Linear dependence and linear independence of vectors.
The basis of vectors. Affine system coordinates

There is a cart with chocolates in the audience, and each visitor today 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 coexist in one wrapper. Pause, eat Twix! ... damn it, well, and argued nonsense. Although, okay, I will not 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 not always the "ordinary" vector that we can depict on a plane or in space. You don't have to go far for proof, try to draw 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 with a vector. Breath of Autumn….

No, I'm not going to load 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 geometric examples will be given. Thus, everything is simple, accessible and clear. In addition to the tasks of analytical 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, whoever likes what). The task will consist of the following actions:

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

2) Based on the selected basis set coordinate system(coordinate grid) to assign coordinates to all objects 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 countertop so that it looks into the monitor. This will be a vector. Now put little finger right hand on the edge of the table in the same way - so that it is directed towards the monitor screen. This will be a vector. Smile, you look great! What about vectors? Data vectors collinear, which means linearly expressed through each other:
, well, or vice versa:, where is some nonzero number.

A picture of this action can be seen in the lesson Vectors for dummies where I explained the rule of multiplying a vector by a number.

Will your fingers set a baseline on the plane of the computer desk? Obviously not. Collinear vectors travel back and forth along one direction, and the 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 degrees, 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 dependent if and only if they are not collinear... So, the basis is obtained. There is no need to be embarrassed that the basis turned out to be "oblique" with non-perpendicular vectors of different lengths. We will see very soon that not only an angle of 90 degrees is suitable for its construction, and not only unit vectors of equal length

Any vector plane unique way decomposed on the basis:
, where are real numbers. The numbers are called vector coordinates in this basis.

It is also said that vectorpresented in the form linear combination basis vectors... That is, the expression is called decomposition of the vectoron the basis or linear combination basis vectors.

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

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

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

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

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

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

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

origin, and orthonormal basis is given cartesian rectangular plane coordinate system ... That is, the rectangular coordinate system unequivocally 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 not always) drawn.

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

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

Such a basis is called orthogonal... The origin of coordinates with vectors sets the coordinate grid, and any point of the plane, any vector have their coordinates in this basis. For example, or. An obvious inconvenience is that the coordinate vectors in general have different lengths other than one. 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, the units along the axes are CONDITIONAL... For example, one unit along the abscissa contains 4 cm, and one unit along the ordinate is 2 cm. This information is enough to convert "non-standard" coordinates into "our usual centimeters" if necessary.

And the second question, which has actually been answered, is whether the angle between the basis vectors must necessarily be 90 degrees? No! As the definition says, the basis vectors must be only non-collinear... Accordingly, the angle can be any other than 0 and 180 degrees.

The point of the plane called origin, and non-collinear vectors, , set affine plane coordinate system :

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

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 associated with dot product of vectors... But the rules for adding vectors and multiplying a vector by a number, formulas for dividing a segment in this respect, as well as some other types of problems, which we will soon consider, are true.

And the conclusion is that the most convenient particular case of the affine coordinate system is the Cartesian rectangular system. Therefore, her, dear, most often you have to contemplate. ... However, everything in this life is relative - there are many situations in which it is appropriate to oblique (or some other, for example, polar) coordinate system. And humanoids may like such systems =)

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

How to determine collinearity of plane vectors?

Typical thing. In order for the two vectors of the plane are collinear, it is necessary and sufficient that their corresponding coordinates are proportional to Essentially, this is a coordinate-wise detailing of the obvious relationship.

Example 1

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

Solution:
a) Let us find out whether there exists for vectors proportionality coefficient, such that the equalities are fulfilled:

I will definitely tell you about the "dude" version of the application of this rule, which is quite effective in practice. The idea is to figure out the proportion right away and see if it is correct:

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

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

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

For self-test, you can use the fact that collinear vectors are linearly expressed through each other. In this case, the equalities hold ... Their validity is easily verified through elementary actions with vectors:

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

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

Output: vectors are linearly independent and form a basis.

A simplified version of the solution looks like this:

Let's compose the proportion from the corresponding coordinates of the vectors :
, which means that these vectors are linearly independent and form a basis.

Usually this option is not rejected by reviewers, but a problem arises in cases where some coordinates are equal to zero. Like this: ... Or like this: ... Or like this: ... How to act here through proportion? (indeed, you cannot divide by zero). It is for this reason that I called the simplified solution "dude".

Answer: a), b) form.

A small creative example for independent decision:

Example 2

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

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

There is an elegant algebraic way of checking vectors for collinearity., We systematize our knowledge and add it as the fifth point:

For two vectors of the plane, the following statements are equivalent:

2) vectors form a basis;
3) 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) 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 terms and statements encountered.

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

We will solve Example 1 in the second way:

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

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

Answer: a), b) form.

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

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

Example 3

The vertices of the quadrangle are given. Prove that a quadrilateral is a parallelogram.

Proof: There is no need to build a drawing in the problem, since the solution will be purely analytical. Let's remember the definition of a parallelogram:
Parallelogram called a quadrilateral, 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 vectors:

2) Find vectors:

The result is the same vector ("according to school" - equal vectors). The collinearity is quite obvious, but the decision is still better to draw up properly, with arrangement. Let us calculate the determinant composed of the coordinates of the vectors:
, therefore, these vectors are collinear, and.

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

More good and different shapes:

Example 4

The vertices of the quadrangle are given. Prove that the quadrilateral is a trapezoid.

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

This is an independent task. Complete solution at the end of the lesson.

And now it's time to quietly move from plane to space:

How to determine collinearity of space vectors?

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

Example 5

Find out if the following space vectors are collinear:

a) ;
b)
v)

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

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

"Simplified" is drawn up by checking the proportion. In this case:
- the corresponding coordinates are not proportional, which means that the vectors are not collinear.

Answer: vectors are not collinear.

b-c) These are items for independent decision. Try to design it in two ways.

There is a method for checking spatial vectors for collinearity and through a third-order determinant, this way highlighted in article Vector product of vectors.

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

Welcome to the second section:

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

Many of the patterns that we have considered on the plane will also be valid for space. I tried to minimize the abstract on 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 explore three-dimensional space. First, let's create its basis. Someone is now in the room, someone is on the street, but in any case, we cannot get away from three dimensions: width, length and height. Therefore, to build the basis, three space vectors are required. One or two vectors are not enough, the fourth is superfluous.

And again we warm up on our fingers. Please raise your hand up and spread it apart. thumb, forefinger and middle finger... These will be vectors, they look in different directions, have different lengths and have different angles between themselves. Congratulations, your 3D baseline is ready! By the way, there is no need to demonstrate this to teachers, no matter how you twist your fingers, and you can't get away from the definitions =)

Next, let's ask an important question, do any three vectors form a basis of three-dimensional space? Please press three fingers firmly against the table top of the computer desk. What happened? Three vectors are located in the same plane, and, roughly speaking, one of our measurements has disappeared - height. Such vectors are coplanar and it is quite obvious that the basis of the 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 do not do this with your fingers, so only Salvador Dali came off =)).

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

Three coplanar vectors are always linearly dependent, that is, they are linearly expressed through each other. For simplicity, let's again imagine that they lie in the same plane. Firstly, the vectors are not only coplanar, but in addition they can be collinear, then any vector can be expressed in terms of 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 - it 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 three-dimensional space.

Definition: The basis of three-dimensional space is a triple of linearly independent (non-coplanar) vectors, taken in a certain order, and any vector of space unique way decomposed according to the given basis, where are the coordinates of the vector in the given basis

Let me remind you that we can also say that the vector is represented in the form 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 unequivocally determine the coordinates of any vector and coordinates of any point in space. Similarly to the plane, some formulas, which I have already mentioned, will not work in the affine coordinate system of space.

The most familiar and convenient special case of the affine coordinate system, as everyone guesses, is rectangular space coordinate system:

The point in space called origin, and orthonormal basis is given cartesian rectangular coordinate system of space ... Familiar picture:

Before moving on to the practical tasks, we re-organize the information:

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

Opposite statements, I think, are understandable.

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

Three vectors of space coplanar if and only if the determinant composed of the coordinates of these 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 profitable for solving some practical problems.

For those readers who have a little bit forgotten the methods of calculating determinants, and maybe even poorly guided by them, I recommend one of my oldest lessons: How to calculate the determinant?

Example 6

Check if the following vectors form the basis of the 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):

, hence, vectors are linearly independent (not coplanar) and form a basis of three-dimensional space.

Answer: these vectors form a basis

b) This is a point for an independent decision. Complete solution and answer at the end of the tutorial.

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 these vectors is zero:

Essentially, you need to solve an equation with a determinant. We put on zeros like kites on jerboas - it is most profitable to open the determinant on 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 in the original determinant and make sure that by revealing it again.

In conclusion, we will consider another typical problem, which is more algebraic in nature and is traditionally included in the course of linear algebra. It is so widespread that it deserves a separate topic:

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

Example 8

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

Solution: First, we deal with the condition. By condition, four vectors are given, and, as you can see, they already have coordinates in some basis. We are not interested in what basis it is. And the following thing is of interest: three vectors may well form a new basis. And the first stage completely coincides with the solution of Example 6, you need to check if the vectors are really linearly independent:

Let us calculate the determinant composed of the coordinates of the vectors:

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

! Important : coordinates of vectors necessarily write down into columns determinant rather than into strings. Otherwise, there will be confusion in the further solution algorithm.

Tasks for control work

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

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

This task has 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.

We calculate the determinant of the matrix:

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

The determinant of the matrix is ​​∆ = 37

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

X = α 1 ε 1 + α 2 ε 2 + α 3 ε 3

Let's 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 get 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 Gaussian method or Cramer's method.

X = ε 1 + 2ε 2 -ε 3

The decision was received and formalized using the service:

Vector coordinates in basis

Together with this task, they also solve:

Solving matrix equations

Cramer's method

Gauss method

Matrix inverse by the Jordan-Gauss method

Inverse matrix in terms of algebraic complements

Online matrix multiplication

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 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 row 3 row 1 multiplied by (-1)

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

Let's swap lines 3 and 4.

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

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

Let's expand the vector in terms of vectors of the given basis:, here,? the required 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 Gaussian method:

We write the system in the form of an extended matrix

For the convenience of calculations, let's swap the lines:

Multiply the 3rd row by (-1). Let's add the 3rd row 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

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

Solution: First, we deal with the condition. By condition, four vectors are given, and, as you can see, they already have coordinates in some basis. We are not interested in what basis it is. And the following thing is of interest: three vectors may well form a new basis. And the first stage completely coincides with the solution of Example 6, you need to check if the vectors are really linearly independent:

Let us calculate the determinant composed of the coordinates of the vectors:

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

! Important: coordinates of vectors necessarily write down into columns determinant rather than into strings. Otherwise, there will be confusion in the further solution algorithm.

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

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

According to the condition, it is required to find the coordinates.

For ease of explanation, I will swap the parts: ... In order to find, this equality should be written coordinatewise:

How are the coefficients arranged? All coefficients on the left side are exactly transferred from the determinant , v right side the coordinates of the vector are recorded.

It turned out system of three linear equations in three unknowns. Usually it is solved by Cramer's formulas, there is often such a requirement even in the problem statement.

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

Further - a matter of technology:

Thus:
- expansion of the vector in terms of the basis.

Answer:

As I have already noted, the problem is algebraic in nature. The vectors that were considered are not necessarily vectors that can be drawn in space, but, first of all, abstract vectors of the linear algebra course. For the case of two-dimensional vectors, you can formulate and solve a similar problem, 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

Given vectors. 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.

Complete solution and a rough sample of finishing at the end of the tutorial.

Similarly, you can consider four-dimensional, five-dimensional, etc. vector spaces, where vectors have 4, 5 and more coordinates, respectively. For these vector spaces, there is also the concept of linear dependence, linear independence of vectors, there is a basis, including orthonormal, decomposition of a vector in terms of the 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 trying to talk about philosophical issues in the article Partial derivatives of functions of three variables which appeared before this tutorial.

Love vectors and vectors will love you!

Solutions and Answers:

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

Answer: at

Example 4: Proof: Trapezoid called a quadrilateral, in which two sides are parallel, and the other two sides are not parallel.
1) Let's check the parallelism of the opposite sides and.
Find vectors:


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

Let us calculate the determinant composed of the coordinates of the vectors:
, therefore, these vectors are collinear, and.
Output: Two sides of a quadrilateral are parallel, and the other two sides are not parallel, which means that it is a trapezoid by definition. Q.E.D.

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

The system has no solution, so the vectors are not collinear.
Simpler design:
- the second and third coordinates are not proportional, which means that the vectors are not collinear.
Answer: vectors are not collinear.
c) Let us examine vectors for collinearity ... Let's compose the system:

The corresponding coordinates of the vectors are proportional, which means
This is where the "dude" method of registration does not work.
Answer:

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

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

Example 9: Solution: Let us calculate the determinant composed of the coordinates of the vectors:


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

Coordinate-wise:

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

Answer:The vectors form a basis,

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Vector product of vectors.
Mixed product of vectors

In this lesson, we'll look at two more vector operations: vector 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, it takes more and more. Such is the vector addiction. One might 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 not enough firewood at all, except that there is enough for Buratino. In fact, the material is very common and simple - hardly more complicated than the same scalar product, even the typical tasks will be smaller. The main thing in analytic geometry, as many will be convinced or have already been convinced, is NOT TO BE MISTAKE IN THE CALCULATIONS. Repeat as a spell, and you will be happy =)

If vectors sparkle somewhere far away, like lightning on the horizon, it doesn't matter, start with the lesson Vectors for dummies to recover or regain basic knowledge of 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

How to please you right away? When I was little, I knew how to juggle with two or even three balls. Dexterously it turned out. Now you won't have to juggle at all, since we will consider only spatial vectors, and plane 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. It's already easier!