An important geometric object that is studied in flat space is a straight line. In three-dimensional space, in addition to a straight line, a plane also appears. Both objects are conveniently defined using guide vectors. What is it, how do these vectors apply to determine the equations of a line and a plane? These and other issues are covered in the article.
Direct and ways to set it
Each student is well aware of what kind of geometric object in question. From the point of view of mathematics, a straight line is a set of points that, in the case of their pairwise arbitrary connection with each other, lead to the collection of parallel vectors. This definition of a line is used to write an equation for it in both two-dimensional and three-dimensional space.
To describe the one-dimensional object under consideration, different types of equations are used, which are listed in the list below:
- general view;
- parametric;
- vectorial;
- canonical or symmetric;
- in segments.
Each of these species has some advantages over others. For example, the equation in segments is convenient to use when studying the behavior of a straight line with respect to the coordinate axes, the general equation is convenient when finding a direction perpendicular to a given line, as well as when calculating the angle of its intersection with the x axis (for the flat case).
Since the topic of this article is related to the directing vector of the line, then we will consider only the equation where this vector is fundamental and contains explicitly, that is, a vector expression.
Setting a line through a vector
Suppose that we have some vector v¯ with known coordinates (a; b; c). Since there are three coordinates, the vector is given in space. How to display it in a rectangular coordinate system? This is done very simply: a segment is laid out on each of the three axes, the length of which is equal to the corresponding coordinate of the vector. The intersection point of the three perpendiculars restored to the planes xy, yz and xz will be the end of the vector. The beginning of it is the point (0; 0; 0).
Nevertheless, the reduced position of the vector is not the only one. In a similar way, we can draw v¯ by placing its beginning at an arbitrary point in space. These considerations suggest that it is impossible to specify a specific line using a vector. It defines a family of an infinite number of parallel lines.
Now fix some point P (x 0 ; y 0 ; z 0 ) of the space. And we set the condition: a straight line should pass through P. In this case, the vector v¯ must also contain this point. The latter fact means that one single line can be defined using P and v¯. It will be written in the form of the following equation:
Q = P + λ × v¯
Here Q is any point that belongs to a line. This point can be obtained by selecting the corresponding parameter λ. The written equation is called vector, and v¯ is called the directing vector of the line. Positioning it so that it passes through P, and changing its length using the parameter λ, we get each point Q of the line.
In coordinate form, the equation is written as follows:
(x; y; z) = (x 0 ; y 0 ; z 0 ) + λ × (a; b; c)
And in an explicit (parametric) form, you can write:
x = x 0 + λ × a;
y = y 0 + λ × b;
z = z 0 + λ × c
If the third coordinate is excluded in the above expressions, then we obtain vector equations of the line on the plane.
For what tasks is it useful to know the direction vector?
As a rule, these are tasks for determining the parallelism and perpendicularity of straight lines. Also, the direction vector defining the direction is used in calculating the distance between the lines and the point and the line, to describe the behavior of the line relative to the plane.
Two straight lines will be parallel, if their directing vectors are those. Accordingly, the perpendicularity of the lines is proved by the perpendicularity of their vectors. In these types of problems, it is enough to calculate the scalar product of the vectors in question to get an answer.
In the case of problems on calculating the distances between lines and points, the guiding vector is explicitly included in the corresponding formula. We write it down:
d = | [P 1 P 2 ¯ × v¯] | / | v¯ |
Here P 1 P 2 ¯ is the directed segment constructed at the points P 1 and P 2 . The point P 2 is arbitrary, lying on a line with the vector v¯, the point P 1 is the one to which the distance should be determined. It can be either independent or belong to another line or plane.
Note that it makes sense to calculate the distance between the lines only when they are parallel or intersecting. If they intersect, then d is zero.
The above formula for d is also valid for calculating the distance between a plane and a straight line parallel to it, only in this case P 1 must belong to the plane.
We will solve several problems in order to more clearly show how to use the vector in question.
The task of compiling a vector equation
It is known that the line is described by the following equality:
y = 3 × x - 4
Write the corresponding expression in vector form.
This typical equation of the line, known to every student, is written in a general way. We show how to rewrite it in vector form.
The expression can be represented as:
(x; y) = (x; 3 × x - 4)
It can be seen that if you open it, you get the original equality. Now we divide its right side into two vectors so that only one of them contains x, we have:
(x; y) = (x; 3 × x) + (0; -4)
It remains to put x out of the brackets, designate it with a Greek symbol and swap the vectors of the right side with places:
(x; y) = (0; -4) + λ × (1; 3)
We got a vector form for writing the original expression. The coordinates of the directing vector of the line are (1; 3).
The task of determining the relative position of lines
In space, two lines are given:
(x; y; z) = (1; 0; -2) + λ × (-1; 3; 1);
(x; y; z) = (3; 2; 2) + γ × (1; 2; 0)
Are they parallel, intersecting or intersecting?
Nonzero vectors (-1; 3; 1) and (1; 2; 0) will be guides for these lines. We express these equations in parametric form and substitute the coordinates of the first into the second. We get:
x = 1 - λ;
y = 3 × λ;
z = -2 + λ;
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x = 3 + γ = 1 - λ => γ = -2 - λ;
y = 2 + 2 × γ = 3 × λ => γ = 3/2 × λ - 1;
z = 2 = -2 + λ => λ = 4
Substituting the found parameter λ in the two equations above, we obtain:
γ = -2 - λ = -6;
γ = 3/2 × λ - 1 = 5
The parameter γ cannot simultaneously take two different values. This means that the lines do not have a single common point, that is, they are crossed. They are not parallel, since nonzero vectors are not parallel to each other (for their parallelism, there must be a number that, by multiplying by one vector, leads to the coordinates of the second).
The mathematical description of the plane
To define a plane in space, we present a general equation:
A × x + B × y + C × z + D = 0
Here, Latin capital letters represent specific numbers. The first three of them determine the coordinates of the normal plane vector. If it is denoted by n¯, then:
n¯ = (A; B; C)
This vector is perpendicular to the plane; therefore, it is called a guideline. His knowledge, as well as the known coordinates of a point belonging to the plane, uniquely specify the latter.
If the point P (x 1 ; y 1 ; z 1 ) of the plane belongs, then the free term D is calculated as follows:
D = -1 × (A × x 1 + B × y 1 + C × z 1 )
We solve a couple of problems using the general equation for the plane.
The problem of finding a normal plane vector
The plane is defined as follows:
(y - 3) / 2 + (x + 1) / 3 - z / 4 = 1
How to find a guide vector for her?
It follows from the above theory that the coordinates of the normal vector n¯ are the coefficients facing the variables. In this regard, to find n¯, one should write the equation in general form. We have:
1/3 × x + 1/2 × y - 1/4 × z - 13/6 = 0
Then the normal plane vector is:
n¯ = (1/3; 1/2; -1/4)
The task of compiling a plane equation
Given the coordinates of three points:
M 1 (1; 0; 0);
M 2 (2; -1; 5);
M 3 (0; -2; -2)
What will the equation of the plane containing all these points look like?
Through three points that do not belong to one line, only one plane can be drawn. To find its equation, we first calculate the direction vector of the plane n¯. To do this, proceed as follows: find two arbitrary vectors that belong to the plane, and calculate their vector product. It will give a vector that will be perpendicular to this plane, that is, n¯. We have:
M 1 M 2 ¯ = (1; -1; 5); M 1 M 3 ¯ = (-1; -2; -2);
n¯ = [M 1 M 2 ¯ × M 1 M 3 ¯] = (12; -3; -3)
Take the point M 1 to compose the expression of the plane. We get:
D = -1 × (12 × 1 + (-3) × 0 + (-3) × 0) = -12;
12 × x - 3 × y - 3 × z - 12 = 0 =>
4 × x - y - z - 4 = 0
We got an expression of a general type for a plane in space, first defining a directional vector for it.
The property of a vector product should be remembered when solving problems with planes, because it allows you to easily determine the coordinates of a normal vector in a simple way.