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

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, we do not need it. 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 can be done and what cannot be done.

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 express the previously derived formula for the cosine of the angle between vectors through the coordinates of the vectors:

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 how 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

The cross product and dot product make it easy to calculate the angle between vectors. Let two vectors $\overline(a)$ and $\overline(b)$ be given, the oriented angle between them is equal to $\varphi$. Let's calculate the values ​​$x = (\overline(a),\overline(b))$ and $y = [\overline(a),\overline(b)]$. Then $x=r\cos\varphi$, $y=r\sin\varphi$, where $r=|\overline(a)|\cdot|\overline(b)|$, and $\varphi$ is the desired angle, that is, the point $(x, y)$ has a polar angle equal to $\varphi$, and therefore $\varphi$ can be found as atan2(y, x).

Area of ​​a triangle

Since the cross product contains the product of two vector lengths and the cosine of the angle between them, the cross product can be used to calculate the area of ​​triangle ABC:

$ S_(ABC) = \frac(1)(2)|[\overline(AB),\overline(AC)]| $.

Belonging of a point to a line

Let a point $P$ and a line $AB$ (given by two points $A$ and $B$) be given. It is necessary to check whether a point belongs to the line $AB$.

A point belongs to the line $AB$ if and only if the vectors $AP$ and $AB$ are collinear, that is, if $ [ \overline(AP), \overline(AB)]=0 $.

Belonging of a point to a ray

Let a point $P$ and a ray $AB$ be given (defined by two points - the beginning of the ray $A$ and a point on the ray $B$). It is necessary to check whether a point belongs to the ray $AB$.

To the condition that the point $P$ belongs to the straight line $AB$, it is necessary to add an additional condition - the vectors $AP$ and $AB$ are codirectional, that is, they are collinear and their scalar product is non-negative, that is, $(\overline(AB), \overline(AP ))\ge 0$.

Belonging of a point to a segment

Let a point $P$ and a segment $AB$ be given. It is necessary to check whether a point belongs to the segment $AB$.

In this case, the point must belong to both ray $AB$ and ray $BA$, so the following conditions must be checked:

$[\overline(AP), \overline(AB)]=0$,

$(\overline(AB), \overline(AP))\ge 0$,

$(\overline(BA), \overline(BP))\ge 0$.

Distance from point to line

Let a point $P$ and a line $AB$ (given by two points $A$ and $B$) be given. It is necessary to find the distance from the point of the line $AB$.

Consider triangle ABP. On the one hand, its area is equal to $S_(ABP)=\frac(1)(2)|[\overline(AB),\overline(AP) ]|$.

On the other hand, its area is equal to $S_(ABP)= \frac(1)(2)h |AB|$, where $h$ is the height dropped from the point $P$, that is, the distance from $P$ to $ AB$. Where $h=|[\overline(AB),\overline(AP)]|/|AB|$.

Distance from point to beam

Let a point $P$ and a ray $AB$ be given (defined by two points - the beginning of the ray $A$ and a point on the ray $B$). It is necessary to find the distance from a point to a ray, that is, the length of the shortest segment from point $P$ to any point on the ray.

This distance is equal to either the length $AP$ or the distance from point $P$ to line $AB$. Which of the cases takes place can be easily determined by the relative position of the ray and the point. If the angle PAB is acute, that is, $(\overline(AB),\overline(AP)) > 0$, then the answer will be the distance from the point $P$ to the straight line $AB$, otherwise the answer will be the length of the segment $AB$.

Distance from point to segment

Let a point $P$ and a segment $AB$ be given. It is necessary to find the distance from $P$ to the segment $AB$.

If the base of the perpendicular dropped from $P$ onto the line $AB$ falls on the segment $AB$, which can be verified by the conditions

$(\overline(AP), \overline(AB))\ge 0$,

$(\overline(BP), \overline(BA))\ge 0$,

then the answer will be the distance from point $P$ to line $AB$. Otherwise the distance will be equal to $\min(AP, BP)$.

There will also be problems for you to solve on your own, to which you can see the answers.

If in the problem both the lengths of the vectors and the angle between them are presented “on a silver platter,” then the condition of the problem and its solution look like this:

Example 1. Vectors are given. Find the scalar product of vectors if their lengths and the angle between them are represented by the following values:

Another definition is also valid, completely equivalent to definition 1.

Definition 2. The scalar product of vectors is a number (scalar) equal to the product of the length of one of these vectors and the projection of another vector onto the axis determined by the first of these vectors. Formula according to definition 2:

We will solve the problem using this formula after the next important theoretical point.

Definition of the scalar product of vectors in terms of coordinates

The same number can be obtained if the vectors being multiplied are given their coordinates.

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

On surface

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

then the scalar product of these vectors is equal to the sum of pairwise products of their corresponding coordinates:

.

Example 2. Find the numerical value of the projection of the vector onto the axis parallel to the vector.

Solution. We find the scalar product of vectors by adding the pairwise products of their coordinates:

Now we need to equate the resulting scalar product to the product of the length of the vector and the projection of the vector onto an axis parallel to the vector (in accordance with the formula).

We find the length of the vector as the square root of the sum of the squares of its coordinates:

.

We create an equation and solve it:

Answer. The required numerical value is minus 8.

In space

If two vectors and in space are defined by their three Cartesian rectangular coordinates

,

then the scalar product of these vectors is also equal to the sum of pairwise products of their corresponding coordinates, only there are already three coordinates:

.

The task of finding the scalar product using the method considered is after analyzing the properties of the scalar product. Because in the problem you will need to determine what angle the multiplied vectors form.

Properties of the scalar product of vectors

Algebraic properties

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

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

3. (distributive property relative to the sum of vectors: the scalar product of the sum of two vectors by the third vector is equal to the sum of the scalar products of the first vector by the third vector and the second vector by the third vector).

4. (scalar square of vector greater than zero), if is a nonzero vector, and , if is a zero vector.

Geometric properties

In the definitions of the operation under study, we have already touched on the concept of an angle between two vectors. It's time to clarify this concept.

In the figure above you can see two vectors that are brought to a common origin. And the first thing you need to pay attention to is that there are two angles between these vectors - φ 1 And φ 2 . Which of these angles appears in the definitions and properties of the scalar product of vectors? The sum of the considered angles is 2 π and therefore the cosines of these angles are equal. The definition of a dot product includes only the cosine of the angle, and not the value of its expression. But the properties only consider one angle. And this is the one of the two angles that does not exceed π , that is, 180 degrees. In the figure this angle is indicated as φ 1 .

1. Two vectors are called orthogonal And the angle between these vectors is straight (90 degrees or π /2 ), if the scalar product of these vectors is zero :

.

Orthogonality in vector algebra is the perpendicularity of two vectors.

2. Two non-zero vectors make up sharp corner (from 0 to 90 degrees, or, which is the same - less π dot product is positive .

3. Two non-zero vectors make up obtuse angle (from 90 to 180 degrees, or, what is the same - more π /2) if and only if they dot product is negative .

Example 3. The coordinates are given by the vectors:

.

Calculate the scalar products of all pairs of given vectors. What angle (acute, right, obtuse) do these pairs of vectors form?

Solution. We will calculate by adding the products of the corresponding coordinates.

We got a negative number, so the vectors form an obtuse angle.

We got a positive number, so the vectors form an acute angle.

We got zero, so the vectors form a right angle.

We got a positive number, so the vectors form an acute angle.

.

We got a positive number, so the vectors form an acute angle.

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

Example 4. Given the lengths of two vectors and the angle between them:

.

Determine at what value of the number the vectors and are orthogonal (perpendicular).

Solution. Let's multiply the vectors using the rule for multiplying polynomials:

Now let's calculate each term:

.

Let’s create an equation (the product is equal to zero), add similar terms and solve the equation:

Answer: we got the value λ = 1.8, at which the vectors are orthogonal.

Example 5. Prove that the vector orthogonal (perpendicular) to the vector

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

.

To do this, you need to multiply each term (term) of the first polynomial by each term of the second and add the resulting products:

.

In the resulting result, the fraction is reduced by. The following result is obtained:

Conclusion: as a result of multiplication we got zero, therefore, the orthogonality (perpendicularity) of the vectors is proven.

Solve the problem yourself and then see the solution

Example 6. The lengths of the vectors and are given, and the angle between these vectors is π /4 . Determine at what value μ vectors and are mutually perpendicular.

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

Matrix representation of the dot product of vectors and the product of n-dimensional vectors

Sometimes it is advantageous for clarity to represent two multiplied vectors in the form of matrices. Then the first vector is represented as a row matrix, and the second - as a column matrix:

Then the scalar product of vectors will be the product of these matrices :

The result is the same as that obtained by the method we have already considered. We got one single number, and the product of a row matrix by a column matrix is ​​also one single number.

It is convenient to represent the product of abstract n-dimensional vectors in matrix form. Thus, the product of two four-dimensional vectors will be the product of a row matrix with four elements by a column matrix also with four elements, the product of two five-dimensional vectors will be the product of a row matrix with five elements by a column matrix also with five elements, and so on.

Example 7. Find scalar products of pairs of vectors

,

using matrix representation.

Solution. The first pair of vectors. We represent the first vector as a row matrix, and the second as a column matrix. We find the scalar product of these vectors as the product of a row matrix and a column matrix:

We similarly represent the second pair and find:

As you can see, the results were the same as for the same pairs from example 2.

Angle between two vectors

The derivation of the formula for the cosine of the angle between two vectors is very beautiful and concise.

To express the dot product of vectors

(1)

in coordinate form, we first find the scalar product of the unit vectors. The scalar product of a vector with itself by definition:

What is written in the formula above means: the scalar product of a vector with itself is equal to the square of its length. The cosine of zero is equal to one, so the square of each unit will be equal to one:

Since vectors

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

Now let's perform the multiplication of vector polynomials:

We substitute the values ​​of the corresponding scalar products of the unit vectors into the right side of the equality:

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

Example 8. Three points are given A(1;1;1), B(2;2;1), C(2;1;2).

Find the angle.

Solution. Finding the coordinates of the vectors:

,

.

Using the cosine angle formula we get:

Hence, .

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

Example 9. Two vectors are given

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

2.Difference

Definition 1

The scalar product of vectors is a number equal to the product of the dynes of these vectors and the cosine of the angle between them.

The notation for the product of vectors a → and b → has the form a → , b → . Let's transform it into the formula:

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

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

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

Definition 2

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

Calculated by the formula:

a → , b → = a → · b → · cos a → , b → ^ .

The notation a → , b → = a → · b → · cos a → , b → ^ = a → · n p a → b → = b → · n p b → a → shows that n p b → a → is the numerical projection of a → onto b → , n p a → a → - projection of b → onto a →, respectively.

Let us formulate the definition of a product for two vectors:

The scalar product of two vectors a → by b → is called the product of the length of the vector a → by the projection b → by the direction of a → or the product of the length b → by the projection a →, respectively.

Dot product in coordinates

The scalar product can be calculated through the coordinates of vectors in a given plane or in space.

The scalar product of two vectors on a plane, in three-dimensional space, is called the sum of the coordinates of given vectors a → and b →.

When calculating the scalar product of given vectors a → = (a x , a y) , b → = (b x , b y) on the plane in the Cartesian system, use:

a → , b → = a x b x + a y b y ,

for three-dimensional space the expression is applicable:

a → , b → = a x · b x + a y · b y + a z · b z .

In fact, this is the third definition of the scalar product.

Let's prove it.

Evidence 1

To prove it, we use a → , b → = a → · b → · cos a → , b → ^ = a x · b x + a y · b y for vectors a → = (a x , a y) , b → = (b x , b y) on Cartesian system.

Vectors should be set aside

O A → = a → = a x , a y and O B → = b → = b x , b y .

Then the length of the vector A B → will be equal to A B → = O B → - O A → = b → - a → = (b x - a x , b y - a y) .

Consider triangle O A B .

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

According to the condition, it is clear that O A = a → , O B = b → , A B = b → - a → , ∠ A O B = a → , b → ^ , which means we write the formula for finding the angle between vectors differently

b → - a → 2 = a → 2 + b → 2 - 2 · a → · b → · cos (a → , b → ^) .

Then from the first definition it follows that b → - a → 2 = a → 2 + b → 2 - 2 · (a → , b →) , which means (a → , b →) = 1 2 · (a → 2 + b → 2 - b → - a → 2) .

Applying the formula for calculating the length of vectors, we get:
a → , b → = 1 2 · ((a 2 x + a y 2) 2 + (b 2 x + b y 2) 2 - ((b x - a x) 2 + (b y - a y) 2) 2) = = 1 2 (a 2 x + a 2 y + b 2 x + b 2 y - (b x - a x) 2 - (b y - a y) 2) = = a x b x + a y b y

Let us prove the equalities:

(a → , b →) = a → b → cos (a → , b → ^) = = a x b x + a y b y + a z b z

– respectively for vectors of three-dimensional space.

The scalar product of vectors with coordinates says that the scalar square of a vector is equal to the sum of the squares of its coordinates in space and on the plane, respectively. a → = (a x , a y , a z) , b → = (b x , b y , b z) and (a → , a →) = a x 2 + a y 2 .

Dot product and its properties

There are properties of the dot product that apply to a →, b →, and c →:

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

Example 1

The properties are explainable thanks to the definition of the scalar product on the plane and the properties of addition and multiplication of real numbers.

Prove the commutative property (a → , b →) = (b → , a →) . From the definition we have that (a → , b →) = a y · b y + a y · b y and (b → , a →) = b x · a x + b y · a y .

By the property of commutativity, the equalities a x · b x = b x · a x and a y · b y = b y · a y are true, which means a x · b x + a y · b y = b x · a x + b y · a y .

It follows that (a → , b →) = (b → , a →) . Q.E.D.

Distributivity is valid for any numbers:

(a (1) → + a (2) → + . . . + a (n) → , b →) = (a (1) → , b →) + (a (2) → , b →) + . . . + (a (n) → , b →)

and (a → , b (1) → + b (2) → + . . + b (n) →) = (a → , b (1) →) + (a → , b (2) →) + . . . + (a → , b → (n)) ,

hence we have

(a (1) → + a (2) → + . . . + a (n) → , b (1) → + b (2) → + . . . + b (m) →) = = (a ( 1) → , b (1) →) + (a (1) → , b (2) →) + . . . + (a (1) → , b (m) →) + + (a (2) → , b (1) →) + (a (2) → , b (2) →) + . . . + (a (2) → , b (m) →) + . . . + + (a (n) → , b (1) →) + (a (n) → , b (2) →) + . . . + (a (n) → , b (m) →)

Dot product with examples and solutions

Any problem of this kind is solved using the properties and formulas relating to the scalar product:

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 example solutions.

Example 2

The length of a → is 3, the length of b → is 7. Find the dot product if the angle has 60 degrees.

Solution

By condition, we have all the data, so we calculate it using the formula:

(a → , b →) = a → b → cos (a → , b → ^) = 3 7 cos 60 ° = 3 7 1 2 = 21 2

Answer: (a → , b →) = 21 2 .

Example 3

Given vectors a → = (1 , - 1 , 2 - 3) , b → = (0 , 2 , 2 + 3) . What is the scalar product?

Solution

This example considers the formula for calculating coordinates, since they are specified in the problem statement:

(a → , b →) = a x · b x + a y · b y + a z · b z = = 1 · 0 + (- 1) · 2 + (2 + 3) · (2 ​​+ 3) = = 0 - 2 + ( 2 - 9) = - 9

Answer: (a → , b →) = - 9

Example 4

Find the scalar product of A B → and A C →. Points A (1, - 3), B (5, 4), C (1, 1) are given on the coordinate plane.

Solution

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

A B → = (5 - 1, 4 - (- 3)) = (4, 7) A C → = (1 - 1, 1 - (- 3)) = (0, 4)

Substituting into the formula using coordinates, we get:

(A B →, A C →) = 4 0 + 7 4 = 0 + 28 = 28.

Answer: (A B → , A C →) = 28 .

Example 5

Given vectors a → = 7 · m → + 3 · n → and b → = 5 · m → + 8 · n → , find their product. m → equals 3 and n → equals 2 units, they are perpendicular.

Solution

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

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

We take the coefficient out of the sign of the product and get:

(7 m → , 5 m →) + (7 m → , 8 n →) + (3 n → , 5 m →) + (3 n → , 8 n →) = = 7 · 5 · (m → , m →) + 7 · 8 · (m → , n →) + 3 · 5 · (n → , m →) + 3 · 8 · (n → , n →) = = 35 · (m → , m →) + 56 · (m → , n →) + 15 · (n → , m →) + 24 · (n → , n →)

By the property of commutativity we transform:

35 · (m → , m →) + 56 · (m → , n →) + 15 · (n → , m →) + 24 · (n → , n →) = = 35 · (m → , m →) + 56 · (m → , n →) + 15 · (m → , n →) + 24 · (n → , n →) = = 35 · (m → , m →) + 71 · (m → , n → ) + 24 · (n → , n →)

As a result we get:

(a → , b →) = 35 · (m → , m →) + 71 · (m → , n →) + 24 · (n → , n →).

Now we apply the formula for the scalar product with the angle specified by the condition:

(a → , b →) = 35 · (m → , m →) + 71 · (m → , n →) + 24 · (n → , n →) = = 35 · m → 2 + 71 · m → · n → · cos (m → , n → ^) + 24 · n → 2 = 35 · 3 2 + 71 · 3 · 2 · cos π 2 + 24 · 2 2 = 411 .

Answer: (a → , b →) = 411

If there is a numerical projection.

Example 6

Find the scalar product of a → and b →. Vector a → has coordinates a → = (9, 3, - 3), projection b → with coordinates (- 3, - 1, 1).

Solution

By condition, the vectors a → and the projection b → are oppositely directed, because a → = - 1 3 · n p a → b → → , which means the projection b → corresponds to the length n p a → b → → , and with the “-” sign:

n p a → b → → = - n p a → b → → = - (- 3) 2 + (- 1) 2 + 1 2 = - 11 ,

Substituting into the formula, we get the expression:

(a → , b →) = a → · n p a → b → → = 9 2 + 3 2 + (- 3) 2 · (- 11) = - 33 .

Answer: (a → , b →) = - 33 .

Problems with a known scalar product, where it is necessary to find the length of a vector or a numerical projection.

Example 7

What value should λ take for a given scalar product a → = (1, 0, λ + 1) and b → = (λ, 1, λ) will be equal to -1.

Solution

From the formula it is clear that it is necessary to find the sum of the products of coordinates:

(a → , b →) = 1 λ + 0 1 + (λ + 1) λ = λ 2 + 2 λ .

Given we have (a → , b →) = - 1 .

To find λ, we calculate the equation:

λ 2 + 2 · λ = - 1, hence λ = - 1.

Answer: λ = - 1.

Physical meaning of the scalar product

Mechanics considers the application of the dot product.

When A works with a constant force F → a moving body from a point M to N, you can find the product of the lengths of the vectors F → and M N → with the cosine of the angle between them, which means the work is equal to the product of the force and displacement vectors:

A = (F → , M N →) .

Example 8

The movement of a material point by 3 meters under the influence of a force equal to 5 Ntons is directed at an angle of 45 degrees relative to the axis. Find A.

Solution

Since work is the product of the force vector and displacement, it means that based on the condition F → = 5, S → = 3, (F →, S → ^) = 45 °, we obtain A = (F →, S →) = F → · S → · cos (F → , S → ^) = 5 · 3 · cos (45 °) = 15 2 2 .

Answer: A = 15 2 2 .

Example 9

A material point, moving from M (2, - 1, - 3) to N (5, 3 λ - 2, 4) under the force F → = (3, 1, 2), did work equal to 13 J. Calculate the length of movement.

Solution

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

Using the formula for finding work with vectors F → = (3, 1, 2) and M N → = (3, 3 λ - 1, 7), we obtain A = (F ⇒, M N →) = 3 3 + 1 (3 λ - 1) + 2 7 = 22 + 3 λ.

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

To find the length of movement M N →, apply the formula and substitute the values:

M N → = 3 2 + (- 10) 2 + 7 2 = 158.

Answer: 158.

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