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, since in order to assimilate the material, you need to be guided in the terms and notation I use, have basic knowledge of vectors and be able to solve elementary problems. This lesson is a logical continuation of the topic, and in it I will analyze in detail typical tasks that use the scalar product of vectors. This is a VERY IMPORTANT job.. Try not to skip the examples, they 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 higher mathematics. The topics are simple, the algorithm for solving many problems is stereotyped and understandable. The only thing. There is a decent amount of information, so it is undesirable to try to master and solve EVERYTHING AND AT ONCE. This is especially true for dummies, believe me, the author absolutely does not want to feel like Chikatilo from mathematics. Well, not from mathematics, of course, either =) More prepared students can use the materials selectively, in a certain sense, “acquire” the missing knowledge, for you I will be a harmless Count Dracula =)

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

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

The concept of dot product

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

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

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

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

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

Now that's a pretty strict definition.

We focus on essential information:

Designation: the scalar product is denoted by or simply .

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

Just a couple of warm-up examples:

Example 1

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, one or another physical unit must be indicated. The canonical example of calculating the work of a force can be found in any textbook (the formula is exactly a dot product). The work of a force is measured in Joules, therefore, the answer will be written quite specifically, for example,.

Example 2

Find if , and the angle between the vectors is .

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

Angle between vectors and dot product value

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

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

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

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

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

The converse statements are also true:

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

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

But the third case is of particular interest:

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

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

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

Dot product properties

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

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

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

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 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 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:

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

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

Example 4

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

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

Example 5

Find the length of the vector if .

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

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

Answer:

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

Example 6

Find the length of the vector if .

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

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

Let's swap the parts:

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

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

Example 7

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

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

Example 7*

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

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

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

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

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

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

Short solution and answer at the end of the lesson.

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

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

Answer:

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

Example 14

Find the scalar product of vectors and if

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

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

Example 15

Find lengths of vectors , if

Solution: again asking for a way previous section: , 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 complete information so that the previously derived formula for the cosine of the angle between vectors express in terms of vector coordinates:

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

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

Example 16

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

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

Set aside these vectors from one point:

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

1. Definition and simple properties. Let us take non-zero vectors a and b and put them aside from an arbitrary point O: OA = a and OB = b. The value of the angle AOB is called the angle between the vectors a and b and is denoted (a,b). If at least one of the two vectors is zero, then the angle between them, by definition, is considered right. Note that, by definition, the angle between vectors is at least 0 and at most . Moreover, the angle between two non-zero vectors is equal to 0 if and only if these vectors are codirectional and equal to if and only if they are in opposite directions.

Let us check that the angle between the vectors does not depend on the choice of point O. This is obvious if the vectors are collinear. Otherwise, we set aside from an arbitrary point O 1 vectors O 1 A 1 = a and o 1 V 1 = b and note that triangles AOB and A 1 O 1 V 1 are equal on three sides, because |OA| = |O 1 A 1 | = |a|, |OB| = |O 1 V 1 | = |b|, |AB| = |A 1 V 1 | = |b–а|. Therefore, the angles AOB and A 1 O 1 V 1 are equal.

Now we can give the main thing in this paragraph

(5.1) Definition. The scalar product of two vectors a and b (denoted by ab) is the number 6 , equal to the product of the lengths of these vectors and the cosine of the angle between the vectors. Briefly speaking:

ab = |a||b|cos (a,b).

The operation of finding the scalar product is called the scalar multiplication of vectors. The scalar product aa of a vector with itself is called the scalar square of this vector and is denoted a 2 .

(5.2) The scalar square of a vector is equal to the square of its length.

 If |a|  0, then (a,a) = 0, whence a 2 = |a||a|cos0 = |a| 2 . If a = 0, then a 2 = |a| 2 = 0. 

(5.3) Cauchy's inequality. The modulus of the scalar product of two vectors does not exceed the product of moduli of factors: |ab| |a||b|. In this case, equality is achieved if and only if the vectors a and b are collinear.

 By definition |ab| = ||a||b|cos (a,b)| = |a||b||cos (a,b)|  |a||b. This proves the Cauchy inequality itself. Now let's notice. that for non-zero vectors a and b equality in it is achieved if and only if |cos (a,b)| = 1, i.e. at (a,b) = 0 or (a,b) =  . The latter is equivalent to the fact that the vectors a and b are co-directed or oppositely directed, i.e. collinear. If at least one of the vectors a and b is zero, then they are collinear and |ab| = |a||b| = 0. 

2. Basic properties of scalar multiplication. These include the following:

(CS1) ab = ba (commutativity);

(CS2) (xa)b = x(ab) (associativity);

(CS3) a(b+c) = ab + ac (distributivity).

The commutativity here is obvious, because ab =  ba. Associativity for x = 0 is also obvious. If x > 0 then

(ha)b = |ha||b|cos (xa,b) = |x||a||b|cos (xa,b) = x|a||b|cos (a,b) = x(ab),

for (xa, b) = (a,b) (from the codirection of the vectors xa and a - Fig. 21). If x< 0, then

(xa)b = |x||a||b|cos (хa,b) = –х|а||b|(–cos (a,b)) = x|a||b|cos (a,b) = x(ab),

for (xa, b) = –  (a,b) (from the opposite direction of the vectors xa and a - Fig.22). Thus, associativity is also proven.

Proving distributivity is more difficult. For this we need such

(5.4) Lemma. Let a be a non-zero vector parallel to the line l and b an arbitrary vector. Then the orthogonal projectionb" of the vector b to the line l is equal to
.



If b = 0, thenb" = 0 and ab = 0, so that in this case the lemma is true. In what follows, we will assume that the vector b" is non-zero. In this case, from an arbitrary point O of the straight line l, we set aside the vectors OA = a and OB = b, and also drop the perpendicular BB "from the point B to the straight line l. By definitionOB" = b" and (a,b) =  AOW. Denote AOB through and prove the lemma separately for each of the following three cases:

1)  <  /2. Then the vectors a and co-directed (Fig. 23) and

b" = =
=
.

2)  >  /2 . Then the vectors a andb"oppositely directed (Fig. 24) and

b" = – = –
= .

3)  =  /2. Thenb" = 0 and ab = 0, whenceb" =
= 0. 

We now prove the distributivity of (CS3). It is obvious if the vector a is zero. Let a  0. Then draw a line l || a, and denote byb" andc" orthogonal projections of the vectors b and c onto it, and throughd" be the orthogonal projection of the vector d = b + c onto it. By Theorem 3.5d" = b"+ c". Applying Lemma 5.4 to the last equality, we obtain the equality
=
. Multiplying it scalarly by a, we find that 2 =
, whence ad = ab+ac, which was to be proved.

Properties we have proven scalar multiplication vectors are similar to the corresponding properties of multiplication of numbers. But not all properties of multiplication of numbers carry over to scalar multiplication of vectors. Here are typical examples:

1

) If ab = 0, then this does not mean that a = 0 or b = 0. Example: two non-zero vectors forming a right angle.

2) If ab = ac, then this does not mean that b = c, even if the vector a is non-zero. Example: b and c are two different vectors of the same length, forming equal angles with the vector a (Fig. 25).

3) It is not true that always a(bc) = (ab)c: if only because the validity of such an equality for bc, ab 0 implies that the vectors a and c are collinear.

3. Orthogonality of vectors. Two vectors are called orthogonal if the angle between them is right. Orthogonality of vectors is indicated by the icon .

When we defined the angle between vectors, we agreed to consider the angle between the zero vector and any other vector as a straight line. Therefore, the zero vector is orthogonal to any. This agreement allows us to prove such

(5.5) Sign of orthogonality of two vectors. Two vectors are orthogonal if and only if their dot product is 0.

 Let a and b be arbitrary vectors. If at least one of them is zero, then they are orthogonal, and their scalar product is equal to 0. Thus, in this case the theorem is true. Let us now assume that both given vectors are non-zero. By definition, ab = |a||b|cos (a,b). Since by our assumption the numbers |a| and |b| are not equal to 0, then ab = 0 cos (a, b) = 0  (a, b) = /2, which was to be proved.

The equality ab = 0 is often taken as the definition of orthogonality of vectors.

(5.6) Corollary. If the vector a is orthogonal to each of the vectors a 1 , …, a P , then it is also orthogonal to any of their linear combinations.

 It suffices to note that from the equality aa 1 = … = aa P = 0 implies the equality a(x 1 a 1 + … +x P a P ) = x 1 (ah 1 ) + … + x P (ah P ) = 0. 

From Corollary 5.6 it is easy to derive the school criterion for the perpendicularity of a line and a plane. Indeed, let some line MN be perpendicular to two intersecting lines AB and AC. Then the vector MN is orthogonal to the vectors AB and AC. Let us take any straight line DE in the plane ABC. The vector DE is coplanar to the noncollinear vectors AB and AC, and therefore expands in them. But then it is also orthogonal to the vector MN, that is, the lines MN and DE are perpendicular. It turns out that the line MN is perpendicular to any line from the plane ABC, which was to be proved.

4. Orthonormal bases. (5.7) Definition. A basis of a vector space is said to be orthonormal if, firstly, all its vectors have unit length and, secondly, any two of its vectors are orthogonal.

Vectors of an orthonormal basis in three-dimensional space are usually denoted by the letters i, j and k, and on the vector plane by the letters i and j. Taking into account the sign of orthogonality of two vectors and the equality of the scalar square of a vector to the square of its length, the orthonormality conditions for the basis (i,j,k) of the space V 3 can be written like this:

(5.8) i 2 = j 2 = k 2 = 1 , ij = ik = jk = 0,

and the basis (i,j) of the vector plane as follows:

(5.9) i 2 = j 2 = 1 , ij = 0.

Let the vectors a and b have in the orthonormal basis (i,j,k) the spaces V 3 coordinates (a 1 , a 2 , a 3 ) and (b 1 b 2 ,b 3 ) respectively. Thenab = (a 1 i+a 2 j+a 3 k)(b 1 i+b 2 j+b 3 k) = a 1 b 1 i 2 +a 2 b 2 j 2 +a 3 b 3 k 2 +a 1 b 2 ij+a 1 b 3 ik+a 2 b 1 ji+a 2 b 3 jk+a 3 b 1 ki+a 3 b 2 kj = a 1 b 1 +a 2 b 2 +a 3 b 3 . This is how the formula for the scalar product of vectors a (a 1 ,a 2 ,a 3 ) and b(b 1 ,b 2 ,b 3 ) given by their coordinates in the orthonormal basis of the space V 3 :

(5.10) ab = a 1 b 1 +a 2 b 2 +a 3 b 3 .

For vectors a(a 1 ,a 2 ) and b(b 1 ,b 2 ) given by their coordinates in an orthonormal basis on the vector plane, it has the form

(5.11) ab = a 1 b 1 +a 2 b 2 .

Let us substitute b = a into formula (5.10). It turns out that in the orthonormal basis a 2 = a 1 2 + a 2 2 + a 3 2 . Because a 2 = |a| 2 , we get such a formula for finding the length of the vector a (a 1 ,a 2 ,a 3 ) defined by its coordinates in the orthonormal basis of the space V 3 :

(5.12) |a| =
.

On the vector plane, by virtue of (5.11), it takes the form

(5.13) |a| =
.

Substituting b = i, b = j, b = k into formula (5.10), we obtain three more useful equalities:

(5.14) ai = a 1 , aj = a 2 , ak = a 3 .

The simplicity of coordinate formulas for finding the scalar product of vectors and vector length is the main advantage of orthonormal bases. For non-orthonormal bases, these formulas are, generally speaking, incorrect, and their application in this case is a gross mistake.

5. Direction cosines. Take in an orthonormal basis (i,j,k) the spaces V 3 vector a(a 1 ,a 2 ,a 3 ). Thenai = |a||i|cos (a,i) = |a|cos (a, i).On the other hand, ai = a 1 according to the formula 5.14. It turns out that

(5.15) a 1 = |a|cos (a, i).

and, likewise,

a 2 = |a|cos (a,j), and 3 = |a|cos (a, k).

If the vector a is unit, these three equalities take on a particularly simple form:

(5.16) a 1 = cos (a, i),a 2 = cos (a, j),a 3 = cos (a, k).

cosines of angles, formed by the vector with vectors of an orthonormal basis are called direction cosines of this vector in the given basis. As formulas 5.16 show, the coordinates of a unit vector in an orthonormal basis are equal to its direction cosines.

From 5.15 it follows that a 1 2 + a 2 2 + a 3 2 = |a| 2 (cos 2  (a,i)+cos 2  (a,j)+cos 2  (a, k)). On the other hand, a 1 2 + a 2 2 + a 3 2 = |a| 2 . It turns out that

(5.17) the sum of the squared direction cosines of a nonzero vector is equal to 1.

This fact is useful for solving some problems.

(5.18) Problem. The diagonal of a rectangular parallelepiped forms with two of its edges coming out of the same vertex angles of 60 . What angle does it form with the third edge coming out of this vertex?

 Consider an orthonormal basis of the space V 3 , whose vectors are represented by the edges of the parallelepiped coming out of the given vertex. Since the diagonal vector forms angles of 60 with two vectors of this basis , the squares of two of its three direction cosines are equal to cos 2 60  = 1/4. Therefore, the square of the third cosine is 1/2, and this cosine itself is 1/
. So the desired angle is 45 . 

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

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

The formula for the scalar product of vectors for spatial problems

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

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

Dot product formula of n-dimensional vectors

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

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

Properties of the Dot Product of Vectors

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

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

a a = 0<=>a = 0

3. The scalar product of a vector by itself is equal to the square of its modulus:

4. The operation of scalar multiplication is communicative:

5. If the scalar product of two non-zero vectors is equal to zero, then these vectors are orthogonal:

a ≠ 0, b ≠ 0, a b = 0<=>a ┴ b

6. (αa) b = α(a b)

7. The operation of scalar multiplication is distributive:

(a + b) c = a c + b c

Examples of tasks for calculating the scalar product of vectors

Examples of calculating the scalar product of vectors for plane problems

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

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

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

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

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

Solution:

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

5 |a| 2 + 12 a · b - 9 |b| 2 \u003d 5 3 2 + 12 3 2 cos 60˚ - 9 2 2 \u003d 45 +36 -36 \u003d 45.

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

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

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

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

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

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

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

2) perpendicular, perpendicular. (one"")

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

Designate: .

physical meaning vector product

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

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

Lecture: Vector coordinates; dot 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 some 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.

Suppose we have some vector whose beginning and end of the vector have the following designations and coordinates: A(A x ; Ay) and B(B x ; By)

To get the coordinates of this vector, it is necessary to subtract the corresponding start coordinates from the coordinates of the end of the vector:

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

Dot product of vectors

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

• Geometric way. According to him, 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 value that results from 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 be non-negative:

• If the scalar product of two identical vectors turned 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 from a permutation of vectors:

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

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

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

Angle between vectors

Definition 1

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

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

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

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

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

Definition 2

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

Calculated according to the formula:

a → , b → = a → b → cos a → , b → ^ .

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

We formulate the definition of the product for two vectors:

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

Dot product in coordinates

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

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

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

a → , b → = a x b x + a y b y ,

for three-dimensional space, the expression is applicable:

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

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

Let's prove it.

Proof 1

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

Vectors should be postponed

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

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

Consider a triangle O A B .

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

By condition, it can be seen that O A = a → , O B = b → , A B = b → - a → , ∠ A O B = a → , b → ^ , so we write the formula for finding the angle between vectors differently

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

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

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

Let's prove the equalities:

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

– respectively for vectors of three-dimensional space.

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

Dot product and its properties

There are dot product properties that apply for a → , b → and c → :

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

Example 1

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

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

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

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

Distributivity is valid for any numbers:

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

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

hence we have

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

Dot product with examples and solutions

Any problem of such a plan is solved using the properties and formulas regarding the scalar product:

1. (a → , b →) = a → b → cos (a → , b → ^) ;
2. (a → , b →) = a → · n p a → b → = b → · n p b → a → ;
3. (a → , b →) = a x b x + a y b y or (a → , b →) = a x b x + a y b y + a z b z ;
4. (a → , a →) = a → 2 .

Let's look at some examples of solutions.

Example 2

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

Solution

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

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

Answer: (a → , b →) = 21 2 .

Example 3

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

Solution

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

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

Answer: (a → , b →) = - 9

Example 4

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

Solution

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

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

Substituting into the formula using coordinates, we get:

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

Answer: (A B → , A C →) = 28 .

Example 5

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

Solution

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

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

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

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

By the property of commutativity, we transform:

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

As a result, we get:

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

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

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

Answer: (a → , b →) = 411

If there is a numerical projection.

Example 6

Find the inner product of a → and b → . The vector a → has coordinates a → = (9 , 3 , - 3) , the projection b → has coordinates (- 3 , - 1 , 1) .

Solution

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

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

Substituting into the formula, we get the expression:

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

Answer: (a → , b →) = - 33 .

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

Example 7

What value should λ take for a given scalar product a → \u003d (1, 0, λ + 1) and b → \u003d (λ, 1, λ) will be equal to -1.

Solution

From the formula it can be seen that it is necessary to find the sum of the products of coordinates:

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

In given we have (a → , b →) = - 1 .

To find λ , we calculate the equation:

λ 2 + 2 · λ = - 1 , hence λ = - 1 .

Answer: λ = - 1 .

The physical meaning of the scalar product

Mechanics considers the application of the dot product.

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

A = (F → , M N →) .

Example 8

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

Solution

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

Answer: A = 15 2 2 .

Example 9

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

Solution

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

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

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

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

M N → = 3 2 + (- 10) 2 + 7 2 = 158 .

Answer: 158 .

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