Canonical equations of a line in space are equations that define a line passing through a given point collinear to the 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., the condition is satisfied for them:
.
The above equations are the canonical equations of the straight 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 simultaneously be equal to zero. But one or two of them may turn out to be zero. In analytical geometry, for example, the following entry is allowed:
,
which means that the projections of the vector on the axis Oy And Oz are equal to zero. Therefore, both the vector and the straight line defined by the canonical equations are perpendicular to the axes Oy And Oz, i.e. planes yOz .
Example 1. Write equations for a line in space perpendicular to a plane 
and passing through the point of intersection of this plane with the axis Oz
.
Solution. Let's find the point of intersection of this plane with the axis Oz. Since any point lying 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 this 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 directing vector of the straight line can be the normal vector 
given plane.
Now let’s write down the required equations of a straight line passing through a point A= (0; 0; 2) in the direction of the vector:
Equations of a 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 determine a line passing through two given points.
Example 2. Write an equation for a line in space passing through the points and .
Solution. Let us write down the required equations of the straight line in the form given above in the theoretical reference:
.
Since , then the desired straight line is perpendicular to the axis Oy .
Straight as the line of intersection of planes
A straight line in space can be defined as the line of intersection of two non-parallel planes and, i.e., as a set of points satisfying 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 line in space given by general equations
Solution. To write the canonical equations of a line or, what is the same, the equations of a line passing through two given points, you need to find the coordinates of any two points on the 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 and 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) the desired line. Then assuming 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 let's write down the equations of the line passing through the points A(0; 2; 6) and B (-2; 0; 0) :
,
or after dividing the denominators by -2:
,
How to write equations of a straight line in space?
Equations of a straight line in space
Similar to 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 line:
If a certain point in space belonging to a line and the direction vector of this line are known, then the canonical equations of this line are expressed by the formulas:
The above notation assumes that the coordinates of the direction vector not equal to zero. We'll look at what to do if one or two coordinates are zero a little later.
Same as in the article Plane equation, for simplicity we will assume that in all problems of the lesson, actions are carried out in an orthonormal basis of space.
Example 1
Compose canonical equations of a line given a point and a direction vector
Solution: We compose the canonical equations of the line using the formula: 
Answer: 
And it’s a no brainer... although, no, it’s a no brainer at all.
What should you note about this very simple example? Firstly, the resulting equations DO NOT need to be reduced by one: 
. To be more precise, it is possible to shorten it, but it unusually hurts the eye and creates inconvenience when solving problems.
And secondly, in analytical geometry two things are inevitable - verification and testing:
Just in case, we look at the denominators of the equations and check - is it right the coordinates of the direction vector are written there. No, don’t think about it, we are not having a lesson at the Brake kindergarten. This advice is very important because it allows you to completely eliminate inadvertent mistakes. No one is insured, what if they copied it incorrectly? Will be awarded the Darwin Prize in 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.
The test is very easy (and quick!) to perform orally.
In a number of problems it is required to find some other point belonging to a given line. How to do it?
We take the resulting equations and mentally “pinch off”, for example, the left piece: . Now let's equate this piece to any number(remember that there was already a zero), for example, to one: . Since , then the other two “pieces” should 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 make the drawing in a rectangular coordinate system. At the same time, let’s remember how to correctly plot points in space: 
Let's build a point:
– from the origin of coordinates in the negative direction of the axis we plot a 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, measure three units upward (purple dotted line).
Construct a point: measure two units “towards you” (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 IN FRONT of the axis.
The straight line itself passes above the axis and, if my eye does not fail me, above the axis. It does not fail, I was convinced analytically. If the straight line passed BEHIND the axis, then you would have to erase with an eraser a piece of the line above and below the crossing point.
A straight line has an infinite number of direction vectors, for example:
(red arrow)
The result was exactly the original vector, but this was purely an accident, that’s how I chose the point. All direction vectors of a straight line are collinear, and their corresponding coordinates are proportional (for more details, see Linear (non) dependence of vectors. Basis of vectors). So, vectors 
will also be direction vectors of this line.
Additional information on creating 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 the points (see drawing) are usually thinly drawn with a simple pencil using the same dotted line.
Let's deal with special cases when one or two coordinates of the direction vector are zero. At the same time, we continue the training of spatial vision, which began at the beginning of the lesson. Plane equation. And again I will tell you the tale of the naked king - I will draw an empty coordinate system and convince you that there are spatial lines there =) 
It’s easier to list all six cases:
1) For a point and a direction vector, the canonical equations of the line break down into three individual equations: .
Or in short:
Example 2: let's create equations of a straight line using a point and a direction vector:
What kind of line is this? The direction vector of the straight line is collinear to the unit vector, which means that this straight line will be parallel to the axis. The canonical equations should be understood as follows:
a) – “y” and “z” permanent, are equal specific numbers;
b) the variable “x” can take any value: (in practice, this equation is usually not written down).
In particular, the equations define the axis itself. Indeed, “x” takes on 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 abscissa axis: . After all, these are equations of two planes! The equation specifies the coordinate plane, and the equation specifies the coordinate plane. You think correctly - these coordinate planes intersect along the axis. We will consider the method when a straight line in space is defined by the intersection of two planes at the very end of the lesson.
Two similar cases:
2) The canonical equations of a line passing through a point parallel to the vector are expressed by the formulas.
Such straight lines will be parallel to the coordinate axis. In particular, the equations specify the coordinate axis itself.
3) The canonical equations of a line passing through a point parallel to the vector are expressed by the formulas.
These straight lines are parallel to the coordinate axis, and the equations define the applicate axis itself.
Let's put the second three in the stall:
4) For a point and a direction vector, the canonical equations of the line break down into proportion and plane equation .
Example 3: let's compose the equations of a straight line using a point and a direction vector.
Properties of a straight line in Euclidean geometry.
An infinite number of straight lines can be drawn through any point.
Through any two non-coinciding points a single straight line can be drawn.
Two divergent lines in a plane either intersect at a single point or are
parallel (follows from the previous one).
In three-dimensional space, there are three options for the relative position of two lines:
- lines intersect;
- lines are parallel;
- straight lines intersect.
Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
and constant A, B are not equal to zero at the same time. This first order equation is called general
equation of a straight line. Depending on the values of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠0, B ≠ 0- a straight line passes through the origin
. A = 0, B ≠0, C ≠0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠0- the straight line coincides with the axis OU
. A = C = 0, B ≠0- the straight line coincides with the axis Oh
The equation of a straight line can be presented in different forms depending on any given
initial conditions.
Equation of a straight line from a point and a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ax + Wu + C = 0.
Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
Let's substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the required equation: 3x - y - 1 = 0.
Equation of a line passing through two points.
Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2, z 2), Then equation of a line,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be set equal to zero. On
plane, the equation of the straight line written above is simplified:
If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .
Fraction = k called slope straight.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Solution. Applying the formula written above, we get:
Equation of a straight line using a point and slope.
If the general equation of the line Ax + Wu + C = 0 lead to:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
Equation of a straight line from a point and a direction vector.
By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task
a straight line through a point and a directing vector of a straight line.
Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition
Aα 1 + Bα 2 = 0 called directing vector of a straight line.
Ax + Wu + C = 0.
Example. Find the equation of a straight line with a direction vector (1, -1) and passing through the point A(1, 2).
Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. According to the definition,
coefficients must satisfy the following conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.
at x = 1, y = 2 we get C/A = -3, i.e. required equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by -С, we get:
or where
The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point
straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis OU.
Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.
C = 1, , a = -1, b = 1.
Normal equation of a line.
If both sides of the equation Ax + Wu + C = 0 divide by number 
which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a line.
The sign ± of the normalizing factor must be chosen so that μ*C< 0.
R- the length of the perpendicular dropped from the origin to the straight line,
A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. The general equation of the line is given 12x - 5y - 65 = 0. Required to write different types of equations
this straight line.
The equation of this line in segments:
The equation of this line with the slope: (divide by 5)
Equation of a line:
cos φ = 12/13; sin φ= -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,
parallel to the axes or passing through the origin.
The angle between straight lines on a plane.
Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, 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 = -1/ k 2 .
Theorem.
Direct Ax + Wu + C = 0 And A 1 x + B 1 y + C 1 = 0 parallel when the coefficients are proportional
A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these lines.
The equation of a line passing through a given point perpendicular to a given line.
Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
Distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the straight line Ax + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of a perpendicular dropped from a point M for a given
direct. Then the distance between points M And M 1:
(1)
Coordinates x 1 And at 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly
given straight line. If we transform the first equation of the system to the form:
A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
3.1. Canonical equations of the line.
Let a straight line be given in the Oxyz coordinate system that passes through the point
(see Fig. 18). Let us denote by 
a vector parallel to a given line. Vector 
called directing vector of a straight line. Let's take a point on a straight line 
and consider the vector Vectors 
are collinear, therefore their corresponding coordinates are proportional:
(3.3.1
)
These equations are called canonical equations straight.
Example: Write the equations of the 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: Turning to zero of one of the denominators means turning to zero of the corresponding numerator, that is, y – 2 = 0; y = 2. This line lies in the y = 2 plane, parallel to the Oxz plane.
3.2. Parametric equations of a straight line.
Let the straight line be given by the canonical equations
Let's 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 a 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 a straight line, we apply the derivation of formulas (3.2.1):
So, 
- parametric equations of a given line.
Answer:
Example 2. Write parametric equations for a 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). Using formulas (3.2.1), we write down the equations of the straight line:
are the required parametric equations of the straight line.
3.3. Equations of a 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 line. Then the equations can be found directly 
them according to formulas (3.1.1): 
).
(3.3.1)
Example 1. Compose canonical and parametric equations of a line passing through points
Solution: We apply formula (3.3.1)
We obtained the canonical equations of the straight line. To obtain parametric equations, we apply the derivation of formulas (3.2.1). We get
are parametric equations of a straight line.
Example 2. Compose canonical and parametric equations of a line passing through points
Solution: Using formulas (3.3.1) we obtain:
These are canonical equations.
Let's move on 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 go to canonical equations (3.1.1) or parametric equations (3.2.1). To do this you need to find a point 
lying on a straight line, and the direction vector 
Point coordinates 
we obtain from system (3.4.1), giving one of the coordinates an arbitrary value (for example, z = 0). Behind the guide vector 
you can take the vector product of vectors, that is
Example 1. Compose the canonical equations of the line 
Solution: Let z = 0. Let us 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, period 
lies on the desired line.
To find the direction vector of a straight line, we write down the normal vectors of the planes: and find their vector product:
We find the equations of the straight line using 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 the system (3.4.1), and then applying formulas (3.3.1) and the derivation of formulas (3.2.1).
Example 2. Compose canonical and parametric equations of the 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 the point 
To find the second point, let's set x = 0. We will have: 
That is 
We obtained the canonical equations of the straight line.
Let's compose the parametric equations of the straight line:
Answer:
; 
.
3.5. The relative position of two lines in space.
Let straight 
are given by the equations:
:
;
: 
.
The angle between these lines is understood as the angle between their direction vectors (see Fig. 22). This angle 
we find using a formula from vector algebra: 
or
(3.5.1)
If straight 
perpendicular ( 
),That 
Hence,
This is the condition of perpendicularity of two lines in space.
If straight 
parallel ( 
), then their direction vectors are collinear ( 
), that is
(3.5.3
)
This is the condition of parallelism of two lines in space.
Example 1. Find the angle between straight lines:
A). 
And 
b). 
And 
Solution: A). Let's write down the direction vector of the straight line 
Let's find the direction vector 
planes included in the system. Then we find their vector product: 
(see example 1 of clause 3.4).
Using formula (3.5.1) we obtain: 
Hence, 
b). Let's write down the direction vectors of these straight lines: Vectors 
are collinear because their corresponding coordinates are proportional:
So it's straight 
parallel ( 
), that is 
Answer: A).
b). 
Example 2. Prove perpendicularity of lines:
And 
Solution: Let's write down the direction vector of the first straight line 
Let's find the direction vector 
second straight line. To do this, we find normal vectors 
planes included in the system: Let us calculate their vector product: 
(See example 1 of paragraph 3.4).
Let us apply the condition of perpendicularity of lines (3.5.2):
The condition is met; therefore, the lines are perpendicular ( 
).
One of the types of equations of a line in space is the canonical equation. We will consider this concept in detail, since knowing it is necessary to solve many practical problems.
In the first paragraph, we will 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 tell you how to construct an equation for a line passing through 2 given points in three-dimensional space, and in the last paragraph we will point out the connections between canonical equations and others. All arguments will be illustrated with examples of problem solving.
We have already discussed 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 in which a straight line is given. As we remember, you can define a straight line in different ways. Let's use the simplest of them - set the point through which the line will pass, and indicate the direction vector. If we denote a line by the letter a and a point by M, then we can write that M 1 (x 1, y 1, z 1) lies on line a and the direction vector of this line will be a → = (a x, a y, a z). In order for the set of points M (x, y, z) to define a straight 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 coordinate form the necessary and sufficient condition for their collinearity. From the initial conditions we already know the coordinates a → . In order to obtain 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 down:
M 1 M → = x - x 1 , y - y 1 , z - z 1
After this, 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 resolve all equations of the system with respect to the parameter λ x - x 1 = λ · a x y - y 1 = λ · a y z - z 1 = λ · a z
After this, it will be possible to put an equal sign between the right sides:
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 = y - y 1 a y = z - z 1 a z, with the help of which we can determine the desired line in three-dimensional space. These are the canonical equations we need.
This notation is used even if one or two parameters a x , a y , a z are zero, since in these cases it will also be correct. All three parameters cannot be equal to 0, since the direction vector a → = (a x, a y, a z) is never 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.
From the definition of the canonical equation of a line in space, several important conclusions can be drawn. Let's look at them.
1) if the original line passes 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 straight 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 create the canonical equation of a line in space
We found out that 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 a guide for it. This means that if we know the equation of a line, we can calculate the coordinates of its direction vector, and given the given coordinates of the vector and some point located on the line, we can write down its canonical equations.
Let's look at a couple of specific problems.
Example 3
We have a line defined in three-dimensional 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 coordinates of the direction vector, we just need to take the denominator values from the equation. We find 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 zero).
Answer: 4 μ, 2 μ, - 5 μ, μ ∈ R, μ ≠ 0
Example 4
Write down the canonical equations if a 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 immediately move on to writing 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 problems are the simplest because they have all or almost all the initial data for writing the equation or vector coordinates. In practice, you can often find those in which you first need to find the required coordinates, and then write down the canonical equations. We analyzed examples of such problems in articles devoted to finding the equations of a line passing through a point in space parallel to a given one, as well as a line passing through a certain point in space perpendicular to a plane.
We have already said earlier 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 notation x - x 1 a x = y - y 1 a y = z - z 1 a z = 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. Let us assume 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 straight lines are located in the planes x - x 1 = 0, y - y 1 = 0 or z - z 1 = 0, which are located parallel to the coordinate planes (if x 1 = 0, y 1 = 0 or z 1 = 0). Examples of such lines are shown in the illustration.
Therefore, we can write the canonical equations a little differently.
- 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 straight lines will coincide with the coordinate axes or 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 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 us show with examples how these rules are applied.
Example 5
Find the 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 straight lines. We also know that our lines will definitely pass through the point O (0, 0, 0), since it is the origin of coordinates. 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 straight line O y: x 0 = y 1 = z 0
For straight line O z: x 0 = y 0 = z 1
Answer: x 1 = y 0 = z 0 , x 0 = y 1 = z 0 , x 0 = y 0 = z 1 .
Example 6
A line is given in space that passes through the point M 1 (3, - 1, 12). It is also known that it is located parallel to the ordinate axis. Write down the canonical equations of this line.
Solution
Taking into account the parallelism condition, we can say that the vector j → = 0, 1, 0 will be a guide for the desired straight line. Therefore, the required equations will look like:
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 assume that we have two divergent 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 a 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 required 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 line passing through two given points. Take a look at the illustration:
Let's 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 a straight 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 = y - y 2 y 2 - y 1 = 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 canonical equations of a line in space into other types of equations
Sometimes using canonical equations of the form x - x 1 a x = y - y 1 a y = z - z 1 a z is not very convenient. 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 we can move from canonical equations to other types, if this is required by the conditions of the problem.
It is not difficult to understand the rules for transition 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 on 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 get:
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 λ z = - 7
Answer: x = 2 + 3 λ y = - 2 λ z = - 7
Our next step will be to transform the canonical equations into an equation of two intersecting planes (for the same line).
The equality x - x 1 a x = y - y 1 a y = 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 this:
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 eliminate one equation from our calculations. Thus, the canonical straight line equations can be transformed into a system of two linear equations that will contain 3 unknowns. They will be the equations of two intersecting planes we need.
The reasoning looks quite 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 equation 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 intersecting, form a straight line defined by the equation x - 1 2 = y 0 = z + 2 0
Answer: y = 0 z + 2 = 0
Example 10
The line is given by the equations x + 1 2 = y - 2 1 = z - 5 - 3 , find the equation of two planes intersecting along this line.
Solution
Equate 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 find 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 second-order minor will not be zero: 1 - 2 3 0 = 1 · 0 - (- 2) · 3 = 6. Then we can accept it as a basic 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. This 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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