Definition. The vector product of vector a (multiplicand) and a non-collinear vector (multiplicand) is the third vector c (product), which is constructed as follows:
1) its module is numerically equal to the area of the parallelogram in Fig. 155), built on vectors, i.e. it is equal to the direction perpendicular to the plane of the mentioned parallelogram;
3) in this case, the direction of the vector c is chosen (from two possible ones) so that the vectors c form a right-handed system (§ 110).
Designation: or
Addition to the definition. If the vectors are collinear, then considering the figure (conditionally) a parallelogram, it is natural to assign zero area. Therefore, the vector product of collinear vectors is considered equal to the null vector.
Since the null vector can be assigned any direction, this agreement does not contradict paragraphs 2 and 3 of the definition.
Remark 1. In the term “vector product” the first word indicates that the result of the action is a vector (as opposed to a scalar product; cf. § 104, remark 1).
Example 1. Find the vector product where are the main vectors of the right coordinate system (Fig. 156).
1. Since the lengths of the main vectors are equal to one scale unit, the area of the parallelogram (square) is numerically equal to one. This means that the modulus of the vector product is equal to one.
2. Since the perpendicular to the plane is an axis, the desired vector product is a vector collinear to the vector k; and since both of them have modulus 1, the desired vector product is equal to either k or -k.
3. Of these two possible vectors, the first one must be chosen, since the vectors k form a right-handed system (and the vectors a left-handed one).
Example 2. Find the cross product
Solution. As in example 1, we conclude that the vector is equal to either k or -k. But now we need to choose -k, since the vectors form a right-handed system (and vectors form a left-handed one). So,
Example 3. Vectors have lengths equal to 80 and 50 cm, respectively, and form an angle of 30°. Taking the meter as the unit of length, find the length of the vector product a
Solution. The area of a parallelogram built on vectors is equal to The length of the desired vector product is equal to
Example 4. Find the length of the vector product of the same vectors, taking centimeters as the unit of length.
Solution. Since the area of a parallelogram built on vectors is equal, the length of the vector product is equal to 2000 cm, i.e.
From a comparison of examples 3 and 4 it is clear that the length of the vector depends not only on the lengths of the factors but also on the choice of the unit of length.
Physical meaning of a vector product. Of the numerous physical quantities represented by the vector product, we will consider only the moment of force.
Let A be the point of application of force. The moment of force relative to point O is called a vector product. Since the modulus of this vector product is numerically equal to the area of the parallelogram (Fig. 157), then the modulus of the moment is equal to the product of the base and the height, i.e., the force multiplied by the distance from point O to the straight line along which the force acts.
In mechanics, it is proven that for a rigid body to be in equilibrium, it is necessary that not only the sum of vectors representing the forces applied to the body be equal to zero, but also the sum of the moments of forces. In the case where all forces are parallel to one plane, the addition of vectors representing moments can be replaced by addition and subtraction of their magnitudes. But with arbitrary directions of forces, such a replacement is impossible. In accordance with this, the vector product is defined precisely as a vector, and not as a number.
7.1. Definition of cross product
Three non-coplanar vectors a, b and c, taken in the indicated order, form a right-handed triplet if, from the end of the third vector c, the shortest turn from the first vector a to the second vector b is seen to be counterclockwise, and a left-handed triplet if clockwise (see Fig. . 16).
The vector product of vector a and vector b is called vector c, which:
1. Perpendicular to vectors a and b, i.e. c ^ a and c ^ b ;
2. Has a length numerically equal to the area of a parallelogram constructed on vectors a andb as on the sides (see Fig. 17), i.e.
3. Vectors a, b and c form a right-handed triple.
The cross product is denoted a x b or [a,b]. The following relations between the unit vectors i directly follow from the definition of the vector product, j And k(see Fig. 18):
i x j = k, j x k = i, k x i = j.
Let us prove, for example, that i xj =k.
1) k ^ i, k ^ j ;
2) |k |=1, but | i x j| = |i | |J | sin(90°)=1;
3) vectors i, j and k form a right triple (see Fig. 16).
7.2. Properties of a cross product
1. When rearranging the factors, the vector product changes sign, i.e. and xb =(b xa) (see Fig. 19).
Vectors a xb and b xa are collinear, have the same modules (the area of the parallelogram remains unchanged), but are oppositely directed (triples a, b, a xb and a, b, b x a of opposite orientation). That is axb = -(b xa).
2. The vector product has a combining property with respect to the scalar factor, i.e. l (a xb) = (l a) x b = a x (l b).
Let l >0. Vector l (a xb) is perpendicular to vectors a and b. Vector ( l a)x b is also perpendicular to vectors a and b(vectors a, l but lie in the same plane). This means that the vectors l(a xb) and ( l a)x b collinear. It is obvious that their directions coincide. They have the same length:
That's why l(a xb)= l a xb. It is proved in a similar way for l<0.
3. Two non-zero vectors a and b are collinear if and only if their vector product is equal to the zero vector, i.e. a ||b<=>and xb =0.
In particular, i *i =j *j =k *k =0 .
4. The vector product has the distribution property:
(a+b) xc = a xc + b xs.
We will accept without proof.
7.3. Expressing the cross product in terms of coordinates
We will use the cross product table of vectors i, j and k:
if the direction of the shortest path from the first vector to the second coincides with the direction of the arrow, then the product is equal to the third vector; if it does not coincide, the third vector is taken with a minus sign.
Let two vectors a =a x i +a y be given j+a z k and b =b x i+b y j+b z k. Let's find the vector product of these vectors by multiplying them as polynomials (according to the properties of the vector product):
The resulting formula can be written even more briefly:
since the right side of equality (7.1) corresponds to the expansion of the third-order determinant in terms of the elements of the first row. Equality (7.2) is easy to remember.
7.4. Some applications of cross product
Establishing collinearity of vectors
Finding the area of a parallelogram and a triangle
According to the definition of the vector product of vectors A and b |a xb | =|a | * |b |sin g, i.e. S pairs = |a x b |. And, therefore, D S =1/2|a x b |.
Determination of the moment of force about a point
Let a force be applied at point A F =AB let it go ABOUT- some point in space (see Fig. 20).
It is known from physics that moment of force F relative to the point ABOUT called a vector M, which passes through the point ABOUT And:
1) perpendicular to the plane passing through the points O, A, B;
2) numerically equal to the product of force per arm
3) forms a right triple with vectors OA and A B.
Therefore, M = OA x F.
Finding linear rotation speed
Speed v point M of a rigid body rotating with angular velocity w around a fixed axis, is determined by Euler’s formula v =w xr, where r =OM, where O is some fixed point of the axis (see Fig. 21).
Definition An ordered collection of (x 1 , x 2 , ... , x n) n real numbers is called n-dimensional vector, and numbers x i (i = ) - components, or coordinates,
Example. If, for example, a certain automobile plant must produce 50 cars, 100 trucks, 10 buses, 50 sets of spare parts for cars and 150 sets for trucks and buses per shift, then the production program of this plant can be written as a vector (50, 100 , 10, 50, 150), having five components.
Notation. Vectors are denoted by bold lowercase letters or letters with a bar or arrow at the top, e.g. a or. The two vectors are called equal, if they have the same number of components and their corresponding components are equal.
Vector components cannot be swapped, for example, (3, 2, 5, 0, 1) and (2, 3, 5, 0, 1) different vectors.
Operations on vectors. The work
x= (x 1 , x 2 , ... ,x n) by a real numberλ called a vectorλ x= (λ x 1, λ x 2, ..., λ x n).
Amountx= (x 1 , x 2 , ... ,x n) and y= (y 1 , y 2 , ... ,y n) is called a vector x+y= (x 1 + y 1 , x 2 + y 2 , ... , x n + + y n).
Vector space. N -dimensional vector space R n is defined as the set of all n-dimensional vectors for which the operations of multiplication by real numbers and addition are defined.
Economic illustration. Economic illustration of n-dimensional vector space: space of goods (goods). Under goods we will understand some good or service that went on sale at a certain time in a certain place. Suppose there is a finite number n of available goods; the quantities of each of them purchased by the consumer are characterized by a set of goods
x= (x 1 , x 2 , ..., x n),
where x i denotes the amount of the i-th good purchased by the consumer. We will assume that all goods have the property of arbitrary divisibility, so that any non-negative quantity of each of them can be purchased. Then all possible sets of goods are vectors of the goods space C = ( x= (x 1 , x 2 , ... , x n) x i ≥ 0, i = ).
Linear independence.
System e 1 , e 2 , ... , e m n-dimensional vectors are called linearly dependent, if there are such numbersλ 1 , λ 2 , ... , λ m , of which at least one is non-zero, such that the equalityλ 1 e 1 + λ 2 e 2 +... + λ m e m = 0; otherwise, this system of vectors is called linearly independent, that is, the indicated equality is possible only in the case when all 
. The geometric meaning of the linear dependence of vectors in R 3, interpreted as directed segments, explain the following theorems.
Theorem 1. A system consisting of one vector is linearly dependent if and only if this vector is zero.
Theorem 2. In order for two vectors to be linearly dependent, it is necessary and sufficient that they be collinear (parallel).
Theorem 3 . In order for three vectors to be linearly dependent, it is necessary and sufficient that they be coplanar (lie in the same plane).
Left and right triples of vectors. Triple of non-coplanar vectors a, b, c called right, if the observer from their common origin bypasses the ends of the vectors a, b, c in the order given appears to occur clockwise. Otherwise a, b, c -left three. All right (or left) triples of vectors are called the same oriented.
Basis and coordinates. Troika e 1, e 2 , e 3 non-coplanar vectors in R 3 is called basis, and the vectors themselves e 1, e 2 , e 3 - basic. Any vector a can be uniquely expanded into basis vectors, that is, represented in the form
A= x 1 e 1+x2 e 2 + x 3 e 3, (1.1)
the numbers x 1 , x 2 , x 3 in expansion (1.1) are called coordinatesa in the basis e 1, e 2 , e 3 and are designated a(x 1, x 2, x 3).
Orthonormal basis. If the vectors e 1, e 2 , e 3 are pairwise perpendicular and the length of each of them is equal to one, then the basis is called orthonormal, and the coordinates x 1 , x 2 , x 3 - rectangular. The basis vectors of an orthonormal basis will be denoted by i, j, k.
We will assume that in space R 3 the right system of Cartesian rectangular coordinates is selected (0, i, j, k}.
Vector artwork. Vector artwork A to vector b called a vector c, which is determined by the following three conditions:
1. Vector length c numerically equal to the area of a parallelogram built on vectors a And b, i.e.
c=
|a||b| sin( a^b).
2. Vector c perpendicular to each of the vectors a And b.
3. Vectors a, b And c, taken in the indicated order, form a right triple.
For a cross product c the designation is introduced c =[ab] or
c = a
× b.
If the vectors a And b are collinear, then sin( a^b) = 0 and [ ab] = 0, in particular, [ aa] = 0. Vector products of unit vectors: [ ij]=k, [jk] = i, [ki]=j.
If the vectors a And b specified in the basis i, j, k coordinates a(a 1, a 2, a 3), b(b 1, b 2, b 3), then
Mixed work. If the vector product of two vectors A And b scalarly multiplied by the third vector c, then such a product of three vectors is called mixed work and is indicated by the symbol a b c.
If the vectors a, b And c in the basis i, j, k given by their coordinates
a(a 1, a 2, a 3), b(b 1, b 2, b 3), c(c 1, c 2, c 3), then
.
The mixed product has a simple geometric interpretation - it is a scalar, equal in absolute value to the volume of a parallelepiped built on three given vectors.
If the vectors form a right triple, then their mixed product is a positive number equal to the indicated volume; if it's a three a, b, c - left, then a b c<0 и V = - a b c, therefore V =|a b c|.
The coordinates of the vectors encountered in the problems of the first chapter are assumed to be given relative to a right orthonormal basis. Unit vector codirectional with vector A, indicated by the symbol A O. Symbol r=OM denoted by the radius vector of point M, symbols a, AB or|a|, | AB|modules of vectors are denoted A And AB.
Example 1.2. Find the angle between the vectors a= 2m+4n And b= m-n, Where m And n- unit vectors and angle between m And n equal to 120 o.
Solution. We have: cos φ = ab/ab ab =(2m+4n) (m-n) = 2m 2 - 4n 2 +2mn=
= 2 - 4+2cos120 o = - 2 + 2(-0.5) = -3; a = ; a 2 = (2m+4n) (2m+4n) =
= 4m 2 +16mn+16n 2 = 4+16(-0.5)+16=12, which means a = . b = ; b 2 =
= (m-n)(m-n) = m 2 -2mn+n 2 =
1-2(-0.5)+1 = 3, which means b = . Finally we have: cosφ = = -1/2, φ = 120 o.
Example 1.3.Knowing the vectors AB(-3,-2.6) and B.C.(-2,4,4),calculate the length of the altitude AD of triangle ABC.
Solution. Denoting the area of triangle ABC by S, we get:
S = 1/2 BC AD. Then AD=2S/BC, BC= = 
= 6,
S = 1/2| AB ×AC|.
AC=AB+BC, which means vector A.C. has coordinates
.
.
Example 1.4 . Two vectors are given a(11,10,2) and b(4,0,3). Find the unit vector c, orthogonal to vectors a And b and directed so that the ordered triple of vectors a, b, c was right.
Solution.Let us denote the coordinates of the vector c with respect to a given right orthonormal basis in terms of x, y, z.
Because the c ⊥ a, c ⊥b, That ca= 0,cb= 0. According to the conditions of the problem, it is required that c = 1 and a b c >0.
We have a system of equations for finding x,y,z: 11x +10y + 2z = 0, 4x+3z=0, x 2 + y 2 + z 2 = 0.
From the first and second equations of the system we obtain z = -4/3 x, y = -5/6 x. Substituting y and z into the third equation, we have: x 2 = 36/125, whence
x =±
. Using the condition a b c > 0, we get the inequality
Taking into account the expressions for z and y, we rewrite the resulting inequality in the form: 625/6 x > 0, which implies that x>0. So, x = , y = - , z =- .
In this lesson we will look at two more operations with vectors: vector product of vectors And mixed product of vectors (immediate link for those who need it). It’s okay, sometimes it happens that for complete happiness, in addition to scalar product of vectors, more and more are required. This is vector addiction. It may seem that we are getting into the jungle of analytical geometry. This is wrong. In this section of higher mathematics there is generally little wood, except perhaps enough for Pinocchio. In fact, the material is very common and simple - hardly more complicated than the same scalar product, there will even be fewer typical tasks. The main thing in analytical geometry, as many will be convinced or have already been convinced, is NOT TO MAKE MISTAKES IN CALCULATIONS. Repeat like a spell and you will be happy =)
If vectors sparkle somewhere far away, like lightning on the horizon, it doesn’t matter, start with the lesson Vectors for dummies to restore or reacquire basic knowledge about vectors. More prepared readers can get acquainted with the information selectively; I tried to collect the most complete collection of examples that are often found in practical work
What will make you happy right away? When I was little, I could juggle two or even three balls. It worked out well. Now you won't have to juggle at all, since we will consider only spatial vectors, and flat vectors with two coordinates will be left out. Why? This is how these actions were born - the vector and mixed product of vectors are defined and work in three-dimensional space. It's already easier!
This operation, just like the scalar product, involves two vectors. Let these be imperishable letters.
The action itself denoted by in the following way: . There are other options, but I’m used to denoting the vector product of vectors this way, in square brackets with a cross.
And right away question: if in scalar product of vectors two vectors are involved, and here two vectors are also multiplied, then what is the difference? The obvious difference is, first of all, in the RESULT:
The result of the scalar product of vectors is NUMBER:
The result of the cross product of vectors is VECTOR: , that is, we multiply the vectors and get a vector again. Closed club. Actually, this is where the name of the operation comes from. In different educational literature, designations may also vary; I will use the letter.
Definition of cross product
First there will be a definition with a picture, then comments.
Definition: Vector product non-collinear vectors, taken in this order, called VECTOR, length which is numerically equal to the area of the parallelogram, built on these vectors; vector orthogonal to vectors, and is directed so that the basis has a right orientation: 
Let’s break down the definition piece by piece, there’s a lot of interesting stuff here!
So, the following significant points can be highlighted:
1) The original vectors, indicated by red arrows, by definition not collinear. It will be appropriate to consider the case of collinear vectors a little later.
2) Vectors are taken in a strictly defined order: – "a" is multiplied by "be", and not “be” with “a”. The result of vector multiplication is VECTOR, which is indicated in blue. If the vectors are multiplied in reverse order, we obtain a vector equal in length and opposite in direction (raspberry color). That is, the equality is true 
.
3) Now let's get acquainted with the geometric meaning of the vector product. This is a very important point! The LENGTH of the blue vector (and, therefore, the crimson vector) is numerically equal to the AREA of the parallelogram built on the vectors. In the figure, this parallelogram is shaded black.
Note : the drawing is schematic, and, naturally, the nominal length of the vector product is not equal to the area of the parallelogram.
Let us recall one of the geometric formulas: The area of a parallelogram is equal to the product of adjacent sides and the sine of the angle between them. Therefore, based on the above, the formula for calculating the LENGTH of a vector product is valid:
I emphasize that the formula is about the LENGTH of the vector, and not about the vector itself. What is the practical meaning? And the meaning is that in problems of analytical geometry, the area of a parallelogram is often found through the concept of a vector product:
Let us obtain the second important formula. The diagonal of a parallelogram (red dotted line) divides it into two equal triangles. Therefore, the area of a triangle built on vectors (red shading) can be found using the formula:
4) An equally important fact is that the vector is orthogonal to the vectors, that is 
. Of course, the oppositely directed vector (raspberry arrow) is also orthogonal to the original vectors.
5) The vector is directed so that basis It has right orientation. In the lesson about transition to a new basis I spoke in sufficient detail about plane orientation, and now we will figure out what space orientation is. I will explain on your fingers right hand. Mentally combine forefinger with vector and middle finger with vector. Ring finger and little finger press it into your palm. As a result thumb– the vector product will look up. This is a right-oriented basis (it is this one in the figure). Now change the vectors ( index and middle fingers) in some places, as a result the thumb will turn around, and the vector product will already look down. This is also a right-oriented basis. You may have a question: which basis has left orientation? “Assign” to the same fingers left hand vectors, and get the left basis and left orientation of space (in this case, the thumb will be located in the direction of the lower vector). Figuratively speaking, these bases “twist” or orient space in different directions. And this concept should not be considered something far-fetched or abstract - for example, the orientation of space is changed by the most ordinary mirror, and if you “pull the reflected object out of the looking glass,” then in the general case it will not be possible to combine it with the “original.” By the way, hold three fingers up to the mirror and analyze the reflection ;-)
...how good it is that you now know about right- and left-oriented bases, because the statements of some lecturers about a change in orientation are scary =)
Cross product of collinear vectors
The definition has been discussed in detail, it remains to find out what happens when the vectors are collinear. If the vectors are collinear, then they can be placed on one straight line and our parallelogram also “folds” into one straight line. The area of such, as mathematicians say, degenerate parallelogram is equal to zero. The same follows from the formula - the sine of zero or 180 degrees is equal to zero, which means the area is zero
Thus, if , then 
. Strictly speaking, the vector product itself is equal to the zero vector, but in practice this is often neglected and they are written that it is simply equal to zero.
A special case is the vector product of a vector with itself:
Using the vector product, you can check the collinearity of three-dimensional vectors, and we will also analyze this problem, among others.
To solve practical examples you may need trigonometric table to find the values of sines from it.
Well, let's light the fire:
Example 1
a) Find the length of the vector product of vectors if 
b) Find the area of a parallelogram built on vectors if 
Solution: No, this is not a typo, I deliberately made the initial data in the clauses of the condition the same. Because the design of the solutions will be different!
a) According to the condition, you need to find length vector (cross product). According to the corresponding formula:
Answer:
Since the question was about length, we indicate the dimension in the answer - units.
b) According to the condition, you need to find square parallelogram built on vectors. The area of this parallelogram is numerically equal to the length of the vector product:
Answer:
Please note that the answer does not talk about the vector product at all; we were asked about area of the figure, accordingly, the dimension is square units.
We always look at WHAT we need to find according to the condition, and, based on this, we formulate clear answer. It may seem literal, but there are plenty of literal teachers among them, and the assignment has a good chance of being returned for revision. Although this is not a particularly far-fetched quibble - if the answer is incorrect, then one gets the impression that the person does not understand simple things and/or has not understood the essence of the task. This point must always be kept under control when solving any problem in higher mathematics, and in other subjects too.
Where did the big letter “en” go? In principle, it could have been additionally attached to the solution, but in order to shorten the entry, I did not do this. I hope everyone understands that and is a designation for the same thing.
A popular example for a DIY solution:
Example 2
Find the area of a triangle built on vectors if 
The formula for finding the area of a triangle through the vector product is given in the comments to the definition. The solution and answer are at the end of the lesson.
In practice, the task is really very common; triangles can generally torment you.
To solve other problems we will need:
Properties of the vector product of vectors
We have already considered some properties of the vector product, however, I will include them in this list.
For arbitrary vectors and an arbitrary number, the following properties are true:
1) In other sources of information, this item is usually not highlighted in the properties, but it is very important in practical terms. So let it be.
2) 
– the property is also discussed above, sometimes it is called anticommutativity. In other words, the order of the vectors matters.
3) – associative or associative vector product laws. Constants can be easily moved outside the vector product. Really, what should they do there?
4) – distribution or distributive vector product laws. There are no problems with opening the brackets either.
To demonstrate, let's look at a short example:
Example 3
Find if 
Solution: The condition again requires finding the length of the vector product. Let's paint our miniature: 
(1) According to associative laws, we take the constants outside the scope of the vector product.
(2) We move the constant outside the module, and the module “eats” the minus sign. The length cannot be negative.
(3) The rest is clear.
Answer: 
It's time to add more wood to the fire:
Example 4
Calculate the area of a triangle built on vectors if 
Solution: Find the area of the triangle using the formula 
. The catch is that the vectors “tse” and “de” are themselves presented as sums of vectors. The algorithm here is standard and somewhat reminiscent of examples No. 3 and 4 of the lesson Dot product of vectors. For clarity, we will divide the solution into three stages:
1) At the first step, we express the vector product through the vector product, in fact, let's express a vector in terms of a vector. No word yet on lengths!
(1) Substitute the expressions of the vectors.
(2) Using distributive laws, we open the brackets according to the rule of multiplication of polynomials.
(3) Using associative laws, we move all constants beyond the vector products. With a little experience, steps 2 and 3 can be performed simultaneously.
(4) The first and last terms are equal to zero (zero vector) due to the nice property. In the second term we use the property of anticommutativity of a vector product:
(5) We present similar terms.
As a result, the vector turned out to be expressed through a vector, which is what was required to be achieved: 
2) In the second step, we find the length of the vector product we need. This action is similar to Example 3: 
3) Find the area of the required triangle: 
Stages 2-3 of the solution could have been written in one line.
Answer:
The problem considered is quite common in tests, here is an example for solving it yourself:
Example 5
Find if
A short solution and answer at the end of the lesson. Let's see how attentive you were when studying the previous examples ;-)
Cross product of vectors in coordinates
, specified in an orthonormal basis, expressed by the formula:
The formula is really simple: in the top line of the determinant we write the coordinate vectors, in the second and third lines we “put” the coordinates of the vectors, and we put in strict order– first the coordinates of the “ve” vector, then the coordinates of the “double-ve” vector. If the vectors need to be multiplied in a different order, then the rows should be swapped: 
Example 10
Check whether the following space vectors are collinear:
A)
b) 
Solution: The check is based on one of the statements in this lesson: if the vectors are collinear, then their vector product is equal to zero (zero vector): 
.
a) Find the vector product: 
Thus, the vectors are not collinear.
b) Find the vector product: 
Answer: a) not collinear, b)
Here, perhaps, is all the basic information about the vector product of vectors.
This section will not be very large, since there are few problems where the mixed product of vectors is used. In fact, everything will depend on the definition, geometric meaning and a couple of working formulas.
A mixed product of vectors is the product of three vectors:
So they lined up like a train and can’t wait to be identified.
First, again, a definition and a picture:
Definition: Mixed work non-coplanar vectors, taken in this order, called parallelepiped volume, built on these vectors, equipped with a “+” sign if the basis is right, and a “–” sign if the basis is left.
Let's do the drawing. Lines invisible to us are drawn with dotted lines: 
Let's dive into the definition:
2) Vectors are taken in a certain order, that is, the rearrangement of vectors in the product, as you might guess, does not occur without consequences.
3) Before commenting on the geometric meaning, I will note an obvious fact: the mixed product of vectors is a NUMBER: . In educational literature, the design may be slightly different; I am used to denoting a mixed product by , and the result of calculations by the letter “pe”.
A-priory the mixed product is the volume of the parallelepiped, built on vectors (the figure is drawn with red vectors and black lines). That is, the number is equal to the volume of a given parallelepiped.
Note : The drawing is schematic.
4) Let’s not worry again about the concept of orientation of the basis and space. The meaning of the final part is that a minus sign can be added to the volume. In simple words, a mixed product can be negative: .
Directly from the definition follows the formula for calculating the volume of a parallelepiped built on vectors.
Yandex.RTB R-A-339285-1Before giving the concept of a vector product, let us turn to the question of the orientation of an ordered triple of vectors a →, b →, c → in three-dimensional space.
To begin with, let’s set aside the vectors a → , b → , c → from one point. The orientation of the triple a → , b → , c → can be right or left, depending on the direction of the vector c → itself. The type of triple a → , b → , c → will be determined from the direction in which the shortest turn is made from vector a → to b → from the end of vector c → .
If the shortest turn is carried out counterclockwise, then the triple of vectors a → , b → , c → is called right, if clockwise – left.
Next, take two non-collinear vectors a → and b →. Let us then plot the vectors A B → = a → and A C → = b → from point A. Let's construct a vector A D → = c →, which is simultaneously perpendicular to both A B → and A C →. Thus, when constructing the vector itself A D → = c →, we can do two things, giving it either one direction or the opposite (see illustration).
An ordered triple of vectors a → , b → , c → can be, as we found out, right or left depending on the direction of the vector.
From the above we can introduce the definition of a vector product. This definition is given for two vectors defined in a rectangular coordinate system of three-dimensional space.
Definition 1
The vector product of two vectors a → and b → we will call such a vector defined in a rectangular coordinate system of three-dimensional space such that:
- if the vectors a → and b → are collinear, it will be zero;
- it will be perpendicular to both vector a → and vector b → i.e. ∠ a → c → = ∠ b → c → = π 2 ;
- its length is determined by the formula: c → = a → · b → · sin ∠ a → , b → ;
- the triple of vectors a → , b → , c → has the same orientation as the given coordinate system.
The vector product of vectors a → and b → has the following notation: a → × b →.
Coordinates of the vector product
Since any vector has certain coordinates in the coordinate system, we can introduce a second definition of a vector product, which will allow us to find its coordinates using the given coordinates of the vectors.
Definition 2
In a rectangular coordinate system of three-dimensional space vector product of two vectors a → = (a x ; a y ; a z) and b → = (b x ; b y ; b z) is called a vector c → = a → × b → = (a y b z - a z b y) i → + (a z b x - a x b z) j → + (a x b y - a y b x) k → , where i → , j → , k → are coordinate vectors.
The vector product can be represented as the determinant of a third-order square matrix, where the first row contains the vector vectors i → , j → , k → , the second row contains the coordinates of the vector a → , and the third row contains the coordinates of the vector b → in a given rectangular coordinate system, this is the determinant of the matrix looks like this: c → = a → × b → = i → j → k → a x a y a z b x b y b z
Expanding this determinant into the elements of the first row, we obtain the equality: c → = a → × b → = i → j → k → a x a y a z b x b y b z = a y a z b y b z · i → - a x a z b x b z · j → + a x a y b x b y · k → = = a → × b → = (a y b z - a z b y) i → + (a z b x - a x b z) j → + (a x b y - a y b x) k →
Properties of a cross product
It is known that the vector product in coordinates is represented as the determinant of the matrix c → = a → × b → = i → j → k → a x a y a z b x b y b z , then on the basis properties of the matrix determinant the following are displayed properties of a vector product:
- anticommutativity a → × b → = - b → × a → ;
- distributivity a (1) → + a (2) → × b = a (1) → × b → + a (2) → × b → or a → × b (1) → + b (2) → = a → × b (1) → + a → × b (2) → ;
- associativity λ a → × b → = λ a → × b → or a → × (λ b →) = λ a → × b →, where λ is an arbitrary real number.
These properties have simple proofs.
As an example, we can prove the anticommutative property of a vector product.
Proof of anticommutativity
By definition, a → × b → = i → j → k → a x a y a z b x b y b z and b → × a → = i → j → k → b x b y b z a x a y a z . And if two rows of the matrix are swapped, then the value of the determinant of the matrix should change to the opposite, therefore, a → × b → = i → j → k → a x a y a z b x b y b z = - i → j → k → b x b y b z a x a y a z = - b → × a → , which and proves that the vector product is anticommutative.
Vector product - examples and solutions
In most cases, there are three types of problems.
In problems of the first type, the lengths of two vectors and the angle between them are usually given, and you need to find the length of the vector product. In this case, use the following formula c → = a → · b → · sin ∠ a → , b → .
Example 1
Find the length of the vector product of vectors a → and b → if you know a → = 3, b → = 5, ∠ a →, b → = π 4.
Solution
By determining the length of the vector product of vectors a → and b →, we solve this problem: a → × b → = a → · b → · sin ∠ a → , b → = 3 · 5 · sin π 4 = 15 2 2 .
Answer: 15 2 2 .
Problems of the second type have a connection with the coordinates of vectors, in them the vector product, its length, etc. are searched through the known coordinates of given vectors a → = (a x; a y; a z) And b → = (b x ; b y ; b z) .
For this type of problem, you can solve a lot of task options. For example, not the coordinates of the vectors a → and b → can be specified, but their expansions into coordinate vectors of the form b → = b x · i → + b y · j → + b z · k → and c → = a → × b → = (a y b z - a z b y) i → + (a z b x - a x b z) j → + (a x b y - a y b x) k →, or vectors a → and b → can be specified by the coordinates of their start and end points.
Consider the following examples.
Example 2
In a rectangular coordinate system, two vectors are given: a → = (2; 1; - 3), b → = (0; - 1; 1). Find their cross product.
Solution
By the second definition, we find the vector product of two vectors in given coordinates: a → × b → = (a y · b z - a z · b y) · i → + (a z · b x - a x · b z) · j → + (a x · b y - a y · b x) · k → = = (1 · 1 - (- 3) · (- 1)) · i → + ((- 3) · 0 - 2 · 1) · j → + (2 · (- 1) - 1 · 0) · k → = = - 2 i → - 2 j → - 2 k → .
If we write the vector product through the determinant of the matrix, then the solution to this example looks like this: a → × b → = i → j → k → a x a y a z b x b y b z = i → j → k → 2 1 - 3 0 - 1 1 = - 2 i → - 2 j → - 2 k → .
Answer: a → × b → = - 2 i → - 2 j → - 2 k → .
Example 3
Find the length of the vector product of vectors i → - j → and i → + j → + k →, where i →, j →, k → are the unit vectors of the rectangular Cartesian coordinate system.
Solution
First, let's find the coordinates of a given vector product i → - j → × i → + j → + k → in a given rectangular coordinate system.
It is known that the vectors i → - j → and i → + j → + k → have coordinates (1; - 1; 0) and (1; 1; 1), respectively. Let's find the length of the vector product using the determinant of the matrix, then we have i → - j → × i → + j → + k → = i → j → k → 1 - 1 0 1 1 1 = - i → - j → + 2 k → .
Therefore, the vector product i → - j → × i → + j → + k → has coordinates (- 1 ; - 1 ; 2) in the given coordinate system.
We find the length of the vector product using the formula (see the section on finding the length of a vector): i → - j → × i → + j → + k → = - 1 2 + - 1 2 + 2 2 = 6.
Answer: i → - j → × i → + j → + k → = 6 . .
Example 4
In a rectangular Cartesian coordinate system, the coordinates of three points A (1, 0, 1), B (0, 2, 3), C (1, 4, 2) are given. Find some vector perpendicular to A B → and A C → at the same time.
Solution
Vectors A B → and A C → have the following coordinates (- 1 ; 2 ; 2) and (0 ; 4 ; 1) respectively. Having found the vector product of the vectors A B → and A C →, it is obvious that it is a perpendicular vector by definition to both A B → and A C →, that is, it is a solution to our problem. Let's find it A B → × A C → = i → j → k → - 1 2 2 0 4 1 = - 6 i → + j → - 4 k → .
Answer: - 6 i → + j → - 4 k → . - one of the perpendicular vectors.
Problems of the third type are focused on using the properties of the vector product of vectors. After applying which, we will obtain a solution to the given problem.
Example 5
Vectors a → and b → are perpendicular and their lengths are 3 and 4, respectively. Find the length of the vector product 3 a → - b → × a → - 2 b → = 3 a → × a → - 2 b → + - b → × a → - 2 b → = = 3 a → × a → + 3 · a → × - 2 · b → + - b → × a → + - b → × - 2 · b → .
Solution
By the distributive property of a vector product, we can write 3 a → - b → × a → - 2 b → = 3 a → × a → - 2 b → + - b → × a → - 2 b → = = 3 a → × a → + 3 a → × - 2 b → + - b → × a → + - b → × - 2 b →
By the property of associativity, we take the numerical coefficients out of the sign of the vector products in the last expression: 3 · a → × a → + 3 · a → × - 2 · b → + - b → × a → + - b → × - 2 · b → = = 3 · a → × a → + 3 · (- 2) · a → × b → + (- 1) · b → × a → + (- 1) · (- 2) · b → × b → = = 3 a → × a → - 6 a → × b → - b → × a → + 2 b → × b →
The vector products a → × a → and b → × b → are equal to 0, since a → × a → = a → · a → · sin 0 = 0 and b → × b → = b → · b → · sin 0 = 0, then 3 · a → × a → - 6 · a → × b → - b → × a → + 2 · b → × b → = - 6 · a → × b → - b → × a → . .
From the anticommutativity of the vector product it follows - 6 · a → × b → - b → × a → = - 6 · a → × b → - (- 1) · a → × b → = - 5 · a → × b → . .
Using the properties of the vector product, we obtain the equality 3 · a → - b → × a → - 2 · b → = = - 5 · a → × b → .
By condition, the vectors a → and b → are perpendicular, that is, the angle between them is equal to π 2. Now all that remains is to substitute the found values into the appropriate formulas: 3 a → - b → × a → - 2 b → = - 5 a → × b → = = 5 a → × b → = 5 a → b → · sin (a → , b →) = 5 · 3 · 4 · sin π 2 = 60 .
Answer: 3 a → - b → × a → - 2 b → = 60.
The length of the vector product of vectors by definition is equal to a → × b → = a → · b → · sin ∠ a → , b → . Since it is already known (from the school course) that the area of a triangle is equal to half the product of the lengths of its two sides multiplied by the sine of the angle between these sides. Consequently, the length of the vector product is equal to the area of the parallelogram - a doubled triangle, namely the product of the sides in the form of vectors a → and b →, laid down from one point, by the sine of the angle between them sin ∠ a →, b →.
This is the geometric meaning of a vector product.
Physical meaning of the vector product
In mechanics, one of the branches of physics, thanks to the vector product, you can determine the moment of a force relative to a point in space.
Definition 3
By the moment of force F → applied to point B, relative to point A, we will understand the following vector product A B → × F →.
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