Linear dependency and linear independence of vectors.
Basis of vectors. affine system coordinates

There is a cart with chocolates in the audience, and today each visitor will get a sweet couple - analytical geometry with linear algebra. This article will touch upon two sections of higher mathematics at once, and we will see how they get along in one wrapper. Take a break, eat Twix! ... damn, well, arguing nonsense. Although okay, I won’t score, in the end, there should be a positive attitude to study.

Linear dependence of vectors, linear independence of vectors, vector basis and other terms have not only a geometric interpretation, but, above all, an algebraic meaning. The very concept of "vector" from the point of view of linear algebra is far from always the "ordinary" vector that we can depict on a plane or in space. You don't need to look far for proof, try drawing a vector of five-dimensional space . Or the weather vector, for which I just went to Gismeteo: - temperature and atmospheric pressure, respectively. The example, of course, is incorrect from the point of view of the properties of the vector space, but, nevertheless, no one forbids formalizing these parameters as a vector. Breath of autumn...

No, I'm not going to bore you with theory, linear vector spaces, the task is to understand definitions and theorems. The new terms (linear dependence, independence, linear combination, basis, etc.) are applicable to all vectors from an algebraic point of view, but examples will be given geometrically. Thus, everything is simple, accessible and visual. In addition to the problems of analytic geometry, we will also consider some typical tasks of algebra. To master the material, it is advisable to familiarize yourself with the lessons Vectors for dummies And How to calculate the determinant?

Linear dependence and independence of plane vectors.
Plane basis and affine coordinate system

Consider the plane of your computer desk(just a table, bedside table, floor, ceiling, whatever you like). The task will consist of the following actions:

1) Select plane basis. Roughly speaking, the tabletop has a length and a width, so it is intuitively clear that two vectors are required to build the basis. One vector is clearly not enough, three vectors are too much.

2) Based on the chosen basis set coordinate system(coordinate grid) to assign coordinates to all items on the table.

Do not be surprised, at first the explanations will be on the fingers. Moreover, on yours. Please place forefinger left hand on the edge of the tabletop so that he looks at the monitor. This will be a vector. Now place little finger right hand on the edge of the table in the same way - so that it is directed at the monitor screen. This will be a vector. Smile, you look great! What can be said about vectors? Data Vectors collinear, which means linearly expressed through each other:
, well, or vice versa: , where is a non-zero number.

You can see a picture of this action in the lesson. Vectors for dummies, where I explained the rule for multiplying a vector by a number.

Will your fingers set the basis on the plane of the computer table? Obviously not. Collinear vectors travel back and forth in alone direction, while a plane has a length and a width.

Such vectors are called linearly dependent.

Reference: The words "linear", "linear" denote the fact that there are no squares, cubes, other powers, logarithms, sines, etc. in mathematical equations, expressions. There are only linear (1st degree) expressions and dependencies.

Two plane vectors linearly dependent if and only if they are collinear.

Cross your fingers on the table so that there is any angle between them except 0 or 180 degrees. Two plane vectorslinearly not are dependent if and only if they are not collinear. So, the basis is received. No need to be embarrassed that the basis turned out to be "oblique" with non-perpendicular vectors of various lengths. Very soon we will see that not only an angle of 90 degrees is suitable for its construction, and not only unit vectors of equal length

Any plane vector the only way expanded in terms of the basis:
, where are real numbers . Numbers are called vector coordinates in this basis.

They also say that vectorpresented in the form linear combination basis vectors. That is, the expression is called vector decompositionbasis or linear combination basis vectors.

For example, one can say that a vector is expanded in an orthonormal basis of the plane , or one can say that it is represented as a linear combination of vectors .

Let's formulate basis definition formally: plane basis is a pair of linearly independent (noncollinear) vectors , , wherein any the plane vector is a linear combination of the basis vectors.

The essential point of the definition is the fact that the vectors are taken in a certain order. bases These are two completely different bases! As they say, the little finger of the left hand cannot be moved to the place of the little finger of the right hand.

We figured out the basis, but it is not enough to set the coordinate grid and assign coordinates to each item on your computer desk. Why not enough? The vectors are free and wander over the entire plane. So how do you assign coordinates to those little dirty table dots left over from a wild weekend? A starting point is needed. And such a reference point is a point familiar to everyone - the origin of coordinates. Understanding the coordinate system:

I'll start with the "school" system. Already in the introductory lesson Vectors for dummies I highlighted some of the differences between a rectangular coordinate system and an orthonormal basis. Here is the standard picture:

When talking about rectangular coordinate system, then most often they mean the origin, coordinate axes and scale along the axes. Try typing “rectangular coordinate system” in the search engine, and you will see that many sources will tell you about the coordinate axes familiar from the 5th-6th grade and how to plot points on a plane.

On the other hand, it seems that rectangular system coordinates can be determined in terms of an orthonormal basis . And it almost is. The wording goes like this:

origin, And orthonormal basis set Cartesian coordinate system of the plane . That is, a rectangular coordinate system definitely is defined by a single point and two unit orthogonal vectors. That is why, you see the drawing that I gave above - in geometric problems, both vectors and coordinate axes are often (but far from always) drawn.

I think everyone understands that with the help of a point (origin) and an orthonormal basis ANY POINT of the plane and ANY VECTOR of the plane coordinates can be assigned. Figuratively speaking, "everything on the plane can be numbered."

Do coordinate vectors have to be unit? No, they can have an arbitrary non-zero length. Consider a point and two orthogonal vectors of arbitrary non-zero length:

Such a basis is called orthogonal. The origin of coordinates with vectors define the coordinate grid, and any point of the plane, any vector has its own coordinates in the given basis. For example, or. The obvious inconvenience is that the coordinate vectors in general have different lengths other than unity. If the lengths are equal to one, then the usual orthonormal basis is obtained.

! Note : in the orthogonal basis, as well as below in the affine bases of the plane and space, units along the axes are considered CONDITIONAL. For example, one unit along the abscissa contains 4 cm, one unit along the ordinate contains 2 cm. This information is enough to convert “non-standard” coordinates into “our usual centimeters” if necessary.

And the second question, which has actually already been answered - is the angle between the basis vectors necessarily equal to 90 degrees? Not! As the definition says, basis vectors must be only non-collinear. Accordingly, the angle can be anything except 0 and 180 degrees.

A point on the plane called origin, And non-collinear vectors , , set affine coordinate system of the plane :

Sometimes this coordinate system is called oblique system. Points and vectors are shown as examples in the drawing:

As you understand, the affine coordinate system is even less convenient, the formulas for the lengths of vectors and segments, which we considered in the second part of the lesson, do not work in it. Vectors for dummies, many delicious formulas related to scalar product of vectors. But the rules for adding vectors and multiplying a vector by a number are valid, the formulas for dividing a segment in this respect, as well as some other types of problems that we will soon consider.

And the conclusion is that the most convenient particular case of an affine coordinate system is the Cartesian rectangular system. Therefore, she, her own, most often has to be seen. ... However, everything in this life is relative - there are many situations in which it is appropriate to have an oblique (or some other, for example, polar) coordinate system. Yes, and humanoids such systems may come to taste =)

Let's move on to the practical part. All problems in this lesson are valid both for a rectangular coordinate system and for the general affine case. There is nothing complicated here, all the material is available even to a schoolboy.

How to determine the collinearity of plane vectors?

Typical thing. In order for two plane vectors are collinear, it is necessary and sufficient that their respective coordinates be proportional.Essentially, this is a coordinate-by-coordinate refinement of the obvious relationship .

Example 1

a) Check if the vectors are collinear .
b) Do vectors form a basis? ?

Solution:
a) Find out if there exists for vectors coefficient of proportionality, such that equalities are fulfilled:

I will definitely tell you about the “foppish” version of the application of this rule, which works quite well in practice. The idea is to immediately draw up a proportion and see if it is correct:

Let's make a proportion from the ratios of the corresponding coordinates of the vectors:

We shorten:
, thus the corresponding coordinates are proportional, therefore,

The relation could be made and vice versa, this is an equivalent option:

For self-testing, one can use the fact that collinear vectors are linearly expressed through each other. In this case, there are equalities . Their validity can be easily checked through elementary operations with vectors:

b) Two plane vectors form a basis if they are not collinear (linearly independent). We examine vectors for collinearity . Let's create a system:

From the first equation it follows that , from the second equation it follows that , which means, the system is inconsistent(no solutions). Thus, the corresponding coordinates of the vectors are not proportional.

Output: the vectors are linearly independent and form a basis.

A simplified version of the solution looks like this:

Compose the proportion from the corresponding coordinates of the vectors :
, hence, these vectors are linearly independent and form a basis.

Usually reviewers do not reject this option, but a problem arises in cases where some coordinates are equal to zero. Like this: . Or like this: . Or like this: . How to work through the proportion here? (Really, you can't divide by zero). It is for this reason that I called the simplified solution "foppish".

Answer: a) , b) form.

A small creative example for independent decision:

Example 2

At what value of the parameter vectors will be collinear?

In the sample solution, the parameter is found through the proportion.

There is an elegant algebraic way to check vectors for collinearity. Let's systematize our knowledge and just add it as the fifth point:

For two plane vectors, the following statements are equivalent:

2) vectors form a basis;
3) the vectors are not collinear;

+ 5) the determinant, composed of the coordinates of these vectors, is nonzero.

Respectively, the following opposite statements are equivalent:
1) vectors are linearly dependent;
2) vectors do not form a basis;
3) the vectors are collinear;
4) vectors can be linearly expressed through each other;
+ 5) the determinant, composed of the coordinates of these vectors, is equal to zero.

I really, really hope that this moment you already understand all the met terms and statements.

Let's take a closer look at the new, fifth point: two plane vectors are collinear if and only if the determinant composed of the coordinates of the given vectors is equal to zero:. To use this feature, of course, you need to be able to find determinants.

We will decide Example 1 in the second way:

a) Calculate the determinant, composed of the coordinates of the vectors :
, so these vectors are collinear.

b) Two plane vectors form a basis if they are not collinear (linearly independent). Let us calculate the determinant composed of the coordinates of the vectors :
, hence the vectors are linearly independent and form a basis.

Answer: a) , b) form.

It looks much more compact and prettier than the solution with proportions.

With the help of the considered material, it is possible to establish not only the collinearity of vectors, but also to prove the parallelism of segments, straight lines. Consider a couple of problems with specific geometric shapes.

Example 3

Vertices of a quadrilateral are given. Prove that the quadrilateral is a parallelogram.

Proof: There is no need to build a drawing in the problem, since the solution will be purely analytical. Remember the definition of a parallelogram:
Parallelogram A quadrilateral is called, in which opposite sides are pairwise parallel.

Thus, it is necessary to prove:
1) parallelism of opposite sides and;
2) parallelism of opposite sides and .

We prove:

1) Find the vectors:

2) Find the vectors:

The result is the same vector (“according to school” - equal vectors). Collinearity is quite obvious, but it is better to make the decision properly, with the arrangement. Calculate the determinant, composed of the coordinates of the vectors :
, so these vectors are collinear, and .

Output: Opposite sides of a quadrilateral are pairwise parallel, so it is a parallelogram by definition. Q.E.D.

More good and different figures:

Example 4

Vertices of a quadrilateral are given. Prove that the quadrilateral is a trapezoid.

For a more rigorous formulation of the proof, it is better, of course, to get a definition of a trapezoid, but it is enough just to remember what it looks like.

This is a task for independent decision. Complete Solution at the end of the lesson.

And now it's time to slowly move from the plane into space:

How to determine the collinearity of space vectors?

The rule is very similar. For two space vectors to be collinear, it is necessary and sufficient that their corresponding coordinates be proportional to.

Example 5

Find out if the following space vectors are collinear:

but) ;
b)
in)

Solution:
a) Check if there is a proportionality coefficient for the corresponding coordinates of the vectors:

The system has no solution, which means the vectors are not collinear.

"Simplified" is made out by checking the proportion. In this case:
– the corresponding coordinates are not proportional, which means that the vectors are not collinear.

Answer: the vectors are not collinear.

b-c) These are points for independent decision. Try it out in two ways.

There is a method for checking space vectors for collinearity and through a third-order determinant, this method covered in the article Cross product of vectors.

Similarly to the plane case, the considered tools can be used to study the parallelism of spatial segments and lines.

Welcome to the second section:

Linear dependence and independence of three-dimensional space vectors.
Spatial basis and affine coordinate system

Many of the regularities that we have considered on the plane will also be valid for space. I tried to minimize the summary of the theory, since the lion's share of the information has already been chewed. Nevertheless, I recommend that you carefully read the introductory part, as new terms and concepts will appear.

Now, instead of the plane of the computer table, let's examine the three-dimensional space. First, let's create its basis. Someone is now indoors, someone is outdoors, but in any case, we can’t get away from three dimensions: width, length and height. Therefore, three spatial vectors are required to construct the basis. One or two vectors are not enough, the fourth is superfluous.

And again we warm up on the fingers. Please raise your hand up and spread out in different directions thumb, index and middle finger. These will be vectors, they look in different directions, have different lengths and have different angles between themselves. Congratulations, the basis of the three-dimensional space is ready! By the way, you don’t need to demonstrate this to teachers, no matter how you twist your fingers, but you can’t get away from definitions =)

Next, we ask an important question, whether any three vectors form a basis of a three-dimensional space? Please press three fingers firmly on the computer table top. What happened? Three vectors are located in the same plane, and, roughly speaking, we have lost one of the measurements - the height. Such vectors are coplanar and, quite obviously, that the basis of three-dimensional space is not created.

It should be noted that coplanar vectors do not have to lie in the same plane, they can be in parallel planes (just don't do this with your fingers, only Salvador Dali came off like that =)).

Definition: vectors are called coplanar if there exists a plane to which they are parallel. Here it is logical to add that if such a plane does not exist, then the vectors will not be coplanar.

Three coplanar vectors are always linearly dependent, that is, they are linearly expressed through each other. For simplicity, again imagine that they lie in the same plane. Firstly, vectors are not only coplanar, but can also be collinear, then any vector can be expressed through any vector. In the second case, if, for example, the vectors are not collinear, then the third vector is expressed through them in a unique way: (and why is easy to guess from the materials of the previous section).

The converse is also true: three non-coplanar vectors are always linearly independent, that is, they are in no way expressed through each other. And, obviously, only such vectors can form the basis of a three-dimensional space.

Definition: The basis of three-dimensional space is called a triple of linearly independent (non-coplanar) vectors, taken in a certain order, while any vector of the space the only way expands in the given basis , where are the coordinates of the vector in the given basis

As a reminder, you can also say that a vector is represented as linear combination basis vectors.

The concept of a coordinate system is introduced in exactly the same way as for the plane case, one point and any three linearly independent vectors are sufficient:

origin, And non-coplanar vectors , taken in a certain order, set affine coordinate system of three-dimensional space :

Of course, the coordinate grid is "oblique" and inconvenient, but, nevertheless, the constructed coordinate system allows us to definitely determine the coordinates of any vector and the coordinates of any point in space. Similar to the plane, in the affine coordinate system of space, some formulas that I have already mentioned will not work.

The most familiar and convenient special case of an affine coordinate system, as everyone can guess, is rectangular space coordinate system:

point in space called origin, And orthonormal basis set Cartesian coordinate system of space . familiar picture:

Before proceeding to practical tasks, we systematize the information again:

For three space vectors, the following statements are equivalent:
1) the vectors are linearly independent;
2) vectors form a basis;
3) the vectors are not coplanar;
4) vectors cannot be linearly expressed through each other;
5) the determinant, composed of the coordinates of these vectors, is different from zero.

Opposite statements, I think, are understandable.

Linear dependence / independence of space vectors is traditionally checked using the determinant (item 5). The remaining practical tasks will be of a pronounced algebraic nature. It's time to hang a geometric stick on a nail and wield a linear algebra baseball bat:

Three space vectors are coplanar if and only if the determinant composed of the coordinates of the given vectors is equal to zero: .

I draw your attention to a small technical nuance: the coordinates of vectors can be written not only in columns, but also in rows (the value of the determinant will not change from this - see the properties of the determinants). But it is much better in columns, since it is more beneficial for solving some practical problems.

For those readers who have forgotten the methods for calculating determinants a little, or maybe they are poorly oriented at all, I recommend one of my oldest lessons: How to calculate the determinant?

Example 6

Check if the following vectors form a basis of a three-dimensional space:

Solution: In fact, the whole solution comes down to calculating the determinant.

a) Calculate the determinant, composed of the coordinates of the vectors (the determinant is expanded on the first line):

, which means that the vectors are linearly independent (not coplanar) and form the basis of a three-dimensional space.

Answer: these vectors form the basis

b) This is a point for independent decision. Full solution and answer at the end of the lesson.

There are also creative tasks:

Example 7

At what value of the parameter will the vectors be coplanar?

Solution: Vectors are coplanar if and only if the determinant composed of the coordinates of the given vectors is equal to zero:

Essentially, it is required to solve an equation with a determinant. We fly into zeros like kites into jerboas - it is most profitable to open the determinant in the second line and immediately get rid of the minuses:

We carry out further simplifications and reduce the matter to the simplest linear equation:

Answer: at

It is easy to check here, for this you need to substitute the resulting value into the original determinant and make sure that by reopening it.

In conclusion, let's consider another typical problem, which is more of an algebraic nature and is traditionally included in the course of linear algebra. It is so common that it deserves a separate topic:

Prove that 3 vectors form a basis of a three-dimensional space
and find the coordinates of the 4th vector in the given basis

Example 8

Vectors are given. Show that the vectors form a basis of three-dimensional space and find the coordinates of the vector in this basis.

Solution: Let's deal with the condition first. By condition, four vectors are given, and, as you can see, they already have coordinates in some basis. What is the basis - we are not interested. And the following thing is of interest: three vectors may well form a new basis. And the first step is completely the same as the solution of Example 6, it is necessary to check if the vectors are really linearly independent:

Calculate the determinant, composed of the coordinates of the vectors :

, hence the vectors are linearly independent and form a basis of a three-dimensional space.

! Important : vector coordinates necessarily write down into columns determinant, not strings. Otherwise, there will be confusion in the further solution algorithm.

Tasks for control work

Task 1 - 10. Vectors are given. Show that the vectors form a basis of three-dimensional space and find the coordinates of the vector in this basis:

Vectors ε 1 (3;1;6), ε 2 (-2;2;-3), ε 3 (-4;5;-1), X(3;0;1) are given. Show that the vectors form a basis of three-dimensional space and find the coordinates of the vector X in this basis.

This task consists of two parts. First you need to check whether the vectors form a basis. Vectors form a basis if the determinant composed of the coordinates of these vectors is nonzero, otherwise the vectors are not basic and the vector X cannot be expanded in this basis.

Calculate the matrix determinant:

∆ = 3*(2*(-1) - 5*(-3)) - -2*(1*(-1) - 5*6) + -4*(1*(-3) - 2*6) = 37

The matrix determinant is ∆ =37

Since the determinant is non-zero, the vectors form a basis, therefore, the vector X can be expanded in this basis. Those. there are such numbers α 1 , α 2 , α 3 that the equality takes place:

X = α 1 ε 1 + α 2 ε 2 + α 3 ε 3

We write this equality in coordinate form:

(3;0;1) = α(3;1;6) + α(-2;2;-3) + α(-4;5;-1)

Using the properties of vectors, we obtain the following equality:

(3;0;1) = (3α 1 ;1α 1 ;6α 1 ;) + (-2α 2 ;2α 2 ;-3α 2 ;) + (-4α 3 ;5α 3 ;-1α 3 ;)

(3;0;1) = (3α 1 -2α 2 -4α 3 ;1α 1 + 2α 2 + 5α 3 ;6α 1 -3α 2 -1α 3)

By the property of equality of vectors, we have:

3α 1 -2α 2 -4α 3 = 3

1α 1 + 2α 2 + 5α 3 = 0

6α 1 -3α 2 -1α 3 = 1

We solve the resulting system of equations Gauss method or Cramer's method.

X \u003d ε 1 + 2ε 2 -ε 3

The solution was received and executed using the service:

Vector coordinates in basis

Together with this task, they also solve:

Solution of matrix equations

Cramer method

Gauss method

Inverse Matrix by Jordan-Gauss Method

Inverse Matrix via Algebraic Complements

Matrix multiplication online

1 (1, 2, 0, 1) , 2 (0, 1, 2, 3) , 3 (1, 3, 2, 2) , 4 (0, 1, 3, 1) , (1, 0, 1, 5).

Solution. Let us show that the vectors 1 (1, 2, 0, 1) , 2 (0, 1, 2, 3) , 3 (1, 3, 2, 2) , 4 (0, 1, 3, 1) form a basis. Let's find the determinant composed of the coordinates of these vectors.

We perform elementary transformations:

Subtract from line 3 line 1 multiplied by (-1)

Subtract line 2 from line 3, Subtract line 2 from line 4

Swap lines 3 and 4.

In this case, the determinant will change sign to the opposite:

Because the determinant is not equal to zero, therefore, the vectors are linearly independent and form a basis.

Let's decompose the vector into vectors of the given basis: , here, ? the desired coordinates of the vector in the basis, . In coordinate form, this is the equation (1, 2, 0, 1) + (0, 1, 2, 3) + (1, 3, 2, 2) + (0, 1, 3, 1) = (1, 0, 1, 5) takes the form:

We solve the system using the Gauss method:

We write the system in the form of an augmented matrix

For the convenience of calculations, we swap the lines:

Multiply the 3rd row by (-1). Let's add the 3rd line to the 2nd. Multiply the 3rd row by 2. Add the 4th row to the 3rd:

Multiply the 1st row by 3. Multiply the 2nd row by (-2). Let's add the 2nd line to the 1st:

Multiply the 2nd row by 5. Multiply the 3rd row by 3. Add the 3rd row to the 2nd:

Multiply the 2nd row by (-2). Let's add the 2nd line to the 1st:

From the 1st line we express? 4

From the 2nd line we express? 3

From the 3rd line we express? 2

Example 8

Vectors are given. Show that the vectors form a basis of three-dimensional space and find the coordinates of the vector in this basis.

Solution: Let's deal with the condition first. By condition, four vectors are given, and, as you can see, they already have coordinates in some basis. What is the basis - we are not interested. And the following thing is of interest: three vectors may well form a new basis. And the first step is completely the same as the solution of Example 6, it is necessary to check if the vectors are really linearly independent:

Calculate the determinant, composed of the coordinates of the vectors :

, hence the vectors are linearly independent and form a basis of a three-dimensional space.

! Important: vector coordinates necessarily write down into columns determinant, not strings. Otherwise, there will be confusion in the further solution algorithm.

Now let's recall the theoretical part: if the vectors form a basis, then any vector can be the only way expand over the given basis: , where are the coordinates of the vector in the basis .

Since our vectors form the basis of a three-dimensional space (this has already been proven), the vector can be expanded in a unique way in this basis:
, where are the coordinates of the vector in the basis .

By condition and it is required to find the coordinates .

For ease of explanation, I'll swap the parts: . In order to find it, this equality should be written coordinate-wise:

On what basis are the coefficients arranged? All coefficients of the left side are exactly transferred from the determinant , in right side vector coordinates are written.

It turned out system of three linear equations in three unknowns. It is usually decided by Cramer's formulas, often even in the condition of the problem there is such a requirement.

The main determinant of the system has already been found:
, so the system has a unique solution.

Next is a matter of technology:

In this way:
is the expansion of the vector in terms of the basis .

Answer:

As I have already noted, the problem is algebraic in nature. The vectors that have been considered are not necessarily those vectors that can be drawn in space, but, first of all, the abstract vectors of the linear algebra course. For the case of two-dimensional vectors, a similar problem can be formulated and solved, the solution will be much simpler. However, in practice, I have never encountered such a task, which is why I skipped it in the previous section.

The same problem with three-dimensional vectors for an independent solution:

Example 9

Vectors are given. Show that the vectors form a basis and find the coordinates of the vector in this basis. Solve the system of linear equations by Cramer's method.

A complete solution and an approximate sample of finishing at the end of the lesson.

Similarly, one can consider four-dimensional, five-dimensional, etc. vector spaces, where the vectors have 4, 5 or more coordinates, respectively. For these vector spaces, there is also the concept of linear dependence, linear independence of vectors, there is a basis, including an orthonormal one, expansion of a vector in terms of a basis. Yes, such spaces cannot be drawn geometrically, but all the rules, properties and theorems of two and three dimensional cases work in them - pure algebra. Actually, I was already forced to talk about philosophical issues in the article Partial derivatives of functions of three variables, which appeared before this lesson.

Love vectors and vectors will love you!

Solutions and answers:

Example 2: Solution: compose a proportion from the corresponding coordinates of the vectors:

Answer: at

Example 4: Proof: Trapeze A quadrilateral is called in which two sides are parallel and the other two sides are not parallel.
1) Check the parallelism of opposite sides and .
Let's find the vectors:


, so these vectors are not collinear and the sides are not parallel.
2) Check the parallelism of opposite sides and .
Let's find the vectors:

Calculate the determinant, composed of the coordinates of the vectors :
, so these vectors are collinear, and .
Output: Two sides of a quadrilateral are parallel, but the other two sides are not parallel, so it is a trapezoid by definition. Q.E.D.

Example 5: Solution:
b) Check if there is a coefficient of proportionality for the corresponding coordinates of the vectors:

The system has no solution, which means the vectors are not collinear.
More simple design:
- the second and third coordinates are not proportional, which means that the vectors are not collinear.
Answer: the vectors are not collinear.
c) We examine the vectors for collinearity . Let's create a system:

The corresponding coordinates of the vectors are proportional, so
This is where the “foppish” design method just doesn’t work.
Answer:

Example 6: Solution: b) Calculate the determinant, composed of the coordinates of the vectors (the determinant is expanded on the first line):

, which means that the vectors are linearly dependent and do not form a basis of a three-dimensional space.
Answer : these vectors do not form a basis

Example 9: Solution: Calculate the determinant, composed of the coordinates of the vectors :


Thus, the vectors are linearly independent and form a basis.
Let's represent the vector as a linear combination of basis vectors:

Coordinate:

We solve the system using Cramer's formulas:
, so the system has a unique solution.

Answer:The vectors form a basis,

Higher mathematics for correspondence students and not only >>>

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Vector product of vectors.
Mixed product of vectors

In this lesson, we will look at two more operations with vectors: cross product of vectors And mixed product of vectors. It's okay, it sometimes happens that for complete happiness, in addition to dot product of vectors, more and more is needed. Such is vector addiction. One may get the impression that we are getting into the jungle of analytic geometry. This is not true. In this section of higher mathematics, there is generally little firewood, except perhaps enough for Pinocchio. In fact, the material is very common and simple - hardly more difficult than the same scalar product, even there will be fewer typical tasks. The main thing in analytic geometry, as many will see or have already seen, is NOT TO MISTAKE CALCULATIONS. Repeat like a spell, and you will be happy =)

If the vectors sparkle somewhere far away, like lightning on the horizon, it doesn't matter, start with the lesson Vectors for dummies to restore or reacquire basic knowledge about vectors. More prepared readers can get acquainted with the information selectively, I tried to collect the most complete collection of examples that are often found in practical work

What will make you happy? When I was little, I could juggle two and even three balls. It worked out well. Now there is no need to juggle at all, since we will consider only space vectors, and flat vectors with two coordinates will be left out. Why? This is how these actions were born - the vector and mixed product of vectors are defined and work in three-dimensional space. Already easier!