The canonical equations of a straight line in space are equations that define a straight line passing through a given point collinearly to a direction vector.
Let a point and a direction vector be given. An arbitrary point lies on a line l only if the vectors and are collinear, i.e., they satisfy the condition:
.
The above equations are the canonical equations of the line.
Numbers m , n and p are projections of the direction vector onto the coordinate axes. Since the vector is non-zero, then all numbers m , n and p cannot be zero at the same time. But one or two of them may be zero. In analytical geometry, for example, the following notation is allowed:
,
which means that the projections of the vector on the axes Oy and Oz are equal to zero. Therefore, both the vector and the straight line given by the canonical equations are perpendicular to the axes Oy and Oz, i.e. planes yOz .
Example 1 Compose equations of a straight line in space perpendicular to a plane 
and passing through the point of intersection of this plane with the axis Oz
.
Solution. Find the point of intersection of the given plane with the axis Oz. Since any point on the axis Oz, has coordinates , then, assuming in the given equation of the plane x=y= 0 , we get 4 z- 8 = 0 or z= 2 . Therefore, the point of intersection of the given plane with the axis Oz has coordinates (0; 0; 2) . Since the desired line is perpendicular to the plane, it is parallel to its normal vector. Therefore, the normal vector can serve as the directing vector of the straight line 
given plane.
Now we write the desired equations of the straight line passing through the point A= (0; 0; 2) in the direction of the vector :
Equations of a straight line passing through two given points
A straight line can be defined by two points lying on it 
and 
In this case, the directing vector of the straight line can be the vector . Then the canonical equations of the line take the form
.
The above equations define a straight line passing through two given points.
Example 2 Write the equation of a straight line in space passing through the points and .
Solution. We write the desired equations of the straight line in the form given above in the theoretical reference:
.
Since , then the desired line is perpendicular to the axis Oy .
Straight as a line of intersection of planes
A straight line in space can be defined as a line of intersection of two non-parallel planes and, i.e., as a set of points that satisfy a system of two linear equations
The equations of the system are also called the general equations of a straight line in space.
Example 3 Compose canonical equations of a straight line in the space given by general equations
Solution. To write the canonical equations of a straight line or, which is the same, the equation of a straight line passing through two given points, you need to find the coordinates of any two points on the straight line. They can be the points of intersection of a straight line with any two coordinate planes, for example yOz and xOz .
Point of intersection of a line with a plane yOz has an abscissa x= 0 . Therefore, assuming in this system of equations x= 0 , we get a system with two variables:
Her decision y = 2 , z= 6 together with x= 0 defines a point A(0; 2; 6) of the desired line. Assuming then in the given system of equations y= 0 , we get the system
Her decision x = -2 , z= 0 together with y= 0 defines a point B(-2; 0; 0) intersection of a line with a plane xOz .
Now we write the equations of a straight line passing through the points A(0; 2; 6) and B (-2; 0; 0) :
,
or after dividing the denominators by -2:
,
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 
Solution: 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 
Solution: 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).
Solution: 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
Solution: 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
Solution: 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 
Solution: 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
Solution: 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: 
That is 
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 ( 
), that is
(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 
Solution: 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 ( 
), that is 
Answer: a).
b). 
Example 2 Prove perpendicularity of lines:
and 
Solution: 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 ( 
).
Let's consider an example solution.
Example.
Find the coordinates of any point on a straight line given in space by the equations of two intersecting planes 
.
Solution.
Let us rewrite the system of equations in the following form
As the basis minor of the main matrix of the system, we take the non-zero minor of the second order 
, that is, z is a free unknown variable. Let's transfer the terms containing z to the right parts of the equations: .
Let's accept , where is an arbitrary real number, then .
Let's solve the resulting system of equations: 
Thus, the general solution of the system of equations has the form , where .
If we take a specific value of the parameter, then we get a particular solution of the system of equations, which gives us the desired coordinates of a point lying on a given line. Let's take it then 
, therefore, is the desired point of the line.
You can check the found point coordinates by substituting them into the original equations of two intersecting planes: 
Answer:
The direction vector of the line along which the two planes intersect.
In a rectangular coordinate system, the direction vector of the straight line is inseparable from a straight line. When the line a in a rectangular coordinate system in three-dimensional space is given by the equations of two intersecting planes and , then the coordinates of the directing vector of the line are not visible. We will now show how to determine them.
We know that a line is perpendicular to a plane when it is perpendicular to any line in that plane. Then the normal vector of the plane is perpendicular to any non-zero vector lying in this plane. We will use these facts when finding the directing vector of the straight line.
The line a lies both in the plane and in the plane. Therefore, the directing vector of the straight line a is also perpendicular to the normal vector 
plane , and the normal vector 
planes. Thus, the directing vector of the line a is and :
The set of all direction vectors of the straight line and we can set as 
, where is a parameter that takes any real value other than zero.
Example.
Find the coordinates of any direction vector of a line that is given in the Oxyz rectangular coordinate system in 3D space by the equations of two intersecting planes 
.
Solution.
The normal vectors of the planes and are the vectors 
and 
respectively. The directing vector of the straight line, which is the intersection of two given planes, we will take the vector product of normal vectors: 
Answer:
Transition to parametric and canonical equations of a straight line in space.
There are cases in which the use of the equations of two intersecting planes to describe a straight line is not very convenient. Some problems are easier to solve if the canonical equations of a straight line in space of the form 
or parametric equations of a straight line in space of the form 
, where x 1 , y 1 , z 1 are the coordinates of some point of the line, a x , a y , a z are the coordinates of the directing vector of the line, and is a parameter that takes arbitrary real values. Let us describe the process of transition from the direct equations of the form 
to canonical and parametric equations of a straight line in space.
In the previous paragraphs, we learned how to find the coordinates of a certain point on a straight line, as well as the coordinates of some directing vector of a straight line, which is given by the equations of two intersecting planes. These data are sufficient to write down both the canonical and parametric equations of this line in a rectangular coordinate system in space.
Consider the solution of the example, and after that we will show another way to find the canonical and parametric equations of a straight line in space.
Example.
Solution.
Let us first calculate the coordinates of the directing vector of the straight line. To do this, we find the vector product of normal vectors 
and 
planes 
and 
:
That is, .
Now let's determine the coordinates of some point of the given line. To do this, we find one of the solutions to the system of equations 
.
Determinant 
is different from zero, we take it as the basis minor of the main matrix of the system. Then the variable z is free, we transfer the terms with it to the right-hand sides of the equations, and give the variable z an arbitrary value:
We solve the resulting system of equations by the Cramer method: 
Hence, 
We accept , while we obtain the coordinates of the point of the straight line: 
.
Now we can write the required canonical and parametric equations of the original line in space: 
Answer:
and 
Here is the second way to solve this problem.
When finding the coordinates of a certain point on a straight line, we solve the system of equations . In general, its solutions can be written as .
And these are just the desired parametric equations of a straight line in space. If each of the obtained equations is resolved with respect to the parameter and then the right-hand sides of the equalities are equated, then we obtain the canonical equations of the straight line in space 
Let us show the solution of the previous problem by this method.
Example.
A straight line in three-dimensional space is given by the equations of two intersecting planes . Write the canonical and parametric equations of this line.
Solution.
We solve this system of two equations with three unknowns (the solution is given in the previous example, we will not repeat it). At the same time, we get . These are the desired parametric equations of a straight line in space.
It remains to obtain the canonical equations of a straight line in space: 
The obtained equations of the straight line outwardly differ from the equations obtained in the previous example, but they are equivalent, since they define the same set of points in three-dimensional space (and hence the same straight line).
Answer:
and 
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.
Let l- some line of space. As in planimetry, any vector
a =/= 0, collinear straight line l, is called guide vector this straight line.
The position of a straight line in space is completely determined by specifying a direction vector and a point belonging to the straight line.
Let the line l with guide vector a passes through the point M 0 , and M is an arbitrary point in space. Obviously, the point M (Fig. 197) belongs to the line l if and only if the vector \(\overrightarrow(M_0 M)\) is collinear to the vector a , i.e.
\(\overrightarrow(M_0 M)\) = t a , t\(\in\) R. (1)
If the points M and M 0 are given by their radius vectors r and r 0 (Fig. 198) with respect to some point O of the space, then \(\overrightarrow(M_0 M)\) = r - r 0 , and equation (1) takes the form
r = r 0 + t a , t\(\in\) R. (2)
Equations (1) and (2) are called vector-parametric equations of a straight line. Variable t in vector-parametric equations, a straight line is called parameter.
Let the point M 0 be a straight line l and direction vector a are given by their coordinates:
M 0 ( X 0 ; at 0 , z 0), a = (a 1 ; a 2 ; a 3).
Then if ( X; y; z) - coordinates of an arbitrary point M of the line l, then
\(\overrightarrow(M_0 M) \) = ( x - x 0 ; u - u 0 ; z - z 0)
and the vector equation (1) is equivalent to the following three equations:
x - x 0 = ta 1 , u - u 0 = ta 2 , z - z 0 = ta 3
$$ \begin(cases) x = x_0 + ta_1 \\ y = y_0 + ta_2 \\ z = z_0 + ta_3, \;\;t\in R\end(cases) (3)$$
Equations (3) are called parametric equations of the straight line in space.
Task 1. Write the parametric equations of a straight line passing through a point
M 0 (-3; 2; 4) and having a direction vector a = (2; -5; 3).
In this case X 0 = -3, at 0 = 2, z 0 = 4; a 1 = 2; a 2 = -5; a 3 = 3. Substituting these values into formulas (3), we obtain the parametric equations of this straight line
$$ \begin(cases) x = -3 - 2t \\ y = 2 - 5t \\ z = 4 + 3t, \;\;t\in R\end(cases) $$
Exclude the parameter t from equations (3). This can be done because a =/= 0, and therefore one of the coordinates of the vector a obviously different from zero.
First, let all coordinates be different from zero. Then
$$ t=\frac(x-x_0)(a_1),\;\;t=\frac(y-y_0)(a_2),\;\;t=\frac(z-z_0)(a_3) $$
and hence
$$ \frac(x-x_0)(a_1)=\frac(y-y_0)(a_2)=\frac(z-z_0)(a_3) \;\; (4)$$
These equations are called canonical equations of the line .
Note that equations (4) form a system of two equations with three variables x, y and z.
If in equations (3) one of the coordinates of the vector a , For example a 1 is equal to zero, then, excluding the parameter t, we again obtain a system of two equations with three variables x, y and z:
\(x=x_0, \;\; \frac(y-y_0)(a_2)=\frac(z-z_0)(a_3)\)
These equations are also called the canonical equations of the line. For uniformity, they are also conditionally written in the form (4)
\(\frac(x-x_0)(0)=\frac(y-y_0)(a_2)=\frac(z-z_0)(a_3)\)
considering that if the denominator is equal to zero, then the corresponding numerator is equal to zero. These equations are equations of a straight line passing through the point M 0 ( X 0 ; at 0 , z 0) parallel to the coordinate plane yOz, since this plane is parallel to its direction vector (0; a 2 ; a 3).
Finally, if in equations (3) two coordinates of the vector a , For example a 1 and a 2 are equal to zero, then these equations take the form
X = X 0 , y = at 0 , z = z 0 + t a 3 , t\(\in\) R.
These are the equations of a straight line passing through the point M 0 ( X 0 ; at 0 ; z 0) parallel to the axis Oz. For such a direct X = X 0 , y = at 0 , a z- any number. And in this case, for uniformity, the equations of a straight line can be written (with the same reservation) in the form (4)
\(\frac(x-x_0)(0)=\frac(y-y_0)(0)=\frac(z-z_0)(a_3)\)
Thus, for any line in space, one can write canonical equations (4), and, conversely, any equation of the form (4) provided that at least one of the coefficients a 1 , a 2 , a 3 is not equal to zero, defines some line of space.
Task 2. Write the canonical equations of a straight line passing through the point M 0 (- 1; 1, 7) parallel to the vector a = (1; 2; 3).
Equations (4) in this case are written as follows:
\(\frac(x+1)(1)=\frac(y-1)(2)=\frac(z-7)(3)\)
Let us derive the equations of a straight line passing through two given points M 1 ( X 1 ; at 1 ; z 1) and
M2( X 2 ; at 2 ; z 2). It is obvious that the direction vector of this straight line can be taken as the vector a = (X 2 - X 1 ; at 2 - at 1 ; z 2 - z 1), but beyond the point M 0 through which the line passes, for example, the point M 1. Then equations (4) will be written as follows:
\(\frac(x-x_1)(x_2 - x_1)=\frac(y-y_1)(y_2 - y_1)=\frac(z-z_1)(z_2 - z_1)\) (5)
These are the equations of a straight line passing through two points M 1 ( X 1 ; at 1 ; z 1) and
M2( X 2 ; at 2 ;z 2).
Task 3. Write the equations of a straight line passing through the points M 1 (-4; 1; -3) and M 2 (-5; 0; 3).
In this case X 1 = -4, at 1 = 1, z 1 = -3, X 2 = -5, at 2 = 0, z 2 = 3. Substituting these values into formulas (5), we obtain
\(\frac(x+4)(-1)=\frac(y-1)(-1)=\frac(z+3)(6)\)
Task 4. Write the equations of a straight line passing through the points M 1 (3; -2; 1) and
M 2 (5; -2; 1/2).
After substituting the coordinates of the points M 1 and M 2 into equations (5), we obtain
\(\frac(x-3)(2)=\frac(y+2)(0)=\frac(z-1)(-\frac(1)(2))\)
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 → = (ax , ay , az) : M 1 M → = λ a → ⇔ x - x 1 = λ axy - y 1 = λ ayz - z 1 = λ az
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 axy - y 1 = λ ayz - z 1 = λ az ⇔ λ = x - x 1 ax λ = y - y 1 ay λ = z - z 1 az ⇔ x - x 1 ax = y - y 1 ay = z - z 1 az
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 → = (ax, ay, az) is the direction vector of the original line, then all vectors μ a → = μ ax , μ ay , μ az , 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 ax = y - y 1 ay = z - z 1 az will correspond to a straight line passing through the point M 1 (x 1 , y 1 , z 1) , and the vector a → = ( ax, ay, az) will be a guide for her. 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.
Solution
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 .
Solution
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 ax = y - y 1 ay = z - z 1 az ⇔ 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 ay = z - z 1 az = λ ⇔ x - x 1 = 0 y = y 1 + ay λ z = z 1 + az λ ⇔ x - x 1 = 0 y - y 1 ay = z - z 1 az = λ
In the second case:
x - x 1 ax = y - y 1 0 = z - z 1 az = λ ⇔ x = x 1 + ax λ y - y 1 = 0 z = z 1 + az λ ⇔ y - y 1 = 0 x - x 1 ax = z - z 1 az = λ
In the third case:
x - x 1 ax = y - y 1 ay = z - z 1 0 = λ ⇔ x = x 1 + ax λ y = y 1 + ay λ z - z 1 = 0 ⇔ z - z 1 = 0 x - x 1 ax = y - y 1 ay = 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.
Solution
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.
Solution
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.
Solution
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 ax = y - y 1 ay = z - z 1 az ⇔ x - x 1 ax = y - y 1 ay = z - z 1 az ⇔ ⇔ x - x 1 ax = λ y - y 1 ay = λ z - z 1 az = λ ⇔ x = x 1 + ax λ y = y 1 + ay λ z = z 1 + az 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.
Solution
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 ax = y - y 1 ayx - x 1 ax = z - z 1 azy - y 1 ay = z - z 1 az ⇔ ay (x - x 1) = ax (y - y 1) az (x - x 1) = ax (z - z 1) az (y - y 1) = ay (z - z 1) ⇔ ⇔ ay x - ax y + ax y 1 - ay x 1 = 0 az x - ax z + ax z 1 - az x 1 = 0 az y - ay z + ay z 1 - az y 1 = 0
As a result, we got that:
x - x 1 ax = y - y 1 ay = z - z 1 az ⇔ ay x - ax y + ax y 1 - ay x 1 = 0 az x - ax z + ax z 1 - az x 1 = 0 az y - ay z + ay z 1 - az 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:
ay - axaz 0 = ax az , ay 0 az - ax = ax ay , - ax 0 0 - ax = ax 2 ay - ax 0 az = ay az , ay 0 0 - ay = - ay 2 , - ax 0 az - ay = ax ayaz 0 0 az = az
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.
Solution
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.
Solution
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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