Scalar product of vectors (hereinafter referred to as SP). Dear friends! The mathematics exam includes a group of problems on solving vectors. We have already considered some problems. You can see them in the “Vectors” category. In general, the theory of vectors is not complicated, the main thing is to study it consistently. Calculations and operations with vectors in the school mathematics course are simple, the formulas are not complicated. Look into . In this article we will analyze problems on SP of vectors (included in the Unified State Examination). Now “immersion” in the theory:
H To find the coordinates of a vector, you need to subtract from the coordinates of its endthe corresponding coordinates of its origin
And further:
*Vector length (modulus) is determined as follows:
These formulas must be remembered!!!
Let's show the angle between the vectors:
It is clear that it can vary from 0 to 180 0(or in radians from 0 to Pi).
We can draw some conclusions about the sign of the scalar product. The lengths of vectors have a positive value, this is obvious. This means the sign of the scalar product depends on the value of the cosine of the angle between the vectors.
Possible cases:
1. If the angle between the vectors is acute (from 0 0 to 90 0), then the cosine of the angle will have a positive value.
2. If the angle between the vectors is obtuse (from 90 0 to 180 0), then the cosine of the angle will have a negative value.
*At zero degrees, that is, when the vectors have the same direction, the cosine is equal to one and, accordingly, the result will be positive.
At 180 o, that is, when the vectors have opposite directions, the cosine is equal to minus one,and accordingly the result will be negative.
Now the IMPORTANT POINT!
At 90 o, that is, when the vectors are perpendicular to each other, the cosine is equal to zero, and therefore the SP is equal to zero. This fact (consequence, conclusion) is used in solving many problems where we are talking about the relative position of vectors, including in problems included in the open bank of mathematics tasks.
Let us formulate the statement: the scalar product is equal to zero if and only if these vectors lie on perpendicular lines.
So, the formulas for SP vectors:
If the coordinates of the vectors or the coordinates of the points of their beginnings and ends are known, then we can always find the angle between the vectors:
Let's consider the tasks:
27724 Find the scalar product of the vectors a and b.
We can find the scalar product of vectors using one of two formulas:
The angle between the vectors is unknown, but we can easily find the coordinates of the vectors and then use the first formula. Since the beginnings of both vectors coincide with the origin, the coordinates of these vectors are equal to the coordinates of their ends, that is
How to find the coordinates of a vector is described in.
We calculate:
Answer: 40
Let's find the coordinates of the vectors and use the formula:
To find the coordinates of a vector, it is necessary to subtract the corresponding coordinates of its beginning from the coordinates of the end of the vector, which means
We calculate the scalar product:
Answer: 40
Find the angle between vectors a and b. Give your answer in degrees.
Let the coordinates of the vectors have the form:
To find the angle between vectors, we use the formula for the scalar product of vectors:
Cosine of the angle between vectors:
Hence:
The coordinates of these vectors are equal:
Let's substitute them into the formula:
The angle between the vectors is 45 degrees.
Answer: 45
Thus, the length of the vector is calculated as the square root of the sum of the squares of its coordinates 
. The length of an n-dimensional vector is calculated similarly 
. If we remember that each coordinate of a vector is the difference between the coordinates of the end and the beginning, then we obtain the formula for the length of the segment, i.e. Euclidean distance between points.
Scalar product two vectors on a plane is the product of the lengths of these vectors and the cosine of the angle between them: 
. It can be proven that the scalar product of two vectors 
= (x 1, x 2) and 
= (y 1 , y 2) is equal to the sum of the products of the corresponding coordinates of these vectors: 
= x 1 * y 1 + x 2 * y 2 .
In n-dimensional space, the scalar product of vectors X= (x 1, x 2,...,x n) and Y= (y 1, y 2,...,y n) is defined as the sum of the products of their corresponding coordinates: X*Y = x 1 * y 1 + x 2 * y 2 + ... + x n * y n.
The operation of multiplying vectors by each other is similar to multiplying a row matrix by a column matrix. We emphasize that the result will be a number, not a vector.
The scalar product of vectors has the following properties (axioms):
1) Commutative property: X*Y=Y*X.
2) Distributive property with respect to addition: X(Y+Z) =X*Y+X*Z.
3) For any real number 
.
4)
, if X is not a zero vector; 
if X is a zero vector.
A linear vector space in which a scalar product of vectors is given that satisfies the four corresponding axioms is called Euclidean linear vectorspace.
It is easy to see that when we multiply any vector by itself, we get the square of its length. So it's different length a vector can be defined as the square root of its scalar square:.
The vector length has the following properties:
1) |X| = 0Х = 0;
2) |X| = ||*|X|, where is a real number;
3) |X*Y||X|*|Y| ( Cauchy-Bunyakovsky inequality);
4) |X+Y||X|+|Y| ( triangle inequality).
The angle between vectors in n-dimensional space is determined based on the concept of a scalar product. Indeed, if 
, That 
. This fraction is not greater than one (according to the Cauchy-Bunyakovsky inequality), so from here we can find .
The two vectors are called orthogonal or perpendicular, if their scalar product is equal to zero. From the definition of the scalar product it follows that the zero vector is orthogonal to any vector. If both orthogonal vectors are non-zero, then cos= 0, i.e.=/2 = 90 o.
Let's look again at Figure 7.4. It can be seen from the figure that the cosine of the angle of the inclination of the vector to the horizontal axis can be calculated as 
, and the cosine of the angleinclination of the vector to the vertical axis is as 
. These numbers are usually called direction cosines. It is easy to verify that the sum of the squares of the direction cosines is always equal to one: cos 2 +cos 2 = 1. Similarly, the concepts of direction cosines can be introduced for spaces of higher dimensions.
Vector space basis
For vectors, we can define the concepts linear combination,linear dependence And independence similar to how these concepts were introduced for matrix rows. It is also true that if the vectors are linearly dependent, then at least one of them can be expressed linearly in terms of the others (i.e., it is a linear combination of them). The converse is also true: if one of the vectors is a linear combination of the others, then all these vectors together are linearly dependent.
Note that if among the vectors a l , a 2 ,...a m there is a zero vector, then this set of vectors is necessarily linearly dependent. In fact, we get l a l + 2 a 2 +...+ m a m = 0 if, for example, we equate the coefficient j at the zero vector to one, and all other coefficients to zero. In this case, not all coefficients will be equal to zero ( j ≠ 0).
In addition, if some part of the vectors from a set of vectors are linearly dependent, then all of these vectors are linearly dependent. In fact, if some vectors give a zero vector in their linear combination with coefficients that are not both zero, then the remaining vectors multiplied by the zero coefficients can be added to this sum of products, and it will still be a zero vector.
How to determine whether vectors are linearly dependent?
For example, let's take three vectors: a 1 = (1, 0, 1, 5), a 2 = (2, 1, 3, -2) and a 3 = (3, 1, 4, 3). Let's create a matrix from them, in which they will be columns:
Then the question of linear dependence will be reduced to determining the rank of this matrix. If it turns out to be equal to three, then all three columns are linearly independent, and if it turns out to be less, then this will indicate a linear dependence of the vectors.
Since the rank is 2, the vectors are linearly dependent.
Note that the solution to the problem could also begin with reasoning that is based on the definition of linear independence. Namely, create a vector equation l a l + 2 a 2 + 3 a 3 = 0, which will take the form l *(1, 0, 1, 5) + 2 *(2, 1, 3, -2) + 3 *(3, 1, 4, 3) = (0, 0, 0, 0). Then we get a system of equations:
Solving this system using the Gaussian method will be reduced to obtaining the same step matrix, only it will have one more column - free terms. They will all be zero, since linear transformations of zeros cannot lead to a different result. The transformed system of equations will take the form:
The solution to this system will be (-с;-с; с), where с is an arbitrary number; for example, (-1;-1;1). This means that if we take l = -1; 2 =-1 and 3 = 1, then l a l + 2 a 2 + 3 a 3 = 0, i.e. the vectors are actually linearly dependent.
From the solved example it becomes clear that if we take the number of vectors greater than the dimension of space, then they will necessarily be linearly dependent. In fact, if we took five vectors in this example, we would get a 4 x 5 matrix, the rank of which could not be greater than four. Those. the maximum number of linearly independent columns would still not be more than four. Two, three or four four-dimensional vectors can be linearly independent, but five or more cannot. Consequently, no more than two vectors can be linearly independent on the plane. Any three vectors in two-dimensional space are linearly dependent. In three-dimensional space, any four (or more) vectors are always linearly dependent. And so on.
That's why dimension space can be defined as the maximum number of linearly independent vectors that can be in it.
A set of n linearly independent vectors of an n-dimensional space R is called basis this space.
Theorem. Each vector of linear space can be represented as a linear combination of basis vectors, and in a unique way.
Proof. Let the vectors e l , e 2 ,...e n form a basis-dimensional space R. Let us prove that any vector X is a linear combination of these vectors. Since, together with vector X, the number of vectors will become (n +1), these (n +1) vectors will be linearly dependent, i.e. there are numbers l , 2 ,..., n ,, not simultaneously equal to zero, such that
l e l + 2 e 2 +...+ n e n +Х = 0
In this case, 0, because otherwise we would get l e l + 2 e 2 +...+ n e n = 0, where not all coefficients l , 2 ,..., n are equal to zero. This means that the basis vectors would be linearly dependent. Therefore, we can divide both sides of the first equation by:
( l /)e l + ( 2 /)e 2 +...+ ( n /)e n + X = 0
Х = -( l /)e l - ( 2 /)e 2 -...- ( n /)e n
Х = x l e l +x 2 e 2 +...+x n e n,
where x j = -( j /), 
.
Now we prove that such a representation in the form of a linear combination is unique. Let's assume the opposite, i.e. that there is another representation:
Х = y l e l +y 2 e 2 +...+y n e n
Let us subtract from it term by term the previously obtained expression:
0 = (y l – x 1)e l + (y 2 – x 2)e 2 +...+ (y n – x n)e n
Since the basis vectors are linearly independent, we obtain that (y j - x j) = 0, 
, i.e. y j = x j . So the expression turned out to be the same. The theorem has been proven.
The expression X = x l e l +x 2 e 2 +...+x n e n is called decomposition vector X based on e l, e 2,...e n, and numbers x l, x 2,...x n - coordinates vector x relative to this basis, or in this basis.
It can be proved that if nnonzero vectors of an n-dimensional Euclidean space are pairwise orthogonal, then they form a basis. Indeed, let's multiply both sides of the equation l e l + 2 e 2 +...+ n e n = 0 by any vector e i . We get l (e l *е i) + 2 (e 2 *е i) +...+ n (e n *е i) = 0 i (e i *е i) = 0 i = 0 for i.
Vectors e l , e 2 ,...e n of n-dimensional Euclidean space form orthonormal basis, if these vectors are pairwise orthogonal and the norm of each of them is equal to one, i.e. if e i *e j = 0 for i≠j и |е i | = 1 fori.
Theorem (no proof). In every n-dimensional Euclidean space there is an orthonormal basis.
An example of an orthonormal basis is a system of n unit vectors e i , for which the i-th component is equal to one and the remaining components are equal to zero. Each such vector is called ort. For example, the vector vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) form the basis of three-dimensional space.
Angle between vectors
Consider two given vectors $\overrightarrow(a)$ and $\overrightarrow(b)$. Let us subtract the vectors $\overrightarrow(a)=\overrightarrow(OA)$ and $\overrightarrow(b)=\overrightarrow(OB)$ from an arbitrarily chosen point $O$, then the angle $AOB$ is called the angle between the vectors $\overrightarrow( a)$ and $\overrightarrow(b)$ (Fig. 1).
Picture 1.
Note here that if the vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ are codirectional or one of them is the zero vector, then the angle between the vectors is $0^0$.
Notation: $\widehat(\overrightarrow(a),\overrightarrow(b))$
The concept of dot product of vectors
Mathematically, this definition can be written as follows:
The dot product can be zero in two cases:
If one of the vectors is a zero vector (Since then its length is zero).
If the vectors are mutually perpendicular (that is, $cos(90)^0=0$).
Note also that the scalar product is greater than zero if the angle between these vectors is acute (since $(cos \left(\widehat(\overrightarrow(a),\overrightarrow(b))\right)\ ) >0$), and less than zero if the angle between these vectors is obtuse (since $(cos \left(\widehat(\overrightarrow(a),\overrightarrow(b))\right)\ )
Related to the concept of a scalar product is the concept of a scalar square.
Definition 2
The scalar square of a vector $\overrightarrow(a)$ is the scalar product of this vector with itself.
We find that the scalar square is equal to
\[\overrightarrow(a)\overrightarrow(a)=\left|\overrightarrow(a)\right|\left|\overrightarrow(a)\right|(cos 0^0\ )=\left|\overrightarrow(a )\right|\left|\overrightarrow(a)\right|=(\left|\overrightarrow(a)\right|)^2\]
Calculating the dot product from vector coordinates
In addition to the standard way of finding the value of the scalar product, which follows from the definition, there is another way.
Let's consider it.
Let the vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ have coordinates $\left(a_1,b_1\right)$ and $\left(a_2,b_2\right)$, respectively.
Theorem 1
The scalar product of the vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ is equal to the sum of the products of the corresponding coordinates.
Mathematically this can be written as follows
\[\overrightarrow(a)\overrightarrow(b)=a_1a_2+b_1b_2\]
Proof.
The theorem has been proven.
This theorem has several consequences:
Corollary 1: Vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ are perpendicular if and only if $a_1a_2+b_1b_2=0$
Corollary 2: The cosine of the angle between the vectors is equal to $cos\alpha =\frac(a_1a_2+b_1b_2)(\sqrt(a^2_1+b^2_1)\cdot \sqrt(a^2_2+b^2_2))$
Properties of the scalar product of vectors
For any three vectors and a real number $k$ the following is true:
$(\overrightarrow(a))^2\ge 0$
This property follows from the definition of a scalar square (Definition 2).
Travel law:$\overrightarrow(a)\overrightarrow(b)=\overrightarrow(b)\overrightarrow(a)$.
This property follows from the definition of the scalar product (Definition 1).
Distributive law:
$\left(\overrightarrow(a)+\overrightarrow(b)\right)\overrightarrow(c)=\overrightarrow(a)\overrightarrow(c)+\overrightarrow(b)\overrightarrow(c)$. \end(enumerate)
By Theorem 1, we have:
\[\left(\overrightarrow(a)+\overrightarrow(b)\right)\overrightarrow(c)=\left(a_1+a_2\right)a_3+\left(b_1+b_2\right)b_3=a_1a_3+a_2a_3+ b_1b_3+b_2b_3==\overrightarrow(a)\overrightarrow(c)+\overrightarrow(b)\overrightarrow(c)\]
Combination law:$\left(k\overrightarrow(a)\right)\overrightarrow(b)=k(\overrightarrow(a)\overrightarrow(b))$. \end(enumerate)
By Theorem 1, we have:
\[\left(k\overrightarrow(a)\right)\overrightarrow(b)=ka_1a_2+kb_1b_2=k\left(a_1a_2+b_1b_2\right)=k(\overrightarrow(a)\overrightarrow(b))\]
An example of a problem for calculating the scalar product of vectors
Example 1
Find the scalar product of the vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ if $\left|\overrightarrow(a)\right|=3$ and $\left|\overrightarrow(b)\right|= 2$, and the angle between them is equal to $((30)^0,\ 45)^0,\ (90)^0,\ (135)^0$.
Solution.
Using Definition 1, we get
For $(30)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((30)^0\right)\ )=6\cdot \frac(\sqrt(3))(2)=3\sqrt( 3)\]
For $(45)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((45)^0\right)\ )=6\cdot \frac(\sqrt(2))(2)=3\sqrt( 2)\]
For $(90)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((90)^0\right)\ )=6\cdot 0=0\]
For $(135)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((135)^0\right)\ )=6\cdot \left(-\frac(\sqrt(2))(2)\ right)=-3\sqrt(2)\]
Lecture: Vector coordinates; scalar product of vectors; angle between vectors
Vector coordinates
So, as mentioned earlier, a vector is a directed segment that has its own beginning and end. If the beginning and end are represented by certain points, then they have their own coordinates on the plane or in space.
If each point has its own coordinates, then we can get the coordinates of the whole vector.
Let's say we have a vector whose beginning and end have the following designations and coordinates: A(A x ; Ay) and B(B x ; By)
To obtain the coordinates of a given vector, it is necessary to subtract the corresponding coordinates of the beginning from the coordinates of the end of the vector:
To determine the coordinates of a vector in space, use the following formula:
Dot product of vectors
There are two ways to define the concept of a scalar product:
- Geometric method. According to it, the scalar product is equal to the product of the values of these modules and the cosine of the angle between them.
- Algebraic meaning. From the point of view of algebra, the scalar product of two vectors is a certain quantity that is obtained as a result of the sum of the products of the corresponding vectors.
If the vectors are given in space, then you should use a similar formula:
Properties:
- If you multiply two identical vectors scalarly, then their scalar product will not be negative:
- If the scalar product of two identical vectors turns out to be equal to zero, then these vectors are considered zero:
- If a certain vector is multiplied by itself, then the scalar product will be equal to the square of its modulus:
- The scalar product has a communicative property, that is, the scalar product will not change if the vectors are rearranged:
- The scalar product of non-zero vectors can be equal to zero only if the vectors are perpendicular to each other:
- For a scalar product of vectors, the commutative law is valid in the case of multiplying one of the vectors by a number:
- With a scalar product, you can also use the distributive property of multiplication:
Angle between vectors
In the case of a plane problem, the scalar product of vectors a = (a x; a y) and b = (b x; b y) can be found using the following formula:
a b = a x b x + a y b y
Formula for the scalar product of vectors for spatial problems
In the case of a spatial problem, the scalar product of vectors a = (a x; a y; a z) and b = (b x; b y; b z) can be found using the following formula:
a b = a x b x + a y b y + a z b z
Formula for the scalar product of n-dimensional vectors
In the case of an n-dimensional space, the scalar product of vectors a = (a 1; a 2; ...; a n) and b = (b 1; b 2; ...; b n) can be found using the following formula:
a b = a 1 b 1 + a 2 b 2 + ... + a n b n
Properties of the scalar product of vectors
1. The scalar product of a vector with itself is always greater than or equal to zero:
2. The scalar product of a vector with itself is equal to zero if and only if the vector is equal to the zero vector:
a · a = 0<=>a = 0
3. The scalar product of a vector by itself is equal to the square of its modulus:
4. The operation of scalar multiplication is communicative:
5. If the scalar product of two non-zero vectors is equal to zero, then these vectors are orthogonal:
a ≠ 0, b ≠ 0, a b = 0<=>a ┴ b
6. (αa) b = α(a b)
7. The operation of scalar multiplication is distributive:
(a + b) c = a c + b c
Examples of tasks for calculating the scalar product of vectors
Examples of calculating the scalar product of vectors for plane problems
Find the scalar product of the vectors a = (1; 2) and b = (4; 8).
Solution: a · b = 1 · 4 + 2 · 8 = 4 + 16 = 20.
Find the scalar product of vectors a and b if their lengths |a| = 3, |b| = 6, and the angle between the vectors is 60˚.
Solution: a · b = |a| · |b| cos α = 3 · 6 · cos 60˚ = 9.
Find the scalar product of the vectors p = a + 3b and q = 5a - 3 b if their lengths |a| = 3, |b| = 2, and the angle between vectors a and b is 60˚.
Solution:
p q = (a + 3b) (5a - 3b) = 5 a a - 3 a b + 15 b a - 9 b b =
5 |a| 2 + 12 a · b - 9 |b| 2 = 5 3 2 + 12 3 2 cos 60˚ - 9 2 2 = 45 +36 -36 = 45.
An example of calculating the scalar product of vectors for spatial problems
Find the scalar product of the vectors a = (1; 2; -5) and b = (4; 8; 1).
Solution: a · b = 1 · 4 + 2 · 8 + (-5) · 1 = 4 + 16 - 5 = 15.
An example of calculating the dot product for n-dimensional vectors
Find the scalar product of the vectors a = (1; 2; -5; 2) and b = (4; 8; 1; -2).
Solution: a · b = 1 · 4 + 2 · 8 + (-5) · 1 + 2 · (-2) = 4 + 16 - 5 -4 = 11.
13. The cross product of vectors and a vector is called third vector , defined as follows:
2) perpendicular, perpendicular. (1"")
3) the vectors are oriented in the same way as the basis of the entire space (positive or negative).
Designate: .
Physical meaning of the vector product
— moment of force relative to point O; - radius - vector of the point of application of force, then
Moreover, if we move it to point O, then the triple should be oriented as a basis vector.
