Biographies Characteristics Analysis

From an arbitrary point, set aside a vector equal to the given one. Vectors Vectors Historical reference The concept of a vector Equality of vectors Postponing a vector from a given point The sum of two vectors Laws of addition Subtraction

1. Define the equality of geometric vectors.

Two geometric vector are said to be equal if:

they are collinear and unidirectional;

their lengths are the same.

2. Define the sum of vectors and the multiplication of a vector by a number.

The sum a + b of two vectors a and b is the vector c constructed according to the following triangle rule. Let's match the beginning of the vector b with the end of the vector a. Then the sum of these vectors will be the vector c, the beginning of which coincides with the beginning of a, and the end of which coincides with the end of b.

Along with the triangle rule, there is the parallelogram rule. Choosing for vectors a and b common beginning, we build a parallelogram on these vectors. Then the diagonal of the parallelogram coming out of the common origin of the vectors determines their sum.

When multiplying a vector by a number, the direction of the vector does not change, but the length of the vector is multiplied by the number.

3. Give definitions of collinear and coplanar vectors.

Two geometric vectors are said to be collinear if they lie on the same line or on parallel lines.

Three geometric vectors are called coplanar if these vectors lie on lines parallel to some plane.

4. Define linearly dependent and linearly independent system vectors.

Vectors a 1 , … , a n are called linearly dependent if such a set of coefficients α 1 , . . . , α n such that α 1 a 1 + . . . + α n a n = 0 and, moreover, at least one of these coefficients is nonzero.

If the specified set of coefficients does not exist, then the vectors are called linearly independent.

5. Formulate geometric criteria linear dependence 2 and 3 vectors.

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

6. Define the basis and coordinates of a vector.

Basis is the set of such vectors in vector space that any vector of this space can be uniquely represented as a linear combination of vectors from this set - basis vectors.

Vector coordinates are the coefficients of the only possible linear combination of basis vectors in the chosen coordinate system equal to the given vector.

7. Formulate a theorem on the expansion of a vector in terms of a basis.

Any vector of a vector space can be decomposed in its basis and, moreover, in a unique way.

If = (̅

– basis , ̅

= (1, 2, 3) , then there is a set of numbers(

…) such that

̅ + + ̅̅, where (

…) are the coordinates of the vector in the basis.

8. Define the orthogonal scalar projection of a vector onto a direction.

The orthogonal projection of a vector onto the direction of the vector is called scalar Pr = | | cos() , where angle is the angle between the vectors.

9. Define the scalar product of vectors.

The scalar product of two vectors is called the number equal to cos -

product of lengths | | and | | these vectors by the cosine of the angle between them.

10. Formulate the linearity property of the scalar product.

λ(̅ ̅ ).

= ̅ c̅+ ̅ c̅.

11. Write down a formula for calculating the scalar product of two vectors given in an orthonormal basis.

̅ = { , }, ̅ = { , }

̅ ̅ = + +

12. Write down the formula for the cosine of the angle between vectors given in an orthonormal basis.

̅ ̅ cos =̅ |̅|| |

13. Define the right and left triple of vectors.

An ordered triple of non-coplanar vectors a, b, c is called right if the direction of vector a coincides with the direction of vector b by means of the shortest rotation of vector a in the plane of these vectors, which is performed counterclockwise from vector ac. Otherwise (clockwise rotation), this triple is called left.

14. Define the vector product of vectors.

vector art non-collinear vectors ̅ and ̅ are called a vector с̅ that satisfies the following three conditions:

vector c is orthogonal to vectors a and b ;

the length of the vector c is equal to |с̅ | = |̅ | |̅ |sin ϕ, where ϕ is the angle between the vectors ̅ and ̅ ;

the ordered triple of vectors ̅ ,̅ ,с̅ is right.

15. Formulate the property of commutativity (symmetry) of the scalar product and the property of anticommutativity (antisymmetry) of the vector product.

The scalar product is commutative: ̅ ̅ =̅ ̅ .

The vector product is anticommutative: ̅ x̅ =− ̅ x̅ .

16. Formulate the linearity property of the vector product of vectors.

the property of associativity together with multiplication by the number (λ ̅ )×̅ = λ(̅ ×̅ );

distributive property with respect to addition (̅ +̅ )×с̅ =̅×с̅ +̅×с̅ .

The associativity and distributivity properties of a vector product combine, similarly to the case of an inner product, into vector product linearity property

with respect to the first factor. Due to the anticommutativity property of the vector product, the vector product is also linear with respect to the second factor:

̅ ×(λ̅ ) = −(λ̅ )×̅ = −λ(̅ ×̅ ) = λ(̅ ×̅ )

̅ ×(̅ +̅s ) = −(̅ +̅s )×̅ = −(̅ ×̅ +̅s ×̅ ) =̅ ×̅ +̅ ×̅s ​​.

17. Write down the formula for calculating the cross product in the right orthonormal basis.

̅ = { , }, ̅ = { , }.

18. Define the mixed product of vectors.

mixed product three vectors ̅ ,̅ , c̅ is called a number equal to (̅ ×̅ )c̅ - the scalar product of the vector product of the first two vectors and the third vector.

19. Formulate the permutation property (skew-symmetry) of the mixed product.

For a mixed product, cyclic permutation rule:

̅ с̅ = с̅ ̅

= ̅с ̅= − ̅с̅

= − с̅ ̅= − ̅ ̅с.

20. Formulate the linearity property of the mixed product.

The mixed product satisfies the associativity property with respect to

multiplication of vectors by a number: (λ ̅ )с̅

= λ(̅ с̅ ).

The mixed product satisfies the distributive property: (̅̅̅ +̅̅̅ )с̅

= ̅̅̅

̅s + ̅̅̅

with.

These properties of the mixed product are formulated for the first factor. However, using a cyclic permutation, one can prove similar

statements for both the second and third factors, i.e. equalities are true

̅ (λ̅ )̅с = λ(̅ ̅ ̅с ),̅ ̅ (λ̅с ) = λ(̅ ̅ ̅с ),̅ (̅̅̅ 1 +̅̅̅ 2 )̅с =̅ ̅̅̅ 1 ̅с +̅ ̅̅̅ 2 ̅с ,̅ ̅ (̅ 1 +̅ 2 ) =̅ ̅ ̅ 1 +̅ ̅ ̅ 2 ,

and as a result we have the linearity property of the mixed product for each factor.

21. Write down the formula for calculating the mixed product in the right orthonormal basis.

̅ = { , }, ̅ = { , }, ̅= { , }

22. Record general equation planes and the equation “in segments”. Explain geometric sense parameters included in these equations.

The equation Ax + By + Cz + D = 0 is called the general equation of the plane. The coefficients A, B, C for the unknowns in this equation have a clear geometric meaning: the vector n = (A; B; C) is perpendicular to the plane. He's called normal vector planes. It, like the general equation of the plane, is determined up to a (non-zero) numerical factor.

The equation + + = 1 is called plane equation in segments, where a, b, c are

the corresponding coordinates of the points lying on the axes OX, OY and OZ, respectively.

23. Write down the equation of a plane passing through 3 given points.

Let 1 (1 , 1 , 1 ) ,2 (2 , 2 , 2 ), 3 (3 , 3 , 3 ) be given points, and the point M(x, y, z) be a point belonging to the plane formed by points1 , 2 and 3 , then the plane equation has

− 1

− 1

− 1

| 2 −1

2 − 1

2 −1 | = 0

3 − 1

3 − 1

3 − 1

24. Formulate the conditions for parallelism and perpendicularity of two planes.

two planes perpendicular, if their normal vectors are orthogonal.

Two planes are parallel if their normal vectors are collinear.

25. Write a formula for the distance from a point to a plane given by the general equation.

To find the distance from the point 0 (0 , 0 , 0 ) to the plane

: + + + = 0 the formula is used: (,) = | 0 + 0 + 0 + |

√ 2 +2 +2

26. Write down the canonical and parametric equations straight line in space. Explain the geometric meaning of the parameters included in these equations.

The equation ( = 0 + , where (l; m; n) are the coordinates of the directing vector̅ of the straight line L and

(0 ;0 ;

are the coordinates of the point 0 L in a rectangular coordinate system, are called

parametric equations of a straight line in space.

The equation

− 0

− 0

− 0

called canonical equations straight

space.

27. Write down the equation of a straight line passing through two given points in space.

Equations

− 1

− 1

− 1

called the equations of a straight line passing through two points

1 (1 ,1 ,1 ) and 2 (2 ,2 ,2 ).

28. Write down the condition for two straight lines to belong to the same plane.

Let a and b be the direction vectors of these lines, and the points M1 and M2 belong to the lines and l 1 and l 2 , respectively. Then two lines will belong to the same plane if the mixed product (a, b, M1 M2 ) is equal to 0.

29. Write down the formula for the distance from a point to a line in space.

The distance from point 1 to line L can be calculated using the formula:

30. Write down the formula for the distance between skew lines.

The distance between crossing lines 1 and 2 can be calculated by the formula:

belonging to straight lines.

1. Prove the geometric criterion of linear dependence three vectors.

Three vectors are linearly dependent if and only if they are coplanar.

Proof:

If three vectors ̅ ,̅ ,̅ are linearly dependent, then, according to Theorem 2.1 (on the linear dependence of vectors), one of them, for example ̅ , is a linear combination of the others: ̅ = β̅ + γ̅ . Let us combine the beginnings of the vectors ̅ and ̅ at the point A. Then the vectors β̅ , γ̅ will have a common origin at the point A and, according to the parallelogram rule, their sum, i.e. vector̅ , will be a vector with the beginning A and the end, which is the vertex of the parallelogram built on vectors of terms. Thus, all vectors lie in the same plane, i.e. coplanar.

Let the vectors ̅ ,̅ ,̅ be coplanar. If one of these vectors is zero, then it is obvious that it will be a linear combination of the others. It suffices to take all the coefficients of the linear combination equal to zero. Therefore, we can assume that all three vectors are not zero. Let us combine the beginnings of these vectors in common point O. Let their ends be respectively points A, B, C (Fig. 2.1). Through the point C we draw lines parallel to the lines passing through the pairs of points O, A and O, B. Denoting the intersection points by A' and B', we obtain

parallelogram OA'CB', therefore = ′ + ′ . Vector′ and non-zero vector̅

are collinear, and therefore the first of them can be obtained by multiplying the second by

real number α: ′ = . Similarly′ = , β R. As a result, we get, what

̅̅̅̅̅ ̅̅̅̅̅

= ′ + ′ , i.e. vector̅ is a linear combination of vectors̅ and. According to the theorem

̅ are linearly dependent.

2.1 (on the linear dependence of vectors), the vectors ̅ ,

2. Prove the theorem on the expansion of a vector in terms of a basis.

Theorem on the expansion of a vector in terms of a basis. If = (̅

– basis , ̅

= (1, 2, 3), then

there is a set of numbers (

…) such that ̅= ̅̅̅

̅ + + ̅ ̅, where (

…) – coordinates

vectors in the basis.

Proof: (for i = 2)

(̅1 , ̅2 )– basis 2 , ̅2

By definition of the space V2: x, e1, e2 are coplanar => (criterion of linear dependence of 3 vectors) => ̅ ,̅ 1 , ̅ 2 are linearly dependent => 0 , 1 , 2 .

0 ̅+1 ̅1 +2 ̅2 = 0̅ ,0 2 +1 2 +2 2 ≠ 0

1 case: 0 \u003d 0, then 1 ̅ 1 + 2 ̅ 2 = 0 ̅, 1 2 + 2 2 ≠ 0, then 1, 2 are linearly dependent (̅ 1, ̅ 2) - lin. depend. ̅ 1 and ̅ 2 are collinear.

2nd case: 0 ≠ 0

̅= (− 1 ) ̅1 + (−2 ) ̅2 0 0

Proved to exist.

Let there be 2 representations:

̅= 1 ̅1 +2 ̅2

Difference:

0 ̅ = ̅− ̅= 1 ̅ 1 + 2 ̅ 2 − 1 ̅ 1 − 2 ̅ 2 = (1 − 1 )̅ 1 + (2 − 2 )̅ 2 => are linearly dependent, and this contradicts the definition of a basis.

3. Prove the linearity property of the scalar product.

Together with multiplication by a number, the operation of scalar multiplication is associative: (λ̅ )̅ =

λ(̅ ̅ ).

Scalar multiplication and vector addition are related by the distributive property: (̅ +̅ )с̅

= ̅ c̅+ ̅ c̅.

Q.E.D.

4. Derive a formula for calculating the scalar product of vectors given in an orthonormal basis.

Derivation of a formula for calculating the scalar product of vectors given in an orthonormal basis.

Let the vectors ̅ and ̅ from 3 be given by their coordinates in the orthonormal basis, ̅ ,̅ ̅ :̅ = ( ; ; ),̅ = ( ; ; ). This means that there are expansions ̅ =̅ +̅ +̅ ,

̅ =̅ +̅ +̅ . Using them and the properties of the scalar product, we calculate

̅̅ = (̅+ ̅+̅ )(̅+ ̅+̅ )

= ̅ ̅+ ̅ ̅+ ̅̅ + ̅ ̅+ ̅ ̅+ ̅̅ +̅ ̅+̅ ̅ +̅ ̅ =2 ̅+2 ̅+̅ 2 = + + .

The final answer was obtained taking into account the fact that the orthonormality of the basis,̅ ,̅

̅ means the fulfillment of the equalities ̅̅ = ̅ ̅ = ̅ ̅ = 0, 2 ̅= 2 ̅= 2 = 1 . Thus,

̅ ̅ = + +

5. Derive a formula for calculating the cross product of vectors given in the right orthonormal basis.

Derivation of a formula for calculating the cross product of vectors given in an orthonormal basis.

Consider two vectors ̅

and, given by their coordinates in the right orthonormal basis

̅ = {

). Then there are expansions of these vectors ̅ =̅ +̅

, ̅, ̅:

= ̅ +̅ +

Based on these

representations

algebraic

vector multiplication,

we get

= ̅× ̅+ ̅× ̅+ ̅× +

̅× ̅+ ̅× ̅+ ̅× +

̅ ̅

× ̅+ × ̅+

× = (

)̅+ (

)̅+ (

To simplify the resulting formula, we note that it is similar to the formula for expanding the third-order determinant in the 1st row, only vectors are used instead of numerical coefficients. Therefore, you can write this formula as a determinant, which is calculated according to the usual rules. Two lines of this determinant will consist of numbers, and one - of vectors. So, the formula for calculating the vector product in the right orthonormal basis,̅ ,̅ ̅ can be written as:

6. Prove the linearity property of the mixed product.

Using the properties of the mixed product, one can prove the linearity of the vector

products by the first factor:

(̅ + ̅ , ̅)= (̅,)̅+ (̅ ,)̅

For this we find scalar product the vector on the left side of the equality and the unit vector of the standard basis. Given the linearity of the mixed product with respect to the second factor,

we get

those. the abscissa of the vector on the left side of the equality being proved is equal to the abscissa of the vector on its right side. Similarly, we prove that the ordinates, as well as the applicates, of the vectors in both parts of the equality are respectively equal. Therefore, this equal vectors, since their coordinates with respect to the standard basis are the same.

7. Derive a formula for calculating the mixed products of three vectors in the right orthonormal basis.

Derivation of a formula for calculating the mixed product of three vectors in the right orthonormal basis.

Let the vectors a, b, c be given by their coordinates in the right orthonormal basis: ̅ = ( ;

), = ( ; ; ), ̅с = ( ; ; ). To find their mixed product,

we will use the formulas for calculating the scalar and vector products:

̅̅= ̅(× ̅)= ̅ (|

8. Derive a formula for the distance from a point to a plane given by the general equation.

Derivation of a formula for the distance from a point to a plane given by the general equation.

Consider some plane π and an arbitrary point 0 in space. Let's choose

for the plane, a unit normal vector n with origin at some point 1 π , and let ρ(0 ,

since | ̅ | = 1.

If the plane π is given in rectangular system coordinates by its general equation

Ax + By + Cz + D = 0, then its normal vector is the vector with coordinates (A; B; C).

Let (0 , 0 , 0 ) and (1 , 1 , 1 ) be the coordinates of points0

and 1 . Then the equality

A 1 +B1 +C1 +D = 0, since the point M1 belongs to the plane, and one can find the coordinates

̅̅̅̅̅̅̅̅

̅̅̅̅̅̅̅̅

̅̅̅̅̅̅̅̅

Vectors 1 0 :

1 0 = (0 − 1 ; 0 − 1 ; 0 − 1 ) . Writing down the scalar product ̅ 1 0

coordinate form and transforming (5.8), we obtain

| (0 −1 ) + (0 −1 ) + (0 −1 )|

| 0 +0 +0 − (1 +1 +1 )|

2 + 2+ 2

2 + 2+ 2

= |0 +0 +0 + | √2 +2 +2

since 1 + 1 + 1 = − . So, to calculate the distance from a point to a plane, you need to substitute the coordinates of the point into the general equation of the plane, and then divide the absolute value of the result by the normalizing factor, equal to the length corresponding normal vector.

9. Derive a formula for the distance from a point to a straight line in space.

Derivation of a formula for the distance from a point to a line in space.

The distance from the point 1 (1 , 1 , 1 ) to the line L given by the canonical equations L: − 0 = − 0 = − 0 can be calculated using the cross product. Really,

the canonical equations of the line give us the point 0 (0 , 0 , 0 ) on the line

and the direction vector ̅ = (l; m; n) of this line. Let's build a parallelogram on the vectors ̅ and ̅̅̅̅̅̅̅̅ .

Then the distance from point 1 to the line L will be equal to the height h of the parallelogram (Fig. 6.6).

So, the required distance can be calculated by the formula

̅̅̅̅̅̅̅̅

(1 ,) =

| 0 1 × |

10. Derive a formula for the distance between skew lines.

Derivation of a formula for the distance between skew lines.

The distance between skew lines can be found using the mixed

work. Let the lines 1

and 2

canonical equations. Since they

̅̅̅̅̅̅̅̅

intersect, their direction vectors 1 ,2 and the vector 1 2 connecting the points on the lines are non-coplanar. Therefore, a parallelepiped can be built on them (Fig. 6.7).

Then the distance between the lines is equal to the height h of this parallelepiped. In turn, the height of the parallelepiped can be calculated as the ratio of the volume of the parallelepiped to the area of ​​its base. Volume of the box equal to the modulo mixed product of three specified vectors, and the area of ​​the parallelogram at the base of the parallelepiped is equal to the modulus of the vector product of the directing vectors of the lines. As a result, we obtain the formula for the distance

(1 , 2 ) between lines:

̅ ̅̅̅ ̅̅̅̅̅̅̅̅

(1 ,2 ) =

| 1 2

1 2|

Knowledge and skills acquired at this lesson, will be useful to students not only in geometry lessons, but also in classes in other sciences. During the lesson, schoolchildren will learn to postpone the vector from given point. It can be a regular geometry lesson, as well as an extracurricular or optional lesson mathematics. This development will help the teacher save his time preparing for the lesson on the topic "Delaying a vector from a given point." It will be enough for him to play the video lesson in class, and then consolidate the material with his own selection of exercises.

Lesson duration takes only 1:44 minutes. But this is enough to teach schoolchildren to postpone the vector from a given point.

The lesson begins with a demonstration of a vector whose beginning is at some point. They say that the vector is postponed from it. Then the author proposes to prove with him the assertion according to which a vector equal to the given one and, moreover, unique, can be drawn from any point. In the course of the proof, the author considers each case in detail. First, it takes the situation when the given vector is zero, and secondly, when the vector is non-zero. During the proof, illustrations are used in the form of drawings and constructions, mathematical notation, which form mathematical literacy among schoolchildren. The author talks slowly, which allows students to take notes in parallel while commenting. The construction that the author led in the course of proving the previously formulated statement shows how a vector equal to the given one can be constructed from a certain point.

If students watch the lesson carefully and take notes at the same time, they will easily learn the material. Moreover, the author tells in detail, measuredly and quite fully. If for some reason you didn’t hear something, you can go back and watch the lesson again.

After watching the video tutorial, it is advisable to start fixing the material. The teacher is recommended to choose tasks on this topic in order to work out the skill of postponing the vector from a given point.

This lesson can be used to self-study topics for schoolchildren. But to consolidate, you need to contact the teacher so that he selects the appropriate tasks. Indeed, without consolidating the material, it is difficult to achieve a positive result in training.

ov, first you need to understand such a concept as postponing a vector from a given point.

Definition 1

If the point $A$ is the beginning of some vector $\overrightarrow(a)$, then the vector $\overrightarrow(a)$ is said to be separated from the point $A$ (Fig. 1).

Figure 1. $\overrightarrow(a)$ plotted from point $A$

We introduce the following theorem:

Theorem 1

From any point $K$ one can draw a vector $\overrightarrow(a)$ and only one.

Proof.

Existence: There are two cases to consider here:

    The vector $\overrightarrow(a)$ is zero.

    In this case, it is obvious that the desired vector is the vector $\overrightarrow(KK)$.

    The vector $\overrightarrow(a)$ is nonzero.

    Let the point $A$ denote the beginning of the vector $\overrightarrow(a)$, and the point $B$ denote the end of the vector $\overrightarrow(a)$. Let us draw a line $b$ parallel to the vector $\overrightarrow(a)$ through the point $K$. Let us plot the segments $\left|KL\right|=|AB|$ and $\left|KM\right|=|AB|$ on this straight line. Consider the vectors $\overrightarrow(KL)$ and $\overrightarrow(KM)$. Of these two vectors, the desired one will be the one that will be co-directed with the $\overrightarrow(a)$ vector (Fig. 2)

Figure 2. Illustration of Theorem 1

Uniqueness: uniqueness immediately follows from the construction carried out in the subsection "existence".

The theorem has been proven.

Subtraction of vectors. Rule One

Let us be given vectors $\overrightarrow(a)$ and $\overrightarrow(b)$.

Definition 2

The difference of two vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ is a vector $\overrightarrow(c)$ which, when added to the vector $\overrightarrow(b)$, gives the vector $\overrightarrow(a)$ , i.e

\[\overrightarrow(b)+\overrightarrow(c)=\overrightarrow(a)\]

Designation:$\overrightarrow(a)-\overrightarrow(b)=\overrightarrow(c)$.

We will consider the construction of the difference of two vectors using the problem.

Example 1

Let vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ be given. Construct the vector $\overrightarrow(a)-\overrightarrow(b)$.

Decision.

Let us construct an arbitrary point $O$ and plot the vectors $\overrightarrow(OA)=\overrightarrow(a)$ and $\overrightarrow(OB)=\overrightarrow(b)$ from it. Connecting the point $B$ with the point $A$, we get the vector $\overrightarrow(BA)$ (Fig. 3).

Figure 3. Difference of two vectors

According to the triangle rule for constructing the sum of two vectors, we see that

\[\overrightarrow(OB)+\overrightarrow(BA)=\overrightarrow(OA)\]

\[\overrightarrow(b)+\overrightarrow(BA)=\overrightarrow(a)\]

From Definition 2, we get that

\[\overrightarrow(a)-\overrightarrow(b)=\overrightarrow(BA)\]

Answer:$\overrightarrow(a)-\overrightarrow(b)=\overrightarrow(BA)$.

From this problem, we obtain the following rule for finding the difference of two vectors. To find the difference $\overrightarrow(a)-\overrightarrow(b)$, from an arbitrary point $O$ you need to set aside the vectors $\overrightarrow(OA)=\overrightarrow(a)$ and $\overrightarrow(OB)=\overrightarrow(b )$ and connect the end of the second vector with the end of the first vector.

Subtraction of vectors. Rule Two

Recall the following notion we need.

Definition 3

The vector $\overrightarrow(a_1)$ is called arbitrary for the vector $\overrightarrow(a)$ if these vectors are oppositely directed and have the same length.

Designation: The vector $(-\overrightarrow(a))$ is the opposite of the vector $\overrightarrow(a)$.

In order to introduce the second rule for the difference of two vectors, we must first introduce and prove the following theorem.

Theorem 2

For any two vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ the following equality holds:

\[\overrightarrow(a)-\overrightarrow(b)=\overrightarrow(a)+(-\overrightarrow(b))\]

Proof.

By Definition 2, we have

Add to both parts the vector $\left(-\overrightarrow(b)\right)$, we get

Since the vectors $\overrightarrow(b)$ and $\left(-\overrightarrow(b)\right)$ are opposite, then $\overrightarrow(b)+\left(-\overrightarrow(b)\right)=\overrightarrow (0)$. We have

The theorem has been proven.

From this theorem, we obtain the following rule for the difference of two vectors: To find the difference $\overrightarrow(a)-\overrightarrow(b)$, we need to postpone the vector $\overrightarrow(OA)=\overrightarrow(a)$ from an arbitrary point $O$, then from the obtained point $A$ set aside the vector $\overrightarrow(AB)=-\overrightarrow(b)$ and connect the beginning of the first vector with the end of the second vector.

An example of a problem on the concept of the difference of vectors

Example 2

Let $ADCD$ be a parallelogram whose diagonals intersect at $O$. $\overrightarrow(AB)=\overrightarrow(a)$, $\overrightarrow(AD)=\overrightarrow(b)$ (Fig. 4). Express the following vectors in terms of $\overrightarrow(a)$ and $\overrightarrow(b)$:

a) $\overrightarrow(DC)+\overrightarrow(CB)$

b) $\overrightarrow(BO)-\overrightarrow(OC)$

Figure 4. Parallelogram

Decision.

a) We add according to the triangle rule, we get

\[\overrightarrow(DC)+\overrightarrow(CB)=\overrightarrow(DB)\]

From the first rule for the difference of two vectors, we obtain

\[\overrightarrow(DB)=\overrightarrow(a)-\overrightarrow(b)\]

b) Since $\overrightarrow(OC)=\overrightarrow(AO)$, we get

\[\overrightarrow(BO)-\overrightarrow(OC)=\overrightarrow(BO)-\overrightarrow(AO)\]

By Theorem 2, we have

\[\overrightarrow(BO)-\overrightarrow(AO)=\overrightarrow(BO)+\left(-\overrightarrow(AO)\right)=\overrightarrow(BO)+\overrightarrow(OA)\]

Using the triangle rule, we finally have

\[\overrightarrow(BO)+\overrightarrow(OA)=\overrightarrow(BA)=-\overrightarrow(AB)=-\overrightarrow(a)\]

A vector is a directed segment of a straight line in Euclidean space, in which one end (point A) is called the beginning of the vector, and the other end (point B) is called the end of the vector (Fig. 1). Vectors are denoted:

If the beginning and end of the vector are the same, then the vector is called zero vector and denoted 0 .

Example. Let the beginning of the vector in two-dimensional space have coordinates A(12,6) , and the end of the vector is the coordinates B(12.6). Then the vector is a null vector.

Cut length AB called module (long, the norm) vector and is denoted by | a|. length vector, equal to one, is called unit vector . In addition to the modulus, a vector is characterized by a direction: a vector has a direction from A to B. A vector is called a vector, opposite vector .

The two vectors are called collinear if they lie on the same line or on parallel lines. In Fig. 3 red vectors are collinear since they lie on the same straight line, and the blue vectors are collinear, because they lie on parallel lines. Two collinear vectors called equally directed if their ends lie on the same side of the line joining their beginnings. Two collinear vectors are called opposite directions if their ends lie along different sides from the straight line connecting them. If two collinear vectors lie on the same line, then they are said to be equally directed if one of the rays formed by one vector completely contains the ray formed by the other vector. Otherwise, the vectors are called oppositely directed. In Figure 3, the blue vectors are in the same direction and the red vectors are in the opposite direction.

The two vectors are called equal if they have equal modules and are equally directed. In Fig.2, the vectors are equal because their moduli are equal and have the same direction.

The vectors are called coplanar if they lie on the same plane or in parallel planes.

AT n In a dimensional vector space, consider the set of all vectors whose starting point coincides with the origin. Then the vector can be written in the following form:

(1)

where x 1 , x 2 , ..., x n vector end point coordinates x.

The vector written in the form (1) is called row vector, and the vector written as

(2)

called column vector.

Number n called dimension (in order) vector. If a then the vector is called zero vector(because the starting point of the vector ). Two vectors x and y are equal if and only if their corresponding elements are equal.

Vector is one of the basic geometric concepts. A vector is characterized by a number (length) and a direction. Visually, it can be imagined as a directed segment, although, speaking of a vector, it is more correct to mean a whole class of directed segments, which are all parallel to each other, have the same length and the same direction (Fig. 1). Examples of physical quantities that are vector in nature are the speed (of a progressively moving body), acceleration, force, etc.

The concept of a vector appeared in the works of the German mathematician of the 19th century. G. Grassmann and Irish mathematician W. Hamilton; then it was readily accepted by many mathematicians and physicists. In modern mathematics and its applications, this concept plays essential role. Vectors are used in classical Galileo-Newton mechanics (in its modern presentation), in the theory of relativity, quantum physics, in mathematical economics and many other branches of natural science, not to mention the use of vectors in various areas of mathematics.

Each of the directed segments that make up the vector (Fig. 1) can be called a representative of this vector. A vector whose representative is a directed segment going from point to point is denoted by . On fig. 1 we have , i.e. and is the same vector (represented by both directed segments highlighted in Fig. 1). Sometimes a vector is denoted by a small letter with an arrow: , .

A vector represented by a directed "segment" whose beginning and end coincide is called zero; it is denoted by , i.e. . Two parallel vector having the same length but opposite directions are called opposite. If a vector is denoted by , then the vector opposite to it is denoted by .

Let's name the main operations related to vectors.

I. Postponing a vector from a point. Let be some vector and be a point. Among the directed segments that are representatives of the vector , there is a directed segment starting at the point . The end of this directed segment is called a point, resulting from the postponement of the vector from the point (Fig. 2). This operation has the following property:

I1. For any point and any vector, there exists, and only one, point for which .

Addition of vectors. Let and be two vectors. Let's take an arbitrary point and set aside the vector from the point , i.e. find a point such that (Fig. 3). Then we set aside the vector from the point, i.e., we find a point such that . The vector is called the sum of the vectors and and is denoted by . It can be proved that the sum does not depend on the choice of the point , i.e. if we replace with another point , then we get a vector equal to (Fig. 3). From the definition of the sum of vectors it follows that for any three points the equality

I2:

(rule of three points). If non-zero vectors and are not parallel, then it is convenient to find their sum using the parallelogram rule (Fig. 4).

II. The main properties of the sum of vectors express the following 4 equalities (valid for any vectors , , ):

II2. .

Note also that the sum of several vectors is found by successively finding the sum of two of them. For example: .

At the same time, in whatever order we add given vectors, the result (as follows from the properties named in items II1 and II2) will always be the same. For example:

Further, geometrically, the sum of several vectors can be obtained as follows: it is necessary to put the directed segments, which are representatives of these vectors, sequentially one after another (i.e., so that the beginning of the second directed segment coincides with the end of the first, the beginning of the third - with the end of the second and etc.); then the vector will have as its representative a “closing” directed segment, going from the beginning of the first to the end of the last (Fig. 5). (Note that if such successive postponement results in a “closed vector broken line”, then .)

III. Multiplying a vector by a number. Let be a non-zero vector and be a non-zero number. A vector is denoted by the following two conditions: a) the length of the vector is ; b) the vector is parallel to the vector , and its direction coincides with the direction of the vector at and opposite to it at (Fig. 6). If at least one of the equalities , is true, then the product is considered equal to . Thus, the product is defined for any vector and any number.

The following 4 equalities (valid for any vectors and any numbers) express the basic properties of the operation of multiplying a vector by a number:

III2. .

III3. .

From these properties follows a series further facts associated with the considered operations on vectors. Let us note some of them, which are often used in solving problems.

a) If is such a point of the segment that , then for any point the equality , in particular, if is the midpoint of the segment , then .

b) If - the point of intersection of the medians of the triangle, then ; moreover, for any point the equality (the inverse theorems are also valid).

c) Let be a point of a straight line and be a non-zero vector parallel to this straight line. A point belongs to the line if and only if (where is a number).

d) Let be a point of the plane and , be non-zero and non-parallel vectors parallel to this plane. A point belongs to the plane if and only if the vector is expressed in terms of and , i.e. .

Finally, we also note the property of dimension, which expresses the fact that space is three-dimensional.

IV. There are three vectors , , , in space such that none of them can be expressed in terms of the other two; any fourth vector is expressed in terms of these three vectors: . is defined by the equality: the scalar product of the vector is denoted (and then the angle between them is not defined).

The properties of vector operations listed above are in many ways similar to the properties of addition and multiplication of numbers. At the same time, a vector is a geometric object, and in the definition of vector operations, such geometric concepts like length and angle; this explains the usefulness of vectors for geometry (and its applications to physics and other fields of knowledge). However, in order to solve geometric problems with the help of vectors, it is necessary first of all to learn how to "translate" the condition of a geometric problem into a vector "language". After such a “translation”, algebraic calculations with vectors are carried out, and then the resulting vector solution is again “translated” into a geometric “language”. This is the vector solution of geometric problems.

When presenting a geometry course at school, a vector is given as a defined concept (see Definition), and therefore the axiomatics adopted in a school textbook (see Axiomatics and the axiomatic method) of geometry does not say anything about the properties of vectors, i.e. all these properties must be proved as theorems.

There is, however, another way of presenting geometry, in which the vector and point are considered to be the initial (undefined) concepts, and the properties I1, I2, II1-II4, III1-III4, IV, V1-V4 noted above are taken as axioms. This way of constructing geometry was proposed in 1917 by the German mathematician G. Weil. Here lines and planes are defined concepts. The advantage of such a construction is its brevity and organic connection with a modern understanding of geometry, both in mathematics itself and in other areas of knowledge. In particular, axioms II1-II4, III1-III4 introduce the so-called vector space used in modern mathematics, physics, mathematical economics, etc.