Angle between vectors

Consider two given vectors $\overrightarrow(a)$ and $\overrightarrow(b)$. Let us set aside 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 a zero vector, then the angle between the vectors is equal to $0^0$.

Notation: $\widehat(\overrightarrow(a),\overrightarrow(b))$

The concept of the scalar product of vectors

Mathematically, this definition can be written as follows:

Scalar product can be zero in two cases:

If one of the vectors will be a zero vector (Since then its length is zero).

If the vectors are mutually perpendicular (i.e. $cos(90)^0=0$).

Note also that the inner product is greater than zero if the angle between these vectors is acute (because $(cos \left(\widehat(\overrightarrow(a),\overrightarrow(b))\right)\ ) >0$), and less than zero if the angle between these vectors is obtuse (since $(cos \left(\widehat(\overrightarrow(a),\overrightarrow(b))\right)\ )

The concept of the scalar square is related to the concept of the scalar product.

Definition 2

The scalar square of the vector $\overrightarrow(a)$ is the scalar product of this vector with itself.

We get that the scalar square is

\[\overrightarrow(a)\overrightarrow(a)=\left|\overrightarrow(a)\right|\left|\overrightarrow(a)\right|(cos 0^0\ )=\left|\overrightarrow(a )\right|\left|\overrightarrow(a)\right|=(\left|\overrightarrow(a)\right|)^2\]

Calculation of the scalar product by the coordinates of vectors

In addition to the standard way of finding the value of the dot 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 implications:

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 $cos\alpha =\frac(a_1a_2+b_1b_2)(\sqrt(a^2_1+b^2_1)\cdot \sqrt(a^2_2+b^2_2))$

Properties of the Dot 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).

displacement law:$\overrightarrow(a)\overrightarrow(b)=\overrightarrow(b)\overrightarrow(a)$.

This property follows from the definition of the inner 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 inner product of vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ if $\left|\overrightarrow(a)\right|=3$ and $\left|\overrightarrow(b)\right|= 2$, and the angle between them is $((30)^0,\ 45)^0,\ (90)^0,\ (135)^0$.

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 $(135)^0:$

\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((135)^0\right)\ )=6\cdot \left(-\frac(\sqrt(2))(2)\ right)=-3\sqrt(2)\]

Thus, the length of a vector is calculated as the square root of the sum of the squares of its coordinates
. Similarly, the length of the n-dimensional vector is calculated
. If we recall that each coordinate of the vector is the difference between the coordinates of the end and the beginning, then we will get 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 proved 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:
\u003d x 1 * y 1 + x 2 * y 2.

In n-dimensional space, the dot product of vectors X= (x 1 , x 2 ,...,x n) and Y= (y 1 , y 2 ,...,yn) is defined as the sum of the products of their respective coordinates: X*Y \u003d x 1 * y 1 + x 2 * y 2 + ... + x n * yn.

The operation of multiplying vectors with 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 the scalar product of vectors is given that satisfies the four corresponding axioms is called Euclidean linear vectorspace.

It is easy to see that when multiplying any vector by itself, we get the square of its length. So it's different length vector can be defined as the square root of its scalar square:.

The length of a vector has the following properties:

1) |X| = 0Х = 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 the scalar product. Indeed, if
, then
. This fraction is not greater than one (according to the Cauchy-Bunyakovsky inequality), so from here you can find .

The two vectors are called orthogonal or perpendicular if their dot product is zero. It follows from the definition of the dot product that the zero vector is orthogonal to any vector. If both orthogonal vectors are non-zero, then necessarily cos= 0, i.e.=/2 = 90 o.

Consider Figure 7.4 again. 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 angle  of the inclination of the vector to the vertical axis as
. These numbers are called direction cosines. It is easy to see that the sum of the squares of the direction cosines is always equal to one: cos 2 +cos 2 = 1. Similarly, we can introduce the concept of direction cosines for spaces of higher dimensions.

Vector space basis

For vectors, one 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 statement is also true: if one of the vectors is a linear combination of the others, then all these vectors in the aggregate are linearly dependent.

Note that if among the vectors a l , a 2 ,...a m there is a zero vector, then this collection of vectors is necessarily linearly dependent. Indeed, we get  l a l +  2 a 2 +...+  m a m = 0, if, for example, we equate the coefficient  j with a 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 of the vectors from the set of vectors are linearly dependent, then all these vectors are linearly dependent. Indeed, if some vectors give a zero vector in their linear combination with coefficients that are not simultaneously zero, then the remaining vectors multiplied by zero coefficients can be added to this sum of products, and it will still be a zero vector.

How to determine if 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 make 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 of the problem could also be started with arguments based on the definition of linear independence. Namely, compose a vector equation  lal + 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:

The solution of this system by the Gauss method will be reduced to obtaining the same step matrix, only it will have one more column - free members. They will all be equal to zero, since linear transformations zeros cannot lead to a different result. The transformed system of equations will take the form:

The solution of this system will be (-s; -s; s), where s is an arbitrary number; for example, (-1;-1;1). This means that if we take  l \u003d -1;  2 \u003d -1 and  3 \u003d 1, then  l a l +  2 a 2 +  3 a 3 \u003d 0, i.e. the vectors are actually linearly dependent.

From the solved example, it becomes clear that if we take the number of vectors more than the dimension of the space, then they will necessarily be linearly dependent. Indeed, 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 may be linearly independent, but five or more may not. Consequently, no more than two vectors can be linearly independent in 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. Etc.

That's why dimension spaces can be defined as the maximum number of linearly independent vectors that can be in it.

The set of n linearly independent vectors of the n-dimensional space R is called basis this space.

Theorem. Each linear space vector can be represented as a linear combination of basis vectors, and moreover in a unique way.

Proof. Let the vectors e l , e 2 ,...e n form a basis of an n-dimensional space R. Let us prove that any vector X is a linear combination of these vectors. Since, together with the 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 into :

( l /)e l + ( 2 /)e 2 +...+ ( n /)e n + Х = 0

X \u003d - ( l / ) e l - ( 2 / ) e 2 -...- ( n / ) e n

X \u003d x l e l + x 2 e 2 + ... + x n e n,

where x j = -( j /),
.

Let us now prove that such a representation as a linear combination is unique. Assume the opposite, i.e. that there is another representation:

X \u003d y l e l + y 2 e 2 + ... + y n e n

Subtract from it term by term the expression obtained earlier:

0 \u003d (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 get that (y j - x j) = 0,
, i.e. y j ​​= x j . So the expression is the same. The theorem has been proven.

The expression X \u003d x l e l + x 2 e 2 + ... + x n e n is called decomposition vector X according to the basis e l , e 2 ,...e n , and the numbers x l , x 2 ,... x n - coordinates vector x with respect 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 (el * e i) +  2 (e 2 * e i) +...+  n (en * e i) = 0   i (ei * e i) = 0   i = 0 for i.

The vectors e l , e 2 ,...e n of the 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≠ji |e i | = 1 for i.

Theorem (without proof). Every n-dimensional Euclidean space has an orthonormal basis.

An example of an orthonormal basis is a system of n unit vectors e i , in 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, vector-orts (1, 0, 0), (0, 1, 0) and (0, 0, 1) form the basis of a three-dimensional space.

Definition 1

The scalar product of vectors is called 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 convert to the formula:

a → , b → = a → b → cos a → , b → ^ . a → and b → denote the lengths of the vectors, a → , b → ^ denote the angle between the given vectors. If at least one vector is zero, that is, it has a value of 0, then the result will be zero, a → , b → = 0

When multiplying a vector by itself, we get the square of its dyne:

a → , b → = a → b → cos a → , a → ^ = a → 2 cos 0 = a → 2

Definition 2

The scalar multiplication of a vector by itself is called a scalar square.

Calculated according to the formula:

a → , b → = a → b → cos a → , b → ^ .

Writing a → , b → = a → b → cos a → , b → ^ = a → npa → b → = b → npb → a → shows that npb → a → is a numerical projection of a → onto b → , npa → a → - projection of b → onto a → respectively.

We formulate the definition of the 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 of b → by the direction a → or the product of the length of b → by the projection of a →, respectively.

Dot product in coordinates

The calculation of the scalar product can be done through the coordinates of the 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 the given vectors a → and b → .

When calculating on the plane the scalar product of given vectors a → = (a x , a y) , b → = (b x , b y) in 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 dot product.

Let's prove it.

Proof 1

To prove it, we use a → , b → = a → b → cos a → , b → ^ = ax bx + ay by for vectors a → = (ax , ay) , b → = (bx , by) on cartesian system.

Vectors should be postponed

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 a triangle O A B .

A B 2 = O A 2 + O B 2 - 2 O A O B cos (∠ A O B) is true, based on the cosine theorem.

By condition, it can be seen that O A = a → , O B = b → , A B = b → - a → , ∠ A O B = a → , b → ^ , so 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 it follows from the first definition that b → - a → 2 = a → 2 + b → 2 - 2 (a → , b →) , so (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 + ay 2) 2 + (b 2 x + by 2) 2 - ((bx - ax) 2 + (by - ay) 2) 2) = = 1 2 (a 2 x + a 2 y + b 2 x + b 2 y - (bx - ax) 2 - (by - ay) 2) = = ax bx + ay by

Let's 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 the vector is equal to the sum 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 dot product properties that apply for a → , b → and c → :

1. commutativity (a → , b →) = (b → , a →) ;
2. distributivity (a → + b → , c →) = (a → , c →) + (b → , c →) , (a → + b → , c →) = (a → , b →) + (a → , c →) ;
3. associative property (λ a → , b →) = λ (a → , b →) , (a → , λ b →) = λ (a → , b →) , λ - any number;
4. the scalar square is always greater than zero (a → , a →) ≥ 0 , where (a → , a →) = 0 when a → zero.

Example 1

The properties are explained by the definition of the dot product in the plane and by the properties of addition and multiplication of real numbers.

Prove the commutativity 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, so 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 such a plan is solved using the properties and formulas regarding the scalar product:

1. (a → , b →) = a → b → cos (a → , b → ^) ;
2. (a → , b →) = a → · n p a → b → = b → · n p b → a → ;
3. (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 ;
4. (a → , a →) = a → 2 .

Let's look at some examples of 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 by 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

In this example, the formula for calculating the coordinates is considered, since they are specified in the problem statement:

(a → , b →) = ax bx + ay by + az bz = = 1 0 + (- 1) 2 + (2 + 3) (2 + 3) = = 0 - 2 + ( 2 - 9) = - 9

Answer: (a → , b →) = - 9

Example 4

Find the inner product of A B → and A C → . On the coordinate plane given points A (1 , - 3) , B (5 , 4) , C (1 , 1) .

Solution

To begin with, the coordinates of the vectors are calculated, since the coordinates of the points are given by condition:

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 → is equal to 3 and n → is equal to 2 units, they are perpendicular.

Solution

(a → , b →) = (7 m → + 3 n → , 5 m → + 8 n →) . Applying the distributive property, we get:

(7 m → + 3 n → , 5 m → + 8 n →) = = (7 m → , 5 m →) + (7 m → , 8 n →) + (3 n n → , 5 m →) + (3 n → , 8 n →)

We take the coefficient outside 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 inner product of a → and b → . The vector a → has coordinates a → = (9 , 3 , - 3) , the projection b → has 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 → → , so 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 → \u003d (1, 0, λ + 1) and b → \u003d (λ, 1, λ) will be equal to -1.

Solution

From the formula it can be seen 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 .

In given we have (a → , b →) = - 1 .

To find λ , we calculate the equation:

λ 2 + 2 · λ = - 1 , hence λ = - 1 .

Answer: λ = - 1 .

The physical meaning of the scalar product

Mechanics considers the application of the dot product.

When working A with a constant force F → a moving body from 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 that the work is equal to the product of the force and displacement vectors:

A = (F → , M N →) .

Example 8

The displacement of a material point by 3 meters under the action of a force equal to 5 Nton 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, then, based on the condition F → = 5 , S → = 3 , (F → , S → ^) = 45 ° , we get A = (F → , S →) = F → S → cos (F → , S → ^) = 5 3 cos (45 °) = 15 2 2 .

Answer: A = 15 2 2 .

Example 9

The 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 the movement.

Solution

For given coordinates of the vector M N → we have M N → = (5 - 2 , 3 λ - 2 - (- 1) , 4 - (- 3)) = (3 , 3 λ - 1 , 7) .

By the formula for finding work with vectors F → = (3 , 1 , 2) and MN → = (3 , 3 λ - 1 , 7) we get A = (F ⇒ , MN →) = 3 3 + 1 (3 λ - 1) + 2 7 = 22 + 3λ.

By condition, it is given that A \u003d 13 J, which means 22 + 3 λ \u003d 13. This implies λ = - 3 , hence M N → = (3 , 3 λ - 1 , 7) = (3 , - 10 , 7) .

To find the travel length M N → , we apply the formula and substitute the values:

M N → = 3 2 + (- 10) 2 + 7 2 = 158 .

Answer: 158 .

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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, which is 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 multiplied vectors are given by their coordinates.

Definition 3. The dot product of vectors is the number equal to the sum of the pairwise products of their respective coordinates.

On surface

If two vectors and in the plane are defined by their two Cartesian coordinates

then the dot product of these vectors is equal to the sum of the pairwise products of their respective 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:

.

Write an equation and solve it:

Answer. The desired 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 the pairwise products of their respective coordinates, only there are already three coordinates:

.

The task of finding the scalar product in the considered way is after analyzing the properties of the scalar product. Because in the task it will be necessary to determine what angle the multiplied vectors form.

Properties of the Dot Product of Vectors

Algebraic properties

1. (commutative property: the value of their scalar product does not change from changing the places of multiplied vectors).

2. (associative property with respect to a numerical factor: the scalar product of a vector multiplied by some factor and another vector is equal to the scalar product of these vectors multiplied by the same factor).

3. (distributive property with respect 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 a 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, two vectors are visible, which are brought to a common beginning. And the first thing you need to pay attention to: 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 the dot product includes only the cosine of the angle, not the value of its expression. But only one corner is considered in the properties. And this is the one of the two angles that does not exceed π ie 180 degrees. This angle is shown in the figure as φ 1 .

1. Two vectors are called orthogonal And the angle between these vectors is a right (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, what 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 dot product is negative .

Example 3 Vectors are given in coordinates:

.

Calculate the dot 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. We multiply the vectors according to the rule of multiplication of polynomials:

Now let's calculate each term:

.

Let's compose an equation (equality of the product to zero), give like 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 vector

Solution. To check orthogonality, we multiply the vectors and as polynomials, substituting the expression given in the problem condition instead of it:

.

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:

.

As a result, the fraction due is reduced. 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 Given the lengths of vectors and , 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 scalar product of vectors and the product of n-dimensional vectors

Sometimes, for clarity, it is advantageous 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 the matrix-row by the matrix-column is also one single number.

In matrix form, it is convenient to represent the product of abstract n-dimensional vectors. 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 Dot 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 the row matrix by the column matrix:

Similarly, we represent the second pair and find:

As you can see, the results are 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 orts. The scalar product of a vector with itself is 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 orth will be equal to one:

Since the vectors

are pairwise perpendicular, then the pairwise products of the orts will be equal to zero:

Now let's perform the multiplication of vector polynomials:

We substitute in the right side of the equality the values ​​of the corresponding scalar products of the orts:

We get the formula for the cosine of the angle between two vectors:

Example 8 Given three points A(1;1;1), B(2;2;1), C(2;1;2).

Find an angle.

Solution. We find the coordinates of the vectors:

,

.

Using the formula for the cosine of an angle, we get:

Consequently, .

For self-test, you can use online calculator Dot product of vectors and cosine of the angle between them .

Example 9 Given two vectors

Find the sum, difference, length, dot product and the angle between them.

2.Difference

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

The 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

Dot product formula 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 Dot 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 inner product of vectors p = a + 3b and q = 5a - 3 b if their lengths |a| = 3, |b| = 2, and the angle between the 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 \u003d 5 3 2 + 12 3 2 cos 60˚ - 9 2 2 \u003d 45 +36 -36 \u003d 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. (one"")

3) the vectors are oriented in the same way as the basis of the entire space (positively or negatively).

Designate: .

physical meaning vector product

is the moment of force relative to the point O; is radius is the vector of force application point, then

moreover, if transferred to the point O, then the triple must be oriented as a vector of the basis.