Thus, the length of the vector is calculated as the square root of the sum of the squares of its coordinates
. The length of an n-dimensional vector is calculated similarly
. If we remember that each coordinate of a vector is the difference between the coordinates of the end and the beginning, then we obtain the formula for the length of the segment, i.e. Euclidean distance between points.

Scalar product two vectors on a plane is the product of the lengths of these vectors and the cosine of the angle between them:
. It can be proven that the scalar product of two vectors = (x 1, x 2) and = (y 1 , y 2) is equal to the sum of the products of the corresponding coordinates of these vectors:
= x 1 * y 1 + x 2 * y 2 .

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

The operation of multiplying vectors by each other is similar to multiplying a row matrix by a column matrix. We emphasize that the result will be a number, not a vector.

The scalar product of vectors has the following properties (axioms):

1) Commutative property: X*Y=Y*X.

2) Distributive property with respect to addition: X(Y+Z) =X*Y+X*Z.

3) For any real number 
.

4)
, ifX is not a zero vector;
ifX is a zero vector.

A linear vector space in which a scalar product of vectors is given that satisfies the four corresponding axioms is called Euclidean linear vectorspace.

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

The vector length has the following properties:

1) |X| = 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 a scalar product. In fact, if
, That
. This fraction is not greater than one (according to the Cauchy-Bunyakovsky inequality), so from here we can find .

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

Let's look again at Figure 7.4. It can be seen from the figure that the cosine of the angle of the inclination of the vector to the horizontal axis can be calculated as
, and the cosine of the angleinclination of the vector to the vertical axis is as
. These numbers are usually called direction cosines. It is easy to verify that the sum of the squares of the direction cosines is always equal to one: cos 2 +cos 2 = 1. Similarly, the concepts of direction cosines can be introduced for spaces of higher dimensions.

Vector space basis

For vectors, we can define the concepts linear combination,linear dependence And independence similar to how these concepts were introduced for matrix rows. It is also true that if the vectors are linearly dependent, then at least one of them can be expressed linearly in terms of the others (i.e., it is a linear combination of them). The converse is also true: if one of the vectors is a linear combination of the others, then all these vectors together are linearly dependent.

Note that if among the vectors a l , a 2 ,...a m there is a zero vector, then this set of vectors is necessarily linearly dependent. In fact, we get l a l + 2 a 2 +...+ m a m = 0 if, for example, we equate the coefficient j at the zero vector to one, and all other coefficients to zero. In this case, not all coefficients will be equal to zero ( j ≠ 0).

In addition, if some part of the vectors from a set of vectors are linearly dependent, then all of these vectors are linearly dependent. In fact, if some vectors give a zero vector in their linear combination with coefficients that are not both zero, then the remaining vectors multiplied by the zero coefficients can be added to this sum of products, and it will still be a zero vector.

How to determine whether vectors are linearly dependent?

For example, let's take three vectors: a 1 = (1, 0, 1, 5), a 2 = (2, 1, 3, -2) and a 3 = (3, 1, 4, 3). Let's create a matrix from them, in which they will be columns:

Then the question of linear dependence will be reduced to determining the rank of this matrix. If it turns out to be equal to three, then all three columns are linearly independent, and if it turns out to be less, then this will indicate a linear dependence of the vectors.

Since the rank is 2, the vectors are linearly dependent.

Note that the solution to the problem could also begin with reasoning that is based on the definition of linear independence. Namely, create a vector equation  l a l + 2 a 2 + 3 a 3 = 0, which will take the form l *(1, 0, 1, 5) + 2 *(2, 1, 3, -2) + 3 *(3, 1, 4, 3) = (0, 0, 0, 0). Then we get a system of equations:

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

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

From the solved example it becomes clear that if we take the number of vectors greater than the dimension of space, then they will necessarily be linearly dependent. In fact, if we took five vectors in this example, we would get a 4 x 5 matrix, the rank of which could not be greater than four. Those. the maximum number of linearly independent columns would still not be more than four. Two, three or four four-dimensional vectors can be linearly independent, but five or more cannot. Consequently, no more than two vectors can be linearly independent on the plane. Any three vectors in two-dimensional space are linearly dependent. In three-dimensional space, any four (or more) vectors are always linearly dependent. And so on.

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

A set of n linearly independent vectors of an n-dimensional space R is called basis this space.

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

Proof. Let the vectors e l , e 2 ,...e n form a basis-dimensional space R. Let us prove that any vector X is a linear combination of these vectors. Since, together with vector X, the number of vectors will become (n +1), these (n +1) vectors will be linearly dependent, i.e. there are numbers l , 2 ,..., n ,, not simultaneously equal to zero, such that

 l e l + 2 e 2 +...+ n e n +Х = 0

In this case, 0, because otherwise we would get l e l + 2 e 2 +...+ n e n = 0, where not all coefficients l , 2 ,..., n are equal to zero. This means that the basis vectors would be linearly dependent. Therefore, we can divide both sides of the first equation by:

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

Х = -( l /)e l - ( 2 /)e 2 -...- ( n /)e n

Х = x l e l +x 2 e 2 +...+x n e n,

where x j = -( j /),
.

Now we prove that such a representation in the form of a linear combination is unique. Let's assume the opposite, i.e. that there is another representation:

X = y l e l +y 2 e 2 +...+y n e n

Let us subtract from it term by term the previously obtained expression:

0 = (y l – x 1)e l + (y 2 – x 2)e 2 +...+ (y n – x n)e n

Since the basis vectors are linearly independent, we obtain that (y j - x j) = 0,
, i.e. y j ​​= x j . So the expression turned out to be the same. The theorem has been proven.

The expression X = x l e l +x 2 e 2 +...+x n e n is called decomposition vector X based on e l, e 2,...e n, and numbers x l, x 2,...x n - coordinates vector x relative to this basis, or in this basis.

It can be proven that if nnonzero vectors of an n-dimensional Euclidean space are pairwise orthogonal, then they form a basis. In fact, let's multiply both sides of the equality l e l + 2 e 2 +...+ n e n = 0 by any vector e i. We get  l (e l *е i) +  2 (e 2 *е i) +...+  n (e n *е i) = 0   i (e i *е i) = 0   i = 0 for  i.

Vectors e l , e 2 ,...e n of n-dimensional Euclidean space form orthonormal basis, if these vectors are pairwise orthogonal and the norm of each of them is equal to one, i.e. if e i *e j = 0 for i≠j и |е i | = 1 fori.

Theorem (no proof). In every n-dimensional Euclidean space there is an orthonormal basis.

An example of an orthonormal basis is a system of n unit vectors e i , for which the i-th component is equal to one and the remaining components are equal to zero. Each such vector is called ort. For example, the vector vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) form the basis of three-dimensional space.

Lecture: Vector coordinates; scalar product of vectors; angle between vectors

Vector coordinates

So, as mentioned earlier, a vector is a directed segment that has its own beginning and end. If the beginning and end are represented by certain points, then they have their own coordinates on the plane or in space.

If each point has its own coordinates, then we can get the coordinates of the whole vector.

Let's say we have a vector whose beginning and end have the following designations and coordinates: A(A x ; Ay) and B(B x ; By)

To obtain the coordinates of a given vector, it is necessary to subtract the corresponding coordinates of the beginning from the coordinates of the end of the vector:

To determine the coordinates of a vector in space, use the following formula:

Dot product of vectors

There are two ways to define the concept of a scalar product:

• Geometric method. According to it, the scalar product is equal to the product of the values ​​of these modules and the cosine of the angle between them.

• Algebraic meaning. From the point of view of algebra, the scalar product of two vectors is a certain quantity that is obtained as a result of the sum of the products of the corresponding vectors.

If the vectors are given in space, then you should use a similar formula:

Properties:

• If you multiply two identical vectors scalarly, then their scalar product will not be negative:

• If the scalar product of two identical vectors turns out to be equal to zero, then these vectors are considered zero:

• If a certain vector is multiplied by itself, then the scalar product will be equal to the square of its modulus:

• The scalar product has a communicative property, that is, the scalar product will not change if the vectors are rearranged:

• The scalar product of non-zero vectors can be equal to zero only if the vectors are perpendicular to each other:

• For a scalar product of vectors, the commutative law is valid in the case of multiplying one of the vectors by a number:

• With a scalar product, you can also use the distributive property of multiplication:

Angle between vectors

In the case of a plane problem, the scalar product of vectors a = (a x; a y) and b = (b x; b y) can be found using the following formula:

a b = a x b x + a y b y

Formula for the scalar product of vectors for spatial problems

In the case of a spatial problem, the scalar product of vectors a = (a x; a y; a z) and b = (b x; b y; b z) can be found using the following formula:

a b = a x b x + a y b y + a z b z

Formula for the scalar product of n-dimensional vectors

In the case of an n-dimensional space, the scalar product of vectors a = (a 1; a 2; ...; a n) and b = (b 1; b 2; ...; b n) can be found using the following formula:

a b = a 1 b 1 + a 2 b 2 + ... + a n b n

Properties of the scalar product of vectors

1. The scalar product of a vector with itself is always greater than or equal to zero:

2. The scalar product of a vector with itself is equal to zero if and only if the vector is equal to the zero vector:

a · a = 0<=>a = 0

3. The scalar product of a vector with 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 problems for calculating the scalar product of vectors

Examples of calculating the scalar product of vectors for plane problems

Find the scalar product of the vectors a = (1; 2) and b = (4; 8).

Solution: a · b = 1 · 4 + 2 · 8 = 4 + 16 = 20.

Find the scalar product of vectors a and b if their lengths |a| = 3, |b| = 6, and the angle between the vectors is 60˚.

Solution: a · b = |a| · |b| cos α = 3 · 6 · cos 60˚ = 9.

Find the scalar product of the vectors p = a + 3b and q = 5a - 3 b if their lengths |a| = 3, |b| = 2, and the angle between vectors a and b is 60˚.

Solution:

p q = (a + 3b) (5a - 3b) = 5 a a - 3 a b + 15 b a - 9 b b =

5 |a| 2 + 12 a · b - 9 |b| 2 = 5 3 2 + 12 3 2 cos 60˚ - 9 2 2 = 45 +36 -36 = 45.

An example of calculating the scalar product of vectors for spatial problems

Find the scalar product of the vectors a = (1; 2; -5) and b = (4; 8; 1).

Solution: a · b = 1 · 4 + 2 · 8 + (-5) · 1 = 4 + 16 - 5 = 15.

An example of calculating the dot product for n-dimensional vectors

Find the scalar product of the vectors a = (1; 2; -5; 2) and b = (4; 8; 1; -2).

Solution: a · b = 1 · 4 + 2 · 8 + (-5) · 1 + 2 · (-2) = 4 + 16 - 5 -4 = 11.

13. The cross product of vectors and a vector is called third vector , defined as follows:

2) perpendicular, perpendicular. (1"")

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

Designate: .

Physical meaning of the vector product

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

Moreover, if we move it to point O, then the triple should be oriented as a basis vector.

Dot product of vectors

We continue to deal with vectors. At the first lesson Vectors for dummies We looked at 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 strongly recommend reading the above introductory article, since in order to master the material you need to be familiar with the terms and notations I use, have basic knowledge about vectors and be able to solve basic 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 activity.. Try not to skip the examples; they come with a useful bonus - practice will help you consolidate the material you have covered and get better at solving common problems in analytical geometry.

Addition of vectors, multiplication of a vector by a number.... It would be naive to think that mathematicians haven't come up with something else. In addition to the actions already discussed, there are a number of other operations with vectors, namely: dot product of vectors, vector product of vectors And mixed product of vectors. The scalar product of vectors is familiar to us from school; the other two products traditionally belong to the course of higher mathematics. The topics are simple, the algorithm for solving many problems is straightforward and understandable. The only thing. There is a decent amount of information, so it is undesirable to try to master and solve EVERYTHING 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 materials selectively, in a certain sense, “get” the missing knowledge, for you I will be a harmless Count Dracula =)

Let’s finally open the door and watch with enthusiasm 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 a 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 detail. Let's consider free nonzero vectors and . If you plot these vectors from an arbitrary point, you will get a picture that many have already imagined mentally:

I admit, 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; for practical problems, in principle, it is of no use to us. Also HERE AND HEREIN I will ignore zero vectors in places due to their low practical significance. I made a reservation specifically for advanced site visitors who may reproach me for the theoretical incompleteness of some subsequent statements.

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

In the literature, the angle symbol is often skipped 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 this is a quite strict definition.

We focus on essential information:

Designation: the scalar product is denoted by or simply.

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

Just a couple of warm-up examples:

Example 1

Solution: We use the formula . In this case:

Answer:

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

From a purely 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, a scalar product always has a certain physical meaning, that is, after the result one or another physical unit must be indicated. A canonical example of calculating the work of a force can be found in any textbook (the formula is exactly a scalar 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 equal to .

This is an example for you to solve on your own, 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's 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 only depend on the value of the cosine.

Note: To better understand the information below, it is better to study the cosine graph in the manual Function graphs and 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 corner between vectors spicy: (from 0 to 90 degrees), then , And the dot product will be positive co-directed, then the angle between them is considered zero, and the scalar product will also be positive. Since , the formula simplifies: .

2) If corner between vectors blunt: (from 90 to 180 degrees), then , and correspondingly, dot product is negative: . Special case: if the vectors opposite directions, then the angle between them is considered expanded: (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 co-directional.

2) If , then the angle between these vectors is obtuse. Alternatively, the vectors are in opposite directions.

But the third case is of particular interest:

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

! Note : Let's repeat basics of mathematical logic: A double-sided logical consequence icon is usually read "if and only if", "if and only if". As you can see, the arrows are directed in both directions - “from this follows this, and vice versa - from that follows this.” What, by the way, is the difference from the one-way follow icon? The icon states only that, that “from this follows this”, and it is not a fact that the opposite is true. For example: , but not every animal is a panther, so in this case you cannot use the icon. 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 an entry will be correct, and even more appropriate than .

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

Properties of the dot product

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 aligned 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 we can obtain a formula for calculating the length of the vector:

So far it seems unclear, but the objectives of the lesson will put everything in its place. To solve the problems we also need properties of the dot product.

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

1) – commutative or commutative scalar product law.

2) – distribution or distributive scalar product law. Simply, you can open the brackets.

3) – associative or associative scalar product law. The constant can be derived from the scalar product.

Often, all kinds of properties (which also need to be proven!) are perceived by students as unnecessary rubbish, 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 rearranging the factors does not change the product: . I must warn you that in higher mathematics it is easy to mess things up with such an approach. So, for example, the commutative property is not true for algebraic matrices. It is also not true for vector product of vectors. Therefore, at a minimum, it is better to delve into any properties that you come across in a higher mathematics course in order to understand what you can do and what you cannot do.

Example 3

.

Solution: First, let's clarify the situation with the vector. What is this anyway? The sum of vectors is a well-defined vector, which is denoted by . A 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 the condition gives similar parameters for vectors, so we will take a different route:

(1) Substitute the expressions of the vectors.

(2) We open the brackets according to the rule for multiplying polynomials; a vulgar tongue twister can be found in the article Complex numbers or Integrating 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) We present similar terms: .

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

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

Answer:

A negative value of the scalar product states the fact that the angle between the vectors is obtuse.

The problem is typical, here is an example for solving it yourself:

Example 4

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

Now another common task, just for the new formula for the length of a vector. The notation here will be a little overlapping, so for clarity I’ll rewrite it with a different letter:

Example 5

Find the length of the vector if .

Solution will be as follows:

(1) We supply the expression for the vector .

(2) We use the length formula: , and the whole expression ve acts as the vector “ve”.

(3) We use the school formula for the square of the sum. Notice how it works here in a curious way: – in fact, it is the square of the difference, and, in fact, that’s how it is. Those who wish can rearrange the vectors: - the same thing happens, up to the 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 an example for you to solve on your own. Full solution and answer at the end of the lesson.

We continue to squeeze useful things out of the dot product. Let's look at our formula again . Using 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 we can calculate the cosine of the angle between these vectors, and, consequently, the angle itself.

Is a dot product a number? Number. Are vector lengths numbers? Numbers. This means that 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 .

Solution: We use the formula:

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

So if , That:

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

Answer:

Again, do not forget to indicate the dimensions - radians and degrees. Personally, in order to obviously “resolve all questions”, I prefer to indicate both (unless the condition, of course, requires presenting the answer only in radians or only in degrees).

Now you can independently cope with a more complex task:

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 it is multi-step.
Let's look at the solution algorithm:

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

2) 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 it’s easy to find the angle itself:

A short solution and answer at the end of the lesson.

The second section of the lesson is devoted to the same scalar 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 an example for you to solve on your own. Here you can use the associativity of the operation, that is, do not count , but immediately take the triple outside the scalar product and multiply it by it last. The solution and answer are at the end of the lesson.

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

Example 15

Find the lengths of vectors , If

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

Let's find the vector:

And its length according to the trivial formula :

The dot product is not relevant here at all!

It is also not useful when calculating the length of a vector:
Stop. Shouldn't we take advantage of the obvious property of vector length? What can you say about the length of the vector? This vector is 5 times longer than the vector. The direction is opposite, but this 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 modulus sign “eats” the possible minus of the number.

Thus:

Answer:

Formula for the cosine of the angle between vectors that are specified by coordinates

Now we have complete information to use the previously derived formula for the cosine of the angle between vectors express through vector coordinates:

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

Cosine of the angle between space vectors, specified in an orthonormal basis, expressed by the formula:

Example 16

Given three vertices of a triangle. Find (vertex angle).

Solution: According to the conditions, the drawing is not required, but still:

The required angle is marked with a green arc. Let us immediately remember the school designation of an angle: – special attention to average letter - this is the vertex of the angle we need. For brevity, you could also write 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 advisable to learn to perform the analysis mentally.

Let's find the vectors:

Let's calculate the scalar product:

And the lengths of the vectors:

Cosine of angle:

This is exactly the order of completing the task that I recommend for 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 little point in getting rid of irrationality in the denominator.

Let's find the angle itself:

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 cover =)

Answer:

In the answer we do not forget that asked about the angle of a 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 using a calculator.

Those who have enjoyed the process can calculate the angles and verify the validity of the canonical equality

Example 17

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

This is an example for you to solve on your own. Full solution and answer at the end of the lesson

A short final section will be devoted to projections, which also involve a scalar product:

Projection of a vector onto a vector. Projection of a vector onto coordinate axes.
Direction cosines of a vector

Consider the vectors and :

Let's project the vector onto the vector; to do this, from the beginning and end of the vector we omit perpendiculars to vector (green dotted lines). Imagine that rays of light fall perpendicularly onto the vector. Then the segment (red line) will be the “shadow” of the vector. In this case, the projection of the vector onto the vector is the LENGTH of the segment. That is, PROJECTION IS A NUMBER.

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

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

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

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

Let us plot these vectors from one point:

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