If in the problem both the lengths of the vectors and the angle between them are presented “on a silver platter,” then the condition of the problem and its solution look like this:
Example 1. Vectors are given. Find the scalar product of vectors if their lengths and the angle between them are represented by the following values:
Another definition is also valid, completely equivalent to definition 1.
Definition 2. The scalar product of vectors is a number (scalar) equal to the product of the length of one of these vectors and the projection of another vector onto the axis determined by the first of these vectors. Formula according to definition 2:
We will solve the problem using this formula after the next important theoretical point.
Definition of the scalar product of vectors in terms of coordinates
The same number can be obtained if the vectors being multiplied are given their coordinates.
Definition 3. The dot product of vectors is a number equal to the sum of the pairwise products of their corresponding coordinates.
On surface
If two vectors and on the plane are defined by their two Cartesian rectangular coordinates
then the scalar product of these vectors is equal to the sum of pairwise products of their corresponding coordinates:
.
Example 2. Find the numerical value of the projection of the vector onto the axis parallel to the vector.
Solution. We find the scalar product of vectors by adding the pairwise products of their coordinates:
Now we need to equate the resulting scalar product to the product of the length of the vector and the projection of the vector onto an axis parallel to the vector (in accordance with the formula).
We find the length of the vector as the square root of the sum of the squares of its coordinates:
.
We create an equation and solve it:
Answer. The required numerical value is minus 8.
In space
If two vectors and in space are defined by their three Cartesian rectangular coordinates
,
then the scalar product of these vectors is also equal to the sum of pairwise products of their corresponding coordinates, only there are already three coordinates:
.
The task of finding the scalar product using the considered method is after analyzing the properties of the scalar product. Because in the problem you will need to determine what angle the multiplied vectors form.
Properties of the scalar product of vectors
Algebraic properties
1. (commutative property: reversing the places of the multiplied vectors does not change the value of their scalar product).
2. 
(associative property with respect to a numerical factor: the scalar product of a vector multiplied by a certain factor and another vector is equal to the scalar product of these vectors multiplied by the same factor).
3. 
(distributive property relative to the sum of vectors: the scalar product of the sum of two vectors by the third vector is equal to the sum of the scalar products of the first vector by the third vector and the second vector by the third vector).
4. (scalar square of vector greater than zero), if is a nonzero vector, and , if is a zero vector.
Geometric properties
In the definitions of the operation under study, we have already touched on the concept of an angle between two vectors. It's time to clarify this concept.
In the figure above you can see two vectors that are brought to a common origin. And the first thing you need to pay attention to is that there are two angles between these vectors - φ 1 And φ 2 . Which of these angles appears in the definitions and properties of the scalar product of vectors? The sum of the considered angles is 2 π and therefore the cosines of these angles are equal. The definition of a dot product includes only the cosine of the angle, and not the value of its expression. But the properties only consider one angle. And this is the one of the two angles that does not exceed π , that is, 180 degrees. In the figure this angle is indicated as φ 1 .
1. Two vectors are called orthogonal And the angle between these vectors is straight (90 degrees or π /2 ), if the scalar product of these vectors is zero :
.
Orthogonality in vector algebra is the perpendicularity of two vectors.
2. Two non-zero vectors make up sharp corner (from 0 to 90 degrees, or, which is the same - less π dot product is positive .
3. Two non-zero vectors make up obtuse angle (from 90 to 180 degrees, or, what is the same - more π /2) if and only if they dot product is negative .
Example 3. The coordinates are given by the vectors:
.
Calculate the scalar products of all pairs of given vectors. What angle (acute, right, obtuse) do these pairs of vectors form?
Solution. We will calculate by adding the products of the corresponding coordinates.
We got a negative number, so the vectors form an obtuse angle.
We got a positive number, so the vectors form an acute angle.
We got zero, so the vectors form a right angle.
We got a positive number, so the vectors form an acute angle.
.
We got a positive number, so the vectors form an acute angle.
For self-test you can use online calculator Dot product of vectors and cosine of the angle between them .
Example 4. Given the lengths of two vectors and the angle between them:
.
Determine at what value of the number the vectors and are orthogonal (perpendicular).
Solution. Let's multiply the vectors using the rule for multiplying polynomials:
Now let's calculate each term:
.
Let’s create an equation (the product is equal to zero), add similar terms and solve the equation:
Answer: we got the value λ = 1.8, at which the vectors are orthogonal.
Example 5. Prove that the vector 
orthogonal (perpendicular) to the vector
Solution. To check orthogonality, we multiply the vectors and as polynomials, substituting instead the expression given in the problem statement:
.
To do this, you need to multiply each term (term) of the first polynomial by each term of the second and add the resulting products:
.
In the resulting result, the fraction is reduced by. The following result is obtained:
Conclusion: as a result of multiplication we got zero, therefore, the orthogonality (perpendicularity) of the vectors is proven.
Solve the problem yourself and then see the solution
Example 6. The lengths of the vectors and are given, and the angle between these vectors is π /4 . Determine at what value μ vectors and are mutually perpendicular.
For self-test you can use online calculator Dot product of vectors and cosine of the angle between them .
Matrix representation of the dot product of vectors and the product of n-dimensional vectors
Sometimes it is advantageous for clarity to represent two multiplied vectors in the form of matrices. Then the first vector is represented as a row matrix, and the second - as a column matrix:
Then the scalar product of vectors will be the product of these matrices :
The result is the same as that obtained by the method we have already considered. We got one single number, and the product of a row matrix by a column matrix is also one single number.
It is convenient to represent the product of abstract n-dimensional vectors in matrix form. Thus, the product of two four-dimensional vectors will be the product of a row matrix with four elements by a column matrix also with four elements, the product of two five-dimensional vectors will be the product of a row matrix with five elements by a column matrix also with five elements, and so on.
Example 7. Find scalar products of pairs of vectors
,
using matrix representation.
Solution. The first pair of vectors. We represent the first vector as a row matrix, and the second as a column matrix. We find the scalar product of these vectors as the product of a row matrix and a column matrix:
We similarly represent the second pair and find:
As you can see, the results were the same as for the same pairs from example 2.
Angle between two vectors
The derivation of the formula for the cosine of the angle between two vectors is very beautiful and concise.
To express the dot product of vectors
(1)
in coordinate form, we first find the scalar product of the unit vectors. The scalar product of a vector with itself by definition:
What is written in the formula above means: the scalar product of a vector with itself is equal to the square of its length. The cosine of zero is equal to one, so the square of each unit will be equal to one:
Since vectors
are pairwise perpendicular, then the pairwise products of the unit vectors will be equal to zero:
Now let's perform the multiplication of vector polynomials:
We substitute the values of the corresponding scalar products of the unit vectors into the right side of the equality:
We obtain the formula for the cosine of the angle between two vectors:
Example 8. Three points are given A(1;1;1), B(2;2;1), C(2;1;2).
Find the angle.
Solution. Finding the coordinates of the vectors:
,
.
Using the cosine angle formula we get:
Hence, .
For self-test you can use online calculator Dot product of vectors and cosine of the angle between them .
Example 9. Two vectors are given
Find the sum, difference, length, dot product and angle between them.
2.Difference
Definition 1
The scalar product of vectors is a number equal to the product of the dynes of these vectors and the cosine of the angle between them.
The notation for the product of vectors a → and b → has the form a → , b → . Let's transform it into the formula:
a → , b → = a → · b → · cos a → , b → ^ . a → and b → denote the lengths of the vectors, a → , b → ^ - designation of the angle between given vectors. If at least one vector is zero, that is, has a value of 0, then the result will be equal to zero, a → , b → = 0
When multiplying a vector by itself, we get the square of its length:
a → , b → = a → b → cos a → , a → ^ = a → 2 cos 0 = a → 2
Definition 2
Scalar multiplication of a vector by itself is called a scalar square.
Calculated by the formula:
a → , b → = a → · b → · cos a → , b → ^ .
The notation a → , b → = a → · b → · cos a → , b → ^ = a → · n p a → b → = b → · n p b → a → shows that n p b → a → is the numerical projection of a → onto b → , n p a → a → - projection of b → onto a →, respectively.
Let us formulate the definition of a product for two vectors:
The scalar product of two vectors a → by b → is called the product of the length of the vector a → by the projection b → by the direction of a → or the product of the length b → by the projection a →, respectively.
Dot product in coordinates
The scalar product can be calculated through the coordinates of vectors in a given plane or in space.
The scalar product of two vectors on a plane, in three-dimensional space, is called the sum of the coordinates of given vectors a → and b →.
When calculating the scalar product of given vectors a → = (a x , a y) , b → = (b x , b y) on the plane in the Cartesian system, use:
a → , b → = a x b x + a y b y ,
for three-dimensional space the expression is applicable:
a → , b → = a x · b x + a y · b y + a z · b z .
In fact, this is the third definition of the scalar product.
Let's prove it.
Evidence 1
To prove it, we use a → , b → = a → · b → · cos a → , b → ^ = a x · b x + a y · b y for vectors a → = (a x , a y) , b → = (b x , b y) on Cartesian system.
Vectors should be set aside
O A → = a → = a x , a y and O B → = b → = b x , b y .
Then the length of the vector A B → will be equal to A B → = O B → - O A → = b → - a → = (b x - a x , b y - a y) .
Consider triangle O A B .
A B 2 = O A 2 + O B 2 - 2 · O A · O B · cos (∠ A O B) is correct based on the cosine theorem.
According to the condition, it is clear that O A = a → , O B = b → , A B = b → - a → , ∠ A O B = a → , b → ^ , which means we write the formula for finding the angle between vectors differently
b → - a → 2 = a → 2 + b → 2 - 2 · a → · b → · cos (a → , b → ^) .
Then from the first definition it follows that b → - a → 2 = a → 2 + b → 2 - 2 · (a → , b →) , which means (a → , b →) = 1 2 · (a → 2 + b → 2 - b → - a → 2) .
Applying the formula for calculating the length of vectors, we get:
a → , b → = 1 2 · ((a 2 x + a y 2) 2 + (b 2 x + b y 2) 2 - ((b x - a x) 2 + (b y - a y) 2) 2) = = 1 2 (a 2 x + a 2 y + b 2 x + b 2 y - (b x - a x) 2 - (b y - a y) 2) = = a x b x + a y b y
Let us prove the equalities:
(a → , b →) = a → b → cos (a → , b → ^) = = a x b x + a y b y + a z b z
– respectively for vectors of three-dimensional space.
The scalar product of vectors with coordinates says that the scalar square of a vector is equal to the sum of the squares of its coordinates in space and on the plane, respectively. a → = (a x , a y , a z) , b → = (b x , b y , b z) and (a → , a →) = a x 2 + a y 2 .
Dot product and its properties
There are properties of the dot product that apply to a →, b →, and c →:
- commutativity (a → , b →) = (b → , a →) ;
- distributivity (a → + b → , c →) = (a → , c →) + (b → , c →) , (a → + b → , c →) = (a → , b →) + (a → , c →) ;
- combinative property (λ · a → , b →) = λ · (a → , b →), (a → , λ · b →) = λ · (a → , b →), λ - any number;
- scalar square is always greater than zero (a → , a →) ≥ 0, where (a → , a →) = 0 in the case when a → zero.
The properties are explainable thanks to the definition of the scalar product on the plane and the properties of addition and multiplication of real numbers.
Prove the commutative property (a → , b →) = (b → , a →) . From the definition we have that (a → , b →) = a y · b y + a y · b y and (b → , a →) = b x · a x + b y · a y .
By the property of commutativity, the equalities a x · b x = b x · a x and a y · b y = b y · a y are true, which means a x · b x + a y · b y = b x · a x + b y · a y .
It follows that (a → , b →) = (b → , a →) . Q.E.D.
Distributivity is valid for any numbers:
(a (1) → + a (2) → + . . . + a (n) → , b →) = (a (1) → , b →) + (a (2) → , b →) + . . . + (a (n) → , b →)
and (a → , b (1) → + b (2) → + . . + b (n) →) = (a → , b (1) →) + (a → , b (2) →) + . . . + (a → , b → (n)) ,
hence we have
(a (1) → + a (2) → + . . . + a (n) → , b (1) → + b (2) → + . . . + b (m) →) = = (a ( 1) → , b (1) →) + (a (1) → , b (2) →) + . . . + (a (1) → , b (m) →) + + (a (2) → , b (1) →) + (a (2) → , b (2) →) + . . . + (a (2) → , b (m) →) + . . . + + (a (n) → , b (1) →) + (a (n) → , b (2) →) + . . . + (a (n) → , b (m) →)
Dot product with examples and solutions
Any problem of this kind is solved using the properties and formulas relating to the scalar product:
- (a → , b →) = a → · b → · cos (a → , b → ^) ;
- (a → , b →) = a → · n p a → b → = b → · n p b → a → ;
- (a → , b →) = a x · b x + a y · b y or (a → , b →) = a x · b x + a y · b y + a z · b z ;
- (a → , a →) = a → 2 .
Let's look at some example solutions.
Example 2
The length of a → is 3, the length of b → is 7. Find the dot product if the angle has 60 degrees.
Solution
By condition, we have all the data, so we calculate it using the formula:
(a → , b →) = a → b → cos (a → , b → ^) = 3 7 cos 60 ° = 3 7 1 2 = 21 2
Answer: (a → , b →) = 21 2 .
Example 3
Given vectors a → = (1 , - 1 , 2 - 3) , b → = (0 , 2 , 2 + 3) . What is the scalar product?
Solution
This example considers the formula for calculating coordinates, since they are specified in the problem statement:
(a → , b →) = a x · b x + a y · b y + a z · b z = = 1 · 0 + (- 1) · 2 + (2 + 3) · (2 + 3) = = 0 - 2 + ( 2 - 9) = - 9
Answer: (a → , b →) = - 9
Example 4
Find the scalar product of A B → and A C →. Points A (1, - 3), B (5, 4), C (1, 1) are given on the coordinate plane.
Solution
To begin with, the coordinates of the vectors are calculated, since by condition the coordinates of the points are given:
A B → = (5 - 1, 4 - (- 3)) = (4, 7) A C → = (1 - 1, 1 - (- 3)) = (0, 4)
Substituting into the formula using coordinates, we get:
(A B →, A C →) = 4 0 + 7 4 = 0 + 28 = 28.
Answer: (A B → , A C →) = 28 .
Example 5
Given vectors a → = 7 · m → + 3 · n → and b → = 5 · m → + 8 · n → , find their product. m → equals 3 and n → equals 2 units, they are perpendicular.
Solution
(a → , b →) = (7 m → + 3 n → , 5 m → + 8 n →) . Applying the distributivity property, we get:
(7 m → + 3 n →, 5 m → + 8 n →) = = (7 m →, 5 m →) + (7 m →, 8 n →) + (3 n → , 5 m →) + (3 n → , 8 n →)
We take the coefficient out of the sign of the product and get:
(7 m → , 5 m →) + (7 m → , 8 n →) + (3 n → , 5 m →) + (3 n → , 8 n →) = = 7 · 5 · (m → , m →) + 7 · 8 · (m → , n →) + 3 · 5 · (n → , m →) + 3 · 8 · (n → , n →) = = 35 · (m → , m →) + 56 · (m → , n →) + 15 · (n → , m →) + 24 · (n → , n →)
By the property of commutativity we transform:
35 · (m → , m →) + 56 · (m → , n →) + 15 · (n → , m →) + 24 · (n → , n →) = = 35 · (m → , m →) + 56 · (m → , n →) + 15 · (m → , n →) + 24 · (n → , n →) = = 35 · (m → , m →) + 71 · (m → , n → ) + 24 · (n → , n →)
As a result we get:
(a → , b →) = 35 · (m → , m →) + 71 · (m → , n →) + 24 · (n → , n →).
Now we apply the formula for the scalar product with the angle specified by the condition:
(a → , b →) = 35 · (m → , m →) + 71 · (m → , n →) + 24 · (n → , n →) = = 35 · m → 2 + 71 · m → · n → · cos (m → , n → ^) + 24 · n → 2 = 35 · 3 2 + 71 · 3 · 2 · cos π 2 + 24 · 2 2 = 411 .
Answer: (a → , b →) = 411
If there is a numerical projection.
Example 6
Find the scalar product of a → and b →. Vector a → has coordinates a → = (9, 3, - 3), projection b → with coordinates (- 3, - 1, 1).
Solution
By condition, the vectors a → and the projection b → are oppositely directed, because a → = - 1 3 · n p a → b → → , which means the projection b → corresponds to the length n p a → b → → , and with the “-” sign:
n p a → b → → = - n p a → b → → = - (- 3) 2 + (- 1) 2 + 1 2 = - 11 ,
Substituting into the formula, we get the expression:
(a → , b →) = a → · n p a → b → → = 9 2 + 3 2 + (- 3) 2 · (- 11) = - 33 .
Answer: (a → , b →) = - 33 .
Problems with a known scalar product, where it is necessary to find the length of a vector or a numerical projection.
Example 7
What value should λ take for a given scalar product a → = (1, 0, λ + 1) and b → = (λ, 1, λ) will be equal to -1.
Solution
From the formula it is clear that it is necessary to find the sum of the products of coordinates:
(a → , b →) = 1 λ + 0 1 + (λ + 1) λ = λ 2 + 2 λ .
Given we have (a → , b →) = - 1 .
To find λ, we calculate the equation:
λ 2 + 2 · λ = - 1, hence λ = - 1.
Answer: λ = - 1.
Physical meaning of the scalar product
Mechanics considers the application of the dot product.
When A works with a constant force F → a moving body from a point M to N, you can find the product of the lengths of the vectors F → and M N → with the cosine of the angle between them, which means the work is equal to the product of the force and displacement vectors:
A = (F → , M N →) .
Example 8
The movement of a material point by 3 meters under the influence of a force equal to 5 Ntons is directed at an angle of 45 degrees relative to the axis. Find A.
Solution
Since work is the product of the force vector and displacement, it means that based on the condition F → = 5, S → = 3, (F →, S → ^) = 45 °, we obtain A = (F →, S →) = F → · S → · cos (F → , S → ^) = 5 · 3 · cos (45 °) = 15 2 2 .
Answer: A = 15 2 2 .
Example 9
A material point, moving from M (2, - 1, - 3) to N (5, 3 λ - 2, 4) under the force F → = (3, 1, 2), did work equal to 13 J. Calculate the length of movement.
Solution
For given vector coordinates M N → we have M N → = (5 - 2, 3 λ - 2 - (- 1) , 4 - (- 3)) = (3, 3 λ - 1, 7) .
Using the formula for finding work with vectors F → = (3, 1, 2) and M N → = (3, 3 λ - 1, 7), we obtain A = (F ⇒, M N →) = 3 3 + 1 (3 λ - 1) + 2 7 = 22 + 3 λ.
According to the condition, it is given that A = 13 J, which means 22 + 3 λ = 13. This implies λ = - 3, which means M N → = (3, 3 λ - 1, 7) = (3, - 10, 7).
To find the length of movement M N →, apply the formula and substitute the values:
M N → = 3 2 + (- 10) 2 + 7 2 = 158.
Answer: 158.
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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. Take a look at. 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 origins of both vectors coincide with the origin of coordinates, 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
): ⟨a |
b ⟩ (\displaystyle \langle a|b\rangle ) b (\displaystyle \mathbf (b) ) is defined as the product of the lengths of these vectors by cosine angle between them:
(a , b) = |a | b | cos (θ) (\displaystyle (\mathbf (a) ,\mathbf (b))=|\mathbf (a) ||\mathbf (b) |\cos(\theta))
Equivalent definition: scalar product is the product of length projections the first vector to the second and the length of the second vector (see figure). If at least one of the vectors is zero, then the product is considered equal to zero. Vector space , , that is, for sets of vectors with the operations of addition and multiplication by scalars. The geometric definition of the scalar product given above is generally unsuitable, since it is not clear what is meant by the lengths of the vectors and the magnitude of the angle between them. Therefore, in modern mathematics, the opposite approach is used: the scalar product is axiomatically determined, and through it - lengths and angles. In particular, the scalar product is defined for And complex vectors Multidimensional spaces Infinite-dimensional spaces.
infinite-dimensional spaces Tensor algebra , The scalar product and its generalizations play an extremely important role in Vector calculus Manifold variety theory
, mechanics and physics. For example,
Work of force
work of force during mechanical movement is equal to the scalar product of the force vector and the displacement vector. Definition during mechanical movement is equal to the scalar product of the force vector and the displacement vector. Definition in Euclidean space n (\displaystyle n)-dimensional real Euclidean space, vectors are defined by their coordinates - sets
real numbers inOrthogonal basis
orthonormal basis . You can define the scalar product of vectors like this: And (a , b) = a 1 b 1 + a 2 b 2 + a 3 b 3 + ⋯ + a n b n (\displaystyle (\mathbf (a) ,\mathbf (b))=a_(1)b_(1)+ a_(2)b_(2)+a_(3)b_(3)+\dots +a_(n)b_(n)) The check shows that all three axioms are satisfied.
( 1 , 3 , − 5 ) ⋅ ( 4 , − 2 , − 1 ) = 1 ⋅ 4 + 3 ⋅ (− 2) + (− 5) ⋅ (− 1) = 4 − 6 + 5 = 3. (\ displaystyle (\begin(aligned)\ \(1,3,-5\)\cdot \(4,-2,-1\)&=1\cdot 4+3\cdot (-2)+(-5) \cdot (-1)\\&=4-6+5\\&=3.\end(aligned)))For complex vectors a = ( a 1 , a 2 … a n ) , b = ( b 1 , b 2 … b n ) (\displaystyle \mathbf (a) =\(a_(1),a_(2)\dots a_(n)\ ),\mathbf (b) =\(b_(1),b_(2)\dots b_(n)\)) let's define similarly:
(a , b) = ∑ k = 1 n a k b k ¯ = a 1 b 1 ¯ + a 2 b 2 ¯ + ⋯ + a n b n ¯ (\displaystyle (\mathbf (a) ,\mathbf (b))=\sum _( k=1)^(n)a_(k)(\overline (b_(k)))=a_(1)(\overline (b_(1)))+a_(2)(\overline (b_(2) ))+\cdots +a_(n)(\overline (b_(n)))).Example (for n = 2 (\displaystyle n=2)): ( 1 + i , 2 ) ⋅ ( 2 + i , i ) = (1 + i) ⋅ (2 + i ¯) + 2 ⋅ i ¯ = (1 + i) ⋅ (2 − i) + 2 ⋅ (− i) = 3 − i .
(\displaystyle \(1+i,2\)\cdot \(2+i,i\)=(1+i)\cdot ((\overline (2+i)))+2\cdot (\overline ( i))=(1+i)\cdot (2-i)+2\cdot (-i)=3-i.)
Related definitions
In the modern axiomatic approach, already on the basis of the concept of the scalar product of vectors, the following derivative concepts are introduced: Length Norm (mathematics) :
norm|
|
a | = (a , a) (\displaystyle |\mathbf (a) |=(\sqrt ((\mathbf (a) ,\mathbf (a)))))(the term "length" is usually applied to finite-dimensional vectors, but in the case of calculating the length of a curved path it is often used in the case of infinite-dimensional spaces). For any elements |
a , b (\displaystyle \mathbf (a) ,\mathbf (b) ) |(a , b) | In case the space is Pseudo-Euclidean space
pseudo-Euclidean- , the concept of an angle is defined only for vectors that do not contain isotropic lines inside the sector formed by the vectors. The angle itself is entered as a number,(perpendicular) are vectors whose scalar product is equal to zero. This definition applies to any space with a positive definite scalar product. For example, orthogonal polynomials are in fact orthogonal (in the sense of this definition) to each other in some Hilbert space.
- A space (real or complex) with a positive definite scalar product is called pre-Hilbert space.
- In this case, a finite-dimensional real space with a positive definite scalar product is also called Euclidean, and complex - hermitian or unitary space.
- The case when the scalar product is not sign-definite leads to the so-called. spaces with indefinite metric. The scalar product in such spaces no longer generates a norm (and it is usually introduced additionally). A finite-dimensional real space with an indefinite metric is called |(the most important special case of such a space is Minkowski space). Among infinite-dimensional spaces with indefinite metric, Pontryagin spaces and Kerin spaces play an important role.
Properties
- Cosine theorem is easily derived using the dot product: |
- B C | 2 = B C → 2 = (A C → − A B →) 2 = ⟨ A C → − A B → , A C → − A B → ⟩ = A C → 2 + A B → 2 − 2 ⟨ A C → , A B → ⟩ = | A B |< 0, если угол между векторами тупой.
- 2 + | A C |: a e = (a , e) = | a | |
- e | cos ∠ (a , e) = | And a | cos ∠ (a , e) (\displaystyle a_(e)=(\mathbf (a) ,\mathbf (e))=|\mathbf (a) ||\mathbf (e) |\cos \angle (( \mathbf (a) ,\mathbf (e)))=|\mathbf (a) |\cos \angle ((\mathbf (a) ,\mathbf (e))))
