Calculation of the cosine of the angle between vectors. Dot product of vectors

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 on 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 come with a useful bonus - the 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, to “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 given fact is written as a double inequality:

(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

Solution: 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 the problems of physics, the scalar product always has a certain physical meaning, that is, after the result, you need to specify one or another physical unit. 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 independent 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: . Lengths 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 corner 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 corner between vectors stupid: (from 90 to 180 degrees), then , and correspondingly, dot product is negative: . A special case: if 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 corner 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 has a large practical significance , 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 .

In this way, 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 junk, 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 the 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

Solution: 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:

Negative meaning 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 .

Solution 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) Use school formula sum squared. 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 .

Solution: 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:

Inverse values 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 can deal with more difficult 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 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

Solution: 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.

In this way:

Answer:

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

now we have full 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 ).

Solution: 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 THE 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 stupid(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

Instruction

Let two nonzero vectors are given on the plane, plotted from one point: vector A with coordinates (x1, y1) B with coordinates (x2, y2). Corner between them is denoted as θ. To find the degree measure of the angle θ, you need to use the definition of the scalar product.

The scalar product of two non-zero vectors is a number equal to the product of the lengths of these vectors and the cosine of the angle between them, that is, (A,B)=|A|*|B|*cos(θ). Now you need to express the cosine of the angle from this: cos(θ)=(A,B)/(|A|*|B|).

The scalar product can also be found using the formula (A,B)=x1*x2+y1*y2, since the product of two non-zero vectors is equal to the sum of the products of the corresponding vectors. If the scalar product of non-zero vectors is equal to zero, then the vectors are perpendicular (the angle between them is 90 degrees) and further calculations can be omitted. If the scalar product of two vectors is positive, then the angle between these vectors acute, and if negative, then the angle is obtuse.

Now calculate the lengths of vectors A and B using the formulas: |A|=√(x1²+y1²), |B|=√(x2²+y2²). The vector length is calculated as Square root from the sum of the squares of its coordinates.

Substitute the found values of the scalar product and the lengths of the vectors into the formula for the angle obtained in step 2, that is, cos(θ)=(x1*x2+y1*y2)/(√(x1²+y1²)+√(x2²+y2²)). Now, knowing the value of , to find the degree measure of the angle between vectors you need to use the Bradis table or take from this: θ=arccos(cos(θ)).

If the vectors A and B are given in three-dimensional space and have coordinates (x1, y1, z1) and (x2, y2, z2), respectively, then one more coordinate is added when finding the cosine of the angle. In this case cosine: cos(θ)=(x1*x2+y1*y2+z1*z2)/(√(x1²+y1²+z1²)+√(x2²+y2²+z2²)).

Useful advice

If two vectors are not plotted from one point, then to find the angle between them by parallel translation, you need to combine the beginnings of these vectors.
The angle between two vectors cannot be greater than 180 degrees.

Sources:

how to calculate angle between vectors
Angle between line and plane

To solve many problems, both applied and theoretical, in physics and linear algebra, it is necessary to calculate the angle between vectors. This seemingly simple task can cause a lot of difficulties if you do not clearly understand the essence of the scalar product and what value appears as a result of this product.

Instruction

The angle between vectors in a linear vector space is the minimum angle at , at which the codirection of the vectors is achieved. One of the vectors is carried around its starting point. From the definition, it becomes obvious that the value of the angle cannot exceed 180 degrees (see the step).

In this case, it is quite rightly assumed that in a linear space, when the vectors are transferred in parallel, the angle between them does not change. Therefore, for the analytical calculation of the angle, the spatial orientation of the vectors does not matter.

The result of the dot product is a number, otherwise a scalar. Remember (this is important to know) in order to prevent errors in further calculations. The formula for the scalar product, located on a plane or in the space of vectors, has the form (see the figure for the step).

If the vectors are located in space, then perform the calculation in a similar way. The only thing will be the appearance of the term in the dividend - this is the term for the applicate, i.e. the third component of the vector. Accordingly, when calculating the modulus of vectors, the z component must also be taken into account, then for vectors located in space, the last expression is transformed as follows (see Figure 6 to the step).

A vector is a line segment with a given direction. The angle between vectors has a physical meaning, for example, when finding the length of the projection of a vector onto an axis.

Instruction

Angle between two non-zero vectors using dot product calculation. By definition, the product is equal to the product of the lengths and the angle between them. On the other hand, the inner product for two vectors a with coordinates (x1; y1) and b with coordinates (x2; y2) is calculated: ab = x1x2 + y1y2. Of these two ways, the dot product is easy to angle between vectors.

Find the lengths or modules of the vectors. For our vectors a and b: |a| = (x1² + y1²)^1/2, |b| = (x2² + y2²)^1/2.

Find the inner product of vectors by multiplying their coordinates in pairs: ab = x1x2 + y1y2. From the definition of the dot product ab = |a|*|b|*cos α, where α is the angle between the vectors. Then we get that x1x2 + y1y2 = |a|*|b|*cos α. Then cos α = (x1x2 + y1y2)/(|a|*|b|) = (x1x2 + y1y2)/((x1² + y1²)(x2² + y2²))^1/2.

Find the angle α using the Bradys tables.

Angle between two vectors.

From the definition of the dot product:

Condition of orthogonality of two vectors:

Collinearity condition for two vectors:

Follows from definition 5 - . Indeed, from the definition of the product of a vector by a number, it follows. Therefore, based on the vector equality rule, we write , , , which implies . But the vector resulting from the multiplication of a vector by a number is collinear to the vector .

Vector-to-vector projection:

Example 4. Given points , , , .

Find the scalar product.

Solution. we find by the formula of the scalar product of vectors given by their coordinates. Because the

, ,

Example 5 Given points , , , .

Find projection.

Solution. Because the

, ,

Based on the projection formula, we have

Example 6 Given points , , , .

Find the angle between the vectors and .

Solution. Note that the vectors

, ,

are not collinear, since their coordinates are not proportional:

These vectors are also not perpendicular, since their dot product is .

Let's find,

Corner find from the formula:

Example 7 Determine for which vectors and collinear.

Solution. In the case of collinearity, the corresponding coordinates of the vectors and must be proportional, that is:

From here and .

Example 8. Determine at what value of the vector and are perpendicular.

Solution. Vector and are perpendicular if their dot product is zero. From this condition we get: . That is, .

Example 9. Find , if , , .

Solution. Due to the properties of the scalar product, we have:

Example 10. Find the angle between the vectors and , where and - unit vectors and the angle between the vectors and is equal to 120o.

Solution. We have: , ,

Finally we have: .

5 B. vector product.

Definition 21.vector art vector to vector is called vector , or , defined by the following three conditions:

1) The module of the vector is , where is the angle between the vectors and , i.e. .

It follows that the module vector product numerically equal to area parallelogram built on vectors and as on sides.

2) The vector is perpendicular to each of the vectors and ( ; ), i.e. perpendicular to the plane of the parallelogram built on the vectors and .

3) The vector is directed in such a way that if viewed from its end, then the shortest turn from vector to vector would be counterclockwise (vectors , , form a right triple).

How to calculate angles between vectors?

When studying geometry, many questions arise on the topic of vectors. The student experiences particular difficulties when it is necessary to find the angles between the vectors.

Basic terms

Before considering the angles between vectors, it is necessary to familiarize yourself with the definition of a vector and the concept of an angle between vectors.

A vector is a segment that has a direction, that is, a segment for which its beginning and end are defined.

The angle between two vectors on a plane that have a common origin is the smaller of the angles, by which it is required to move one of the vectors around a common point, to a position where their directions coincide.

Solution Formula

Once you understand what a vector is and how its angle is determined, you can calculate the angle between vectors. The solution formula for this is quite simple, and the result of its application will be the value of the cosine of the angle. By definition, it is equal to the quotient of the scalar product of vectors and the product of their lengths.

The scalar product of vectors is considered as the sum of the corresponding coordinates of multiplier vectors multiplied by each other. The length of a vector, or its modulus, is calculated as the square root of the sum of the squares of its coordinates.

Having received the value of the cosine of the angle, you can calculate the value of the angle itself using a calculator or using a trigonometric table.

Example

After you figure out how to calculate the angle between vectors, the solution to the corresponding problem becomes simple and straightforward. As an example, consider the simple problem of finding the magnitude of an angle.

First of all, it will be more convenient to calculate the values of the lengths of the vectors and their scalar product necessary for solving. Using the description above, we get:

Substituting the obtained values into the formula, we calculate the value of the cosine of the desired angle:

This number is not one of the five common cosine values, so to get the value of the angle, you will have to use a calculator or the Bradis trigonometric table. But before getting the angle between the vectors, the formula can be simplified to get rid of the extra negative sign:

The final answer can be left in this form to maintain accuracy, or you can calculate the value of the angle in degrees. According to the Bradis table, its value will be approximately 116 degrees and 70 minutes, and the calculator will show a value of 116.57 degrees.

Angle calculation in n-dimensional space

When considering two vectors in three-dimensional space, it is much more difficult to understand which angle we are talking about if they do not lie in the same plane. To simplify perception, you can draw two intersecting segments that form the smallest angle between them, and it will be the desired one. Despite the presence of a third coordinate in the vector, the process of how the angles between vectors are calculated will not change. Calculate the scalar product and modules of vectors, the arccosine of their quotient and will be the answer to this problem.

In geometry, problems often occur with spaces that have more than three dimensions. But for them, the algorithm for finding the answer looks similar.

Difference between 0 and 180 degrees

One of the common mistakes when writing an answer to a problem designed to calculate the angle between vectors is the decision to write that the vectors are parallel, that is, the desired angle turned out to be 0 or 180 degrees. This answer is incorrect.

Having received an angle value of 0 degrees as a result of the solution, the correct answer would be to designate the vectors as co-directional, that is, the vectors will have the same direction. In the case of obtaining 180 degrees, the vectors will be in the nature of opposite directions.

Specific vectors

By finding the angles between the vectors, one of the special types can be found, in addition to the co-directed and oppositely directed ones described above.

Several vectors parallel to one plane are called coplanar.
Vectors that are the same in length and direction are called equal.
Vectors that lie on the same straight line, regardless of direction, are called collinear.
If the length of the vector is zero, that is, its beginning and end coincide, then it is called zero, and if it is one, then it is called one.

How to find the angle between vectors?

help me please! I know the formula but I can't figure it out
vector a (8; 10; 4) vector b (5; -20; -10)

Alexander Titov

The angle between the vectors given by their coordinates is found according to the standard algorithm. First you need to find the scalar product of vectors a and b: (a, b) = x1x2 + y1y2 + z1z2. We substitute here the coordinates of these vectors and consider:
(a,b) = 8*5 + 10*(-20) = 4*(-10) = 40 - 200 - 40 = -200.
Next, we determine the lengths of each of the vectors. The length or modulus of a vector is the square root of the sum of the squares of its coordinates:
|a| = root of (x1^2 + y1^2 + z1^2) = root of (8^2 + 10^2 + 4^2) = root of (64 + 100 + 16) = root of 180 = 6 roots of 5
|b| = square root of (x2^2 + y2^2 + z2^2) = square root of (5^2 + (-20)^2 + (-10)^2) = square root of (25 + 400 + 100) = square root out of 525 = 5 roots out of 21.
We multiply these lengths. We get 30 roots out of 105.
And finally, we divide the scalar product of vectors by the product of the lengths of these vectors. We get -200 / (30 roots out of 105) or
- (4 roots of 105) / 63. This is the cosine of the angle between the vectors. And the angle itself is equal to the arc cosine of this number
f \u003d arccos (-4 roots of 105) / 63.
If I counted correctly.

How to calculate the sine of an angle between vectors from the coordinates of the vectors

Mikhail Tkachev

We multiply these vectors. Their dot product is equal to the product of the lengths of these vectors and the cosine of the angle between them.
The angle is unknown to us, but the coordinates are known.
Let's write it mathematically like this.
Let, given vectors a(x1;y1) and b(x2;y2)
Then

A*b=|a|*|b|*cosA

CosA=a*b/|a|*|b|

We argue.
a*b-scalar product of vectors is equal to the sum of the products of the corresponding coordinates of the coordinates of these vectors, i.e. equal to x1*x2+y1*y2

|a|*|b|-product of vector lengths is equal to √((x1)^2+(y1)^2)*√((x2)^2+(y2)^2).

So the cosine of the angle between the vectors is:

CosA=(x1*x2+y1*y2)/√((x1)^2+(y1)^2)*√((x2)^2+(y2)^2)

Knowing the cosine of an angle, we can calculate its sine. Let's discuss how to do it:

If the cosine of an angle is positive, then this angle lies in 1 or 4 quarters, so its sine is either positive or negative. But since the angle between the vectors is less than or equal to 180 degrees, then its sine is positive. We argue similarly if the cosine is negative.

SinA=√(1-cos^2A)=√(1-((x1*x2+y1*y2)/√((x1)^2+(y1)^2)*√((x2)^2+( y2)^2))^2)

That's it)))) good luck figuring it out)))

Dmitry Levishchev

The fact that it is impossible to directly sine is not true.
In addition to the formula:
(a,b)=|a|*|b|*cos A
There is also this one:
||=|a|*|b|*sin A
That is, instead of the scalar product, you can take the module of the vector product.

At your request!

1. Eliminate irrationality in the denominator:

3. Solve the exponential equation:

4. Solve the inequality:

The arithmetic square root exists only of a non-negative number and is always expressed by a non-negative number, so this inequality will be true for all X, satisfying the condition: 2-х≥0. From here we get: x≤2. We write the answer as a numerical interval: (-∞; 2].

5. Solve the inequality: 7 x > -1.

By definition: an exponential function is called a function of the form y \u003d a x, where a > 0, a ≠ 1, x is any number. The range of the exponential function is the set of all positive numbers, since a positive number to any power will be positive. That's why 7 x >0 for any x, and even more so 7 x > -1, i.e. the inequality is true for all x ∈ (-∞; +∞).

6. Convert to product:

We apply the formula for the sum of sines: the sum of the sines of two angles is equal to twice the product of the sine of the half-sum of these angles and the cosine of their half-difference.

8. It is known that f(x) = -15x+3. For what values of x, f(x)=0?

We substitute the number 0 instead of f (x) and solve the equation:

15x+3=0 ⇒ -15x=-3 ⇒ x=3:15 ⇒ x = 1/5.

11 . In the first and second alloys, copper and zinc are in a ratio of 5:2 and 3:4. How much of each alloy should be taken to get 28 kg of a new alloy with an equal content of copper and zinc.

We understand that the new alloy will contain 14 kg of copper and 14 kg of zinc. Similar problems are all solved in the same way: they make up an equation in the left and right parts of which the same amount of substance (let's take copper), written in different ways (based on the specific conditions of the problem). We have 14 kg of copper in the new alloy will be composed of copper from both these alloys. Let the mass of the first alloy X kg, then the mass of the second alloy is ( 28th)kg. In the first alloy there are 5 parts of copper and 2 parts of zinc, therefore copper will be (5/7) of x kg. To find a fraction of a number, multiply the fraction by the given number. In the second alloy, 3 parts of copper and 4 parts of zinc, i.e. copper contains (3/7) from (28's) kg. So:

12. Solve the equation: log 2 8 x = -1.

By definition of a logarithm:

8 x = 2 -1 ⇒ 2 3x = 2 -1 ⇒ 3x = -1 ⇒ x = -1/3.

15. Find the derivative of the function f(x) = -ln cosx 2 .

20. Find the value of an expression:

The modulus of a number can only be expressed as a non-negative number. If there is a negative expression under the module sign, then when opening the module brackets, all terms are written with opposite signs.

22. Solve the system of inequalities:

First, we solve each inequality separately.

Note that the smallest common period for these functions will be 2π, therefore, both left and right were attributed 2πn. Answer C).

23. Find the area of the figure bounded by the graph of the function y=3-|x-3| and straight line y=0.

The graph of this function will consist of two half-lines coming out of one point. Let's write the equations of the lines. For x≥3 we expand the modular brackets and get: y=3-x+3 ⇒ y=6-x. For x<3 получаем функцию: y=3+x-3 ⇒ y=x.

A triangle bounded by a graph of a function and a segment of the x-axis is a figure whose area must be found. Of course, we will do without integrals here. We find the area of a triangle as half the product of its base and the height drawn to this base. Our base is equal to 6 unit segments, and the height drawn to this base is equal to 3 unit segments. The area will be 9 square meters. units

24. Find the cosine of angle A of a triangle with vertices at points A(1; 4), B(-2; 3), C(4; 2).

To find the coordinates of a vector given by the coordinates of its ends, you need to subtract the coordinates of the beginning from the coordinates of the end.

Angle A is formed by the vectors:

25. There are 23 balls in a box: red, white and black. There are 11 times more white balls than red ones. How many black balls?

Let it be in the box X red balls. Then the whites 11x balls.

Red and white x+11x= 12x balls. Therefore, black balls 23-12h. Since this is an integer number of balls, the only possible value is x=1. It turns out: 1 red ball, 11 white balls and 11 black balls.

When studying geometry, many questions arise on the topic of vectors. The student experiences particular difficulties when it is necessary to find the angles between the vectors.

Basic terms

Before considering the angles between vectors, it is necessary to familiarize yourself with the definition of a vector and the concept of an angle between vectors.

A vector is a segment that has a direction, that is, a segment for which its beginning and end are defined.

Solution Formula

Having received the value of the cosine of the angle, you can calculate the value of the angle itself using a calculator or using a trigonometric table.

Example

First of all, it will be more convenient to calculate the values of the lengths of the vectors and their scalar product necessary for solving. Using the description above, we get:

Substituting the obtained values into the formula, we calculate the value of the cosine of the desired angle:

Angle calculation in n-dimensional space

In geometry, problems often occur with spaces that have more than three dimensions. But for them, the algorithm for finding the answer looks similar.

Difference between 0 and 180 degrees

Specific vectors

By finding the angles between the vectors, one of the special types can be found, in addition to the co-directed and oppositely directed ones described above.

Several vectors parallel to one plane are called coplanar.
Vectors that are the same in length and direction are called equal.
Vectors that lie on the same straight line, regardless of direction, are called collinear.
If the length of the vector is zero, that is, its beginning and end coincide, then it is called zero, and if it is one, then it is called one.

Calculation of the cosine of the angle between vectors. Dot product of vectors

Dot product of vectors

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

The concept of dot product

Angle between vectors and dot product value

Dot product properties

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

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

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

Angle between two vectors.

How to calculate angles between vectors?

Basic terms

Solution Formula

Example

Angle calculation in n-dimensional space

Difference between 0 and 180 degrees

Specific vectors

How to find the angle between vectors?

How to calculate the sine of an angle between vectors from the coordinates of the vectors

Basic terms

Solution Formula

Example

Angle calculation in n-dimensional space

Difference between 0 and 180 degrees

Specific vectors

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

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

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