 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.
Let Oxyz be fixed in three-dimensional space. Let's define a straight line in it. Let's choose the following way of specifying a straight line in space: specify the point through which the straight line a passes, and the directing vector of the straight line a . We will assume that the point lies on the line a and 
- directing vector of a straight line a.
Obviously, the set of points in three-dimensional space defines a straight line if and only if the vectors and are collinear.
Pay attention to the following important facts:
Here are a couple of examples of canonical equations of a straight line in space:
Compilation of canonical equations of a straight line in space.
So, the canonical equations of a straight line in a fixed rectangular coordinate system Oxyz in a three-dimensional space of the form 
correspond to a straight line that passes through the point , and the direction vector of this straight line is the vector . Thus, if we know the form of the canonical equations of a straight line in space, then we can immediately write down the coordinates of the directing vector of this straight line, and if the coordinates of the directing vector of the straight line and the coordinates of some point of this straight line are known, then we can immediately write down its canonical equations.
Let us show solutions to such problems.
Example.
A straight line in a rectangular coordinate system Oxyz in three-dimensional space is given by the canonical equations of a straight line of the form 
. Write the coordinates of all direction vectors of this line.
Decision.
The numbers in the denominators of the canonical equations of the straight line are the corresponding coordinates of the directing vector of this straight line, that is, 
- one of the direction vectors of the original line. Then the set of all direction vectors of the line can be given as 
, where is a parameter that takes any real value except zero.
Answer:
Example.
Write the canonical equations of a straight line, which in the rectangular coordinate system Oxyz in space passes through the point 
, and the directing vector of the straight line has coordinates .
Decision.
From the condition we have . That is, we have all the data to write the required canonical equations of a straight line in space. In our case 
.
Answer:
We have considered the simplest problem of compiling the canonical equations of a straight line in a given rectangular coordinate system in three-dimensional space, when the coordinates of the directing vector of the straight line and the coordinates of some point of the straight line are known. However, tasks are much more common in which you first need to find the coordinates of the directing vector of the straight line, and only then write down the canonical equations of the straight line. As an example, we can cite problems for finding the equations of a straight line passing through a given point in space parallel to a given straight line and problems for finding equations of a straight line passing through a given point in space perpendicular to a given plane.
Particular cases of canonical equations of a straight line in space.
We have already noted that one or two of the numbers in the canonical equations of the line in space of the form 
may be zero. Then the entry is considered formal (since the denominators of one or two fractions will have zeros) and should be understood as 
, where .
Let's take a closer look at all these special cases of the canonical equations of a straight line in space.
Let be 
, or 
, or 
, then the canonical equations of the lines have the form
or
or
In these cases, in the rectangular coordinate system Oxyz in space, the lines lie in the planes , or , respectively, which are parallel to the coordinate planes Oyz , Oxz , or Oxy , respectively (or coincide with these coordinate planes at , or ). The figure shows examples of such lines.
At 
, or 
, or 
the canonical equations of lines are written as 
or 
or 
respectively.
In these cases, the lines are parallel to the coordinate axes Oz , Oy or Ox respectively (or coincide with these axes at , or ). Indeed, the direction vectors of the considered lines have coordinates , or , or , it is obvious that they are collinear to the vectors , or , or respectively, where are the direction vectors of the coordinate lines. Look at the illustrations for these particular cases of the canonical equations of a straight line in space.
It remains to consolidate the material of this paragraph to consider the solutions of examples.
Example.
Write the canonical equations of the coordinate lines Ox , Oy and Oz .
Decision.
The direction vectors of the coordinate lines Ox , Oy and Oz are the coordinate vectors 
and correspondingly. In addition, the coordinate lines pass through the origin of coordinates - through the point. Now we can write the canonical equations of the coordinate lines Ox , Oy and Oz , they have the form 
and correspondingly.
Answer:
Canonical equations of the coordinate line Ox , - canonical equations of the y-axis Oy , - canonical equations of the applicate axis.
Example.
Write the canonical equations of a straight line, which in the rectangular coordinate system Oxyz in space passes through the point 
and parallel to the ordinate axis Oy .
Decision.
Since the straight line, the canonical equations of which we need to compose, is parallel to the coordinate axis Oy, then its directing vector is the vector. Then the canonical equations of this line in space have the form .
Answer:
Canonical equations of a straight line passing through two given points in space.
Let us set ourselves the task: to write the canonical equations of a straight line passing in a rectangular coordinate system Oxyz in three-dimensional space through two non-coinciding points and 
.
As a directing vector of a given straight line, you can take a vector (if you like the vector more, then you can take it). According to the known coordinates of the points M 1 and M 2, you can calculate the coordinates of the vector: . Now we can write down the canonical equations of the straight line, since we know the coordinates of a point on the straight line (in our case, even the coordinates of two points M 1 and M 2 ), and we know the coordinates of its direction vector. Thus, a given line in the rectangular coordinate system Oxyz in three-dimensional space is determined by canonical equations of the form 
or 
. This is the desired canonical equations of a line passing through two given points in space.
Example.
Write the canonical equations of a straight line passing through two points in three-dimensional space 
and 
.
Decision.
From the condition we have . We substitute these data into the canonical equations of a straight line passing through two points :
If we use the canonical direct equations of the form , then we get
.
Answer:
or 
Transition from canonical equations of a straight line in space to other types of equations of a straight line.
To solve some problems, the canonical equations of a straight line in space may turn out to be less convenient than the parametric equations of a straight line in space of the form . And sometimes it is preferable to define a straight line in the rectangular coordinate system Oxyz in space through the equations of two intersecting planes as 
. Therefore, the problem arises of the transition from the canonical equations of a straight line in space to the parametric equations of a straight line or to the equations of two intersecting planes.
It is easy to pass from the equations of a straight line in the canonical form to the parametric equations of this straight line. This requires each of the fractions in the canonical equations of the straight line in space to be taken equal to the parameter and to solve the resulting equations with respect to the variables x, y and z: 
In this case, the parameter can take any real values (since the variables x , y and z can take any real values).
Now we will show how from the canonical equations of the straight line get the equations of two intersecting planes that define the same line.
double equality is essentially a system of three equations of the form 
(we pairwise equated the fractions from the canonical equations of the straight line). Since we understand the proportion as , then 
So we got 
.
Since the numbers a x , a y and a z are not equal to zero at the same time, then the main matrix of the resulting system is equal to two, since 
and at least one of the second-order determinants 
different from zero.
Therefore, an equation that does not participate in the formation of the basis minor can be excluded from the system. Thus, the canonical equations of a straight line in space will be equivalent to a system of two linear equations with three unknowns, which are the equations of intersecting planes, and the line of intersection of these planes will be a straight line, determined by the canonical straight line equations of the form .
For clarity, we give a detailed solution of the example, in practice everything is simpler.
Example.
Write the equations of two intersecting planes that define a line defined in the rectangular coordinate system Oxyz in space by the canonical equations of the line. Write the equations for two planes intersecting along this line.
Decision.
Equate in pairs the fractions that form the canonical equations of the straight line: 
The determinant of the main matrix of the resulting system of linear equations is equal to zero (if necessary, refer to the article), and the second-order minor 
is different from zero, we will take it as the basis minor. Thus, the rank of the main matrix of the system of equations 
is equal to two, and the third equation of the system does not participate in the formation of the basic minor, that is, the third equation can be excluded from the system. Hence, 
. So we got the required equations of two intersecting planes that define the original straight line.
Answer:
Bibliography.
- Bugrov Ya.S., Nikolsky S.M. Higher Mathematics. Volume One: Elements of Linear Algebra and Analytic Geometry.
- Ilyin V.A., Poznyak E.G. Analytic geometry.
