Dot product of vectors

We continue to deal with vectors. At the first lesson Vectors for dummies we have considered the concept of a vector, actions with vectors, vector coordinates and the simplest problems with vectors. If you came to this page for the first time from a search engine, I highly recommend reading the above introductory article, because in order to assimilate the material, you need to be guided in the terms and notation I use, have basic knowledge of vectors and be able to solve elementary problems. This lesson is a logical continuation of the topic, and in it I will analyze in detail typical tasks that use the scalar product of vectors. This is a VERY IMPORTANT job.. Try not to skip the examples, they are accompanied by a useful bonus - practice will help you to consolidate the material covered and "get your hand" on solving common problems of analytical geometry.

Adding vectors, multiplying a vector by a number…. It would be naive to think that mathematicians have not come up with something else. In addition to the actions already considered, there are a number of other operations with vectors, namely: dot product of vectors, cross product of vectors and mixed product of vectors. The scalar product of vectors is familiar to us from school, the other two products are traditionally related to the course of higher mathematics. The topics are simple, the algorithm for solving many problems is stereotyped and understandable. The only thing. There is a decent amount of information, so it is undesirable to try to master and solve EVERYTHING AND AT ONCE. This is especially true for dummies, believe me, the author absolutely does not want to feel like Chikatilo from mathematics. Well, not from mathematics, of course, either =) More prepared students can use the materials selectively, in a certain sense, “acquire” the missing knowledge, for you I will be a harmless Count Dracula =)

Finally, let's open the door a little and take a look at what happens when two vectors meet each other….

Definition of the scalar product of vectors.
Properties of the scalar product. Typical tasks

The concept of dot product

First about angle between vectors. I think everyone intuitively understands what the angle between vectors is, but just in case, a little more. Consider free nonzero vectors and . If we postpone these vectors from an arbitrary point, then we get a picture that many have already presented mentally:

I confess, here I described the situation only at the level of understanding. If you need a strict definition of the angle between vectors, please refer to the textbook, but for practical tasks, we, in principle, do not need it. Also HERE AND FURTHER, I will sometimes ignore zero vectors due to their low practical significance. I made a reservation specifically for advanced visitors to the site, who can reproach me for the theoretical incompleteness of some of the following statements.

can take values ​​from 0 to 180 degrees (from 0 to radians) inclusive. Analytically, this fact is written as a double inequality: or (in radians).

In the literature, the angle icon is often omitted and simply written.

Definition: The scalar product of two vectors is a NUMBER equal to the product of the lengths of these vectors and the cosine of the angle between them:

Now that's a pretty strict definition.

We focus on essential information:

Designation: the scalar product is denoted by or simply .

The result of the operation is a NUMBER: Multiply a vector by a vector to get a number. Indeed, if the lengths of vectors are numbers, the cosine of the angle is a number, then their product will also be a number.

Just a couple of warm-up examples:

Example 1

Decision: We use the formula . In this case:

Answer:

Cosine values ​​can be found in trigonometric table. I recommend printing it - it will be required in almost all sections of the tower and will be required many times.

Purely from a mathematical point of view, the scalar product is dimensionless, that is, the result, in this case, is just a number and that's it. From the point of view of physics problems, the scalar product always has a certain physical meaning, that is, after the result, one or another physical unit must be indicated. The canonical example of calculating the work of a force can be found in any textbook (the formula is exactly a dot product). The work of a force is measured in Joules, therefore, the answer will be written quite specifically, for example,.

Example 2

Find if , and the angle between the vectors is .

This is an example for self-decision, the answer is at the end of the lesson.

Angle between vectors and dot product value

In Example 1, the scalar product turned out to be positive, and in Example 2, it turned out to be negative. Let us find out what the sign of the scalar product depends on. Let's look at our formula: . The lengths of non-zero vectors are always positive: , so the sign can depend only on the value of the cosine.

Note: For a better understanding of the information below, it is better to study the cosine graph in the manual Graphs and function properties. See how the cosine behaves on the segment.

As already noted, the angle between the vectors can vary within , and the following cases are possible:

1) If injection between vectors spicy: (from 0 to 90 degrees), then , and dot product will be positive co-directed, then the angle between them is considered to be zero, and the scalar product will also be positive. Since , then the formula is simplified: .

2) If injection between vectors blunt: (from 90 to 180 degrees), then , and correspondingly, dot product is negative: . Special case: if the vectors directed oppositely, then the angle between them is considered deployed: (180 degrees). The scalar product is also negative, since

The converse statements are also true:

1) If , then the angle between these vectors is acute. Alternatively, the vectors are codirectional.

2) If , then the angle between these vectors is obtuse. Alternatively, the vectors are directed oppositely.

But the third case is of particular interest:

3) If injection between vectors straight: (90 degrees) then and dot product is zero: . The converse is also true: if , then . The compact statement is formulated as follows: The scalar product of two vectors is zero if and only if the given vectors are orthogonal. Short math notation:

! Note : repeat foundations of mathematical logic: double-sided logical consequence icon is usually read "if and only then", "if and only if". As you can see, the arrows are directed in both directions - "from this follows this, and vice versa - from this follows this." What, by the way, is the difference from the one-way follow icon ? Icon claims only that that "from this follows this", and not the fact that the reverse is true. For example: , but not every animal is a panther, so the icon cannot be used in this case. At the same time, instead of the icon can use one-sided icon. For example, while solving the problem, we found out that we concluded that the vectors are orthogonal: - such a record will be correct, and even more appropriate than .

The third case is of great practical importance., since it allows you to check whether the vectors are orthogonal or not. We will solve this problem in the second section of the lesson.

Dot product properties

Let's return to the situation when two vectors co-directed. In this case, the angle between them is zero, , and the scalar product formula takes the form: .

What happens if a vector is multiplied by itself? It is clear that the vector is co-directed with itself, so we use the above simplified formula:

The number is called scalar square vector , and are denoted as .

Thus, the scalar square of a vector is equal to the square of the length of the given vector:

From this equality, you can get a formula for calculating the length of a vector:

While it seems obscure, but the tasks of the lesson will put everything in its place. To solve problems, we also need dot product properties.

For arbitrary vectors and any number, the following properties are true:

1) - displaceable or commutative scalar product law.

2) - distribution or distributive scalar product law. Simply put, you can open parentheses.

3) - combination or associative scalar product law. The constant can be taken out of the scalar product.

Often, all kinds of properties (which also need to be proved!) Are perceived by students as unnecessary trash, which only needs to be memorized and safely forgotten immediately after the exam. It would seem that what is important here, everyone already knows from the first grade that the product does not change from a permutation of factors:. I must warn you, in higher mathematics with such an approach it is easy to mess things up. So, for example, the commutative property is not valid for algebraic matrices. It is not true for cross product of vectors. Therefore, it is at least better to delve into any properties that you will meet in the course of higher mathematics in order to understand what can and cannot be done.

Example 3

.

Decision: First, let's clarify the situation with the vector. What is it all about? The sum of the vectors and is a well-defined vector, which is denoted by . Geometric interpretation of actions with vectors can be found in the article Vectors for dummies. The same parsley with a vector is the sum of the vectors and .

So, according to the condition, it is required to find the scalar product. In theory, you need to apply the working formula , but the trouble is that we do not know the lengths of the vectors and the angle between them. But in the condition, similar parameters are given for vectors, so we will go the other way:

(1) We substitute expressions of vectors .

(2) We open the brackets according to the rule of multiplication of polynomials, a vulgar tongue twister can be found in the article Complex numbers or Integration of a fractional-rational function. I won't repeat myself =) By the way, the distributive property of the scalar product allows us to open the brackets. We have the right.

(3) In the first and last terms, we compactly write the scalar squares of the vectors: . In the second term, we use the commutability of the scalar product: .

(4) Here are similar terms: .

(5) In the first term, we use the scalar square formula, which was mentioned not so long ago. In the last term, respectively, the same thing works: . The second term is expanded according to the standard formula .

(6) Substitute these conditions , and CAREFULLY carry out the final calculations.

Answer:

The negative value of the dot product states the fact that the angle between the vectors is obtuse.

The task is typical, here is an example for an independent solution:

Example 4

Find the scalar product of the vectors and , if it is known that .

Now another common task, just for the new vector length formula. The designations here will overlap a little, so for clarity, I will rewrite it with a different letter:

Example 5

Find the length of the vector if .

Decision will be as follows:

(1) We supply the vector expression .

(2) We use the length formula: , while we have an integer expression as the vector "ve".

(3) We use the school formula for the square of the sum. Pay attention to how it curiously works here: - in fact, this is the square of the difference, and, in fact, it is so. Those who wish can rearrange the vectors in places: - it turned out the same thing up to a rearrangement of the terms.

(4) What follows is already familiar from the two previous problems.

Answer:

Since we are talking about length, do not forget to indicate the dimension - "units".

Example 6

Find the length of the vector if .

This is a do-it-yourself example. Full solution and answer at the end of the lesson.

We continue to squeeze useful things out of the scalar product. Let's look at our formula again . By the rule of proportion, we reset the lengths of the vectors to the denominator of the left side:

Let's swap the parts:

What is the meaning of this formula? If the lengths of two vectors and their scalar product are known, then the cosine of the angle between these vectors can be calculated, and, consequently, the angle itself.

Is the scalar product a number? Number. Are vector lengths numbers? Numbers. So a fraction is also a number. And if the cosine of the angle is known: , then using the inverse function it is easy to find the angle itself: .

Example 7

Find the angle between the vectors and , if it is known that .

Decision: We use the formula:

At the final stage of calculations, a technique was used - the elimination of irrationality in the denominator. In order to eliminate irrationality, I multiplied the numerator and denominator by .

So if , then:

The values ​​of inverse trigonometric functions can be found by trigonometric table. Although this rarely happens. In problems of analytical geometry, some clumsy bear like appears much more often, and the value of the angle has to be found approximately using a calculator. In fact, we will see this picture again and again.

Answer:

Again, do not forget to specify the dimension - radians and degrees. Personally, in order to deliberately “remove all questions”, I prefer to indicate both (unless, of course, by condition, it is required to present the answer only in radians or only in degrees).

Now you will be able to cope with a more difficult task on your own:

Example 7*

Given are the lengths of the vectors , and the angle between them . Find the angle between the vectors , .

The task is not so much difficult as multi-way.
Let's analyze the solution algorithm:

1) According to the condition, it is required to find the angle between the vectors and , so you need to use the formula .

2) We find the scalar product (see Examples No. 3, 4).

3) Find the length of the vector and the length of the vector (see Examples No. 5, 6).

4) The ending of the solution coincides with Example No. 7 - we know the number , which means that it is easy to find the angle itself:

Short solution and answer at the end of the lesson.

The second section of the lesson is devoted to the same dot product. Coordinates. It will be even easier than in the first part.

Dot product of vectors,
given by coordinates in an orthonormal basis

Answer:

Needless to say, dealing with coordinates is much more pleasant.

Example 14

Find the scalar product of vectors and if

This is a do-it-yourself example. Here you can use the associativity of the operation, that is, do not count, but immediately take the triple out of the scalar product and multiply by it last. Solution and answer at the end of the lesson.

At the end of the paragraph, a provocative example of calculating the length of a vector:

Example 15

Find lengths of vectors , if

Decision: again the method of the previous section suggests itself: but there is another way:

Let's find the vector:

And its length according to the trivial formula :

The scalar product is not relevant here at all!

How out of business it is when calculating the length of a vector:
Stop. Why not take advantage of the obvious length property of a vector? What can be said about the length of a vector? This vector is 5 times longer than the vector. The direction is opposite, but it does not matter, because we are talking about length. Obviously, the length of the vector is equal to the product module numbers per vector length:
- the sign of the module "eats" the possible minus of the number.

Thus:

Answer:

The formula for the cosine of the angle between vectors that are given by coordinates

Now we have complete information so that the previously derived formula for the cosine of the angle between vectors express in terms of vector coordinates:

Cosine of the angle between plane vectors and , given in the orthonormal basis , is expressed by the formula:
.

Cosine of the angle between space vectors, given in the orthonormal basis , is expressed by the formula:

Example 16

Three vertices of a triangle are given. Find (vertex angle ).

Decision: By condition, the drawing is not required, but still:

The required angle is marked with a green arc. We immediately recall the school designation of the angle: - special attention to middle letter - this is the vertex of the angle we need. For brevity, it could also be written simply.

From the drawing it is quite obvious that the angle of the triangle coincides with the angle between the vectors and , in other words: .

It is desirable to learn how to perform the analysis performed mentally.

Let's find the vectors:

Let's calculate the scalar product:

And the lengths of the vectors:

Cosine of an angle:

It is this order of the task that I recommend to dummies. More advanced readers can write the calculations "in one line":

Here is an example of a "bad" cosine value. The resulting value is not final, so there is not much point in getting rid of the irrationality in the denominator.

Let's find the angle:

If you look at the drawing, the result is quite plausible. To check the angle can also be measured with a protractor. Do not damage the monitor coating =)

Answer:

In the answer, do not forget that asked about the angle of the triangle(and not about the angle between the vectors), do not forget to indicate the exact answer: and the approximate value of the angle: found with a calculator.

Those who have enjoyed the process can calculate the angles, and make sure the canonical equality is true

Example 17

A triangle is given in space by the coordinates of its vertices. Find the angle between the sides and

This is a do-it-yourself example. Full solution and answer at the end of the lesson

A small final section will be devoted to projections, in which the scalar product is also “involved”:

Projection of a vector onto a vector. Vector projection onto coordinate axes.
Vector direction cosines

Consider vectors and :

We project the vector onto the vector , for this we omit from the beginning and end of the vector perpendiculars per vector (green dotted lines). Imagine that rays of light are falling perpendicularly on a vector. Then the segment (red line) will be the "shadow" of the vector. In this case, the projection of a vector onto a vector is the LENGTH of the segment. That is, PROJECTION IS A NUMBER.

This NUMBER is denoted as follows: , "large vector" denotes a vector WHICH project, "small subscript vector" denotes the vector ON THE which is projected.

The entry itself reads like this: “the projection of the vector “a” onto the vector “be””.

What happens if the vector "be" is "too short"? We draw a straight line containing the vector "be". And the vector "a" will be projected already to the direction of the vector "be", simply - on a straight line containing the vector "be". The same thing will happen if the vector "a" is set aside in the thirtieth kingdom - it will still be easily projected onto the line containing the vector "be".

If the angle between vectors spicy(as in the picture), then

If the vectors orthogonal, then (the projection is a point whose dimensions are assumed to be zero).

If the angle between vectors blunt(in the figure, mentally rearrange the arrow of the vector), then (the same length, but taken with a minus sign).

Set aside these vectors from one point:

Obviously, when moving a vector, its projection does not change

The scalar product of vectors (hereinafter in the text of the joint venture). Dear friends! The mathematics exam includes a group of problems for solving vectors. We have already considered some problems. You can see them in the "Vectors" category. In general, the theory of vectors is simple, the main thing is to study it consistently. Calculations and actions with vectors in the school mathematics course are simple, the formulas are not complicated. Look into . In this article, we will analyze tasks on the joint venture of vectors (included in the exam). Now "immersion" in the theory:

H To find the coordinates of a vector, you need to subtract from the coordinates of its endcorresponding coordinates of its beginning

And further:

*Vector length (modulus) is defined as follows:

These formulas must be memorized!!!

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 are positive, obviously. So 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 sharp (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 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 zero, and hence the joint venture is zero. This fact (consequence, conclusion) is used in solving many problems where we are talking about the mutual arrangement of vectors, including in problems included in the open bank of tasks in mathematics.

We formulate the statement: the scalar product is equal to zero if and only if the given vectors lie on perpendicular lines.

So, the formulas for the SP vectors are:

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:

Consider the tasks:

27724 Find the inner product of 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

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:

Let's plug them into the formula:

The angle between the vectors is 45 degrees.

Answer: 45

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 ,...,y n) 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 * y n.

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  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:

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 of 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.

So 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 (e l * e i) +  2 (e 2 * e i) +...+  n (e n * e i) = 0   i (e i * 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.