Unit vector- This vector, the absolute value (modulus) of which equal to one. To denote a unit vector, we will use the subscript e. So, if a vector is given a, then its unit vector will be the vector a e. This unit vector points in the same direction as the vector itself a, and its modulus is equal to one, that is, a e \u003d 1.

Obviously, a= a a e (a - vector modulus a). This follows from the rule by which the operation of multiplying a scalar by a vector is performed.

Unit vectors often associated with the coordinate axes of the coordinate system (in particular, with the axes of the Cartesian coordinate system). Directions of these vectors coincide with the directions of the corresponding axes, and their origins are often combined with the origin of the coordinate system.

Let me remind you that Cartesian coordinate system in space is traditionally called a triple of mutually perpendicular axes intersecting at a point called the origin. Coordinate axes usually denoted by the letters X, Y, Z and are called respectively the abscissa axis, the ordinate axis and the applicate axis. Descartes himself used only one axis, on which the abscissas were plotted. merit of use systems axes belongs to his students. Therefore the phrase cartesian system coordinates historically wrong. Better talk rectangular coordinate system or orthogonal coordinate system. Nevertheless, we will not change traditions and in the future we will assume that the Cartesian and rectangular (orthogonal) coordinate systems are one and the same.

Unit vector, directed along the X axis, is denoted i, unit vector, directed along the Y axis, is denoted j, a unit vector, directed along the Z axis, is denoted k. Vectors i, j, k called orts(Fig. 12, left), they have single modules, that is
i = 1, j = 1, k = 1.

axes and orts rectangular coordinate system in some cases they have other names and designations. So, the abscissa axis X can be called the tangent axis, and its unit vector is denoted τ (Greek small letter tau), the y-axis is the normal axis, its unit vector is denoted n, the applicate axis is the axis of the binormal, its unit vector is denoted b. Why change the names if the essence remains the same?

The fact is that, for example, in mechanics, when studying the motion of bodies, a rectangular coordinate system is used very often. So, if the coordinate system itself is motionless, and the change in the coordinates of a moving object is tracked in this motionless system, then usually the axes denote X, Y, Z, and their orts respectively i, j, k.

But often, when an object moves along some curvilinear trajectory(for example, along a circle) it is more convenient to consider mechanical processes in a coordinate system moving with this object. It is for such a moving coordinate system that other names of the axes and their unit vectors are used. It's just accepted. In this case, the X-axis is directed tangentially to the trajectory at the point where this moment this object is located. And then this axis is no longer called the X axis, but the tangent axis, and its unit vector is no longer denoted i, a τ . The Y axis is directed along the radius of curvature of the trajectory (in the case of movement in a circle - to the center of the circle). And since the radius is perpendicular to the tangent, the axis is called the axis of the normal (perpendicular and normal are the same thing). The ort of this axis is no longer denoted j, a n. The third axis (the former Z) is perpendicular to the two previous ones. This is a binormal with a vector b(Fig. 12, right). By the way, in this case rectangular system coordinates often referred to as "natural" or natural.

In this lesson, we will look at two more operations with vectors: cross 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, besides dot product of vectors, more and more is needed. Such is vector addiction. It may seem that we are climbing into the wilds analytical geometry. This is not true. In this section of higher mathematics, there is generally little firewood, except perhaps enough for Pinocchio. In fact, the material is very common and simple - hardly more difficult than the same scalar product, even typical tasks will be less. The main thing in analytic geometry, as many will see or have already seen, is NOT TO MISTAKE CALCULATIONS. Repeat like a spell, and you will be happy =)

If the vectors sparkle somewhere far away, like lightning on the horizon, it doesn't matter, start with the lesson Vectors for dummies to restore or repurchase 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? When I was little, I could juggle two and even three balls. It worked out well. Now there is no need to juggle at all, since we will consider only space vectors, and flat vectors with two coordinates will be left out. Why? This is how these actions were born - vector and mixed product vectors are defined and work in three-dimensional space. Already easier!

In this operation, in the same way as in the scalar product, two vectors. Let it be imperishable letters.

The action itself denoted in the following way: . There are other options, but I'm used to designating the cross product of vectors in this way, in square brackets with a cross.

And immediately question: if in dot product of vectors two vectors are involved, and here two vectors are also multiplied, then what is the difference? A clear difference, first of all, in the RESULT:

The result of the scalar product of vectors is a NUMBER:

The result of the cross product of vectors is a VECTOR: , that is, we multiply the vectors and get a vector again. Closed club. Actually, hence the name of the operation. In various educational literature the notation can also vary, I will use the letter .

Definition of cross product

First there will be a definition with a picture, then comments.

Definition: cross product non-collinear vectors , taken in this order, is 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:

We analyze the definition by bones, there is a lot of interesting things!

So, we can highlight the following significant points:

1) Source vectors , indicated by red arrows, by definition not collinear. Happening collinear vectors it will be appropriate to consider a little later.

2) Vectors taken in a strict order: – "a" is multiplied by "be", not "be" to "a". The result of vector multiplication is VECTOR , which is denoted in blue. If the vectors are multiplied by reverse order, then we get a vector equal in length and opposite in direction (crimson color). That is, the equality .

3) Now let's get acquainted with the geometric meaning of the vector product. This is 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 in black.

Note : the drawing is schematic, and, of course, the nominal length of the cross product is not equal to the area of ​​the parallelogram.

We remember one of geometric formulas: the area of ​​a parallelogram is equal to the product adjacent parties by the sine of the angle between them. Therefore, based on the foregoing, the formula for calculating the LENGTH of a vector product is valid:

I emphasize that in the formula we are talking about the LENGTH of the vector, and not about the vector itself. What is the practical meaning? And the meaning is such that in problems of analytic geometry, the area of ​​a parallelogram is often found through the concept of a vector product:

Let's get a second important formula. The diagonal of the parallelogram (red dotted line) divides it into two equal triangle. Therefore, the area of ​​a triangle built on vectors (red shading) can be found by the formula:

4) Not less than important fact is that the vector is orthogonal to the vectors , that is, . Of course, the oppositely directed vector (crimson arrow) is also orthogonal to the original vectors .

5) The vector is directed so that basis It has right orientation. In a lesson about transition to a new basis I have spoken in detail about plane orientation, and now we will figure out what the orientation of space 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 into your palm. As a result thumb - the vector product will look up. This is the right-oriented basis (it is in the figure). Now swap 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. Perhaps you have a question: what basis has a left orientation? "Assign" the same fingers left hand vectors , and get the left basis and left space orientation (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 sides. And this concept should not be considered something far-fetched or abstract - for example, the most ordinary mirror changes the orientation of space, and if you “pull the reflected object out of the mirror”, then it general case cannot be matched with the original. By the way, bring three fingers 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 the change of orientation are terrible =)

Vector product of collinear vectors

The definition has been worked out 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 zero. The same follows from the formula - the sine of zero or 180 degrees zero, and hence the area is zero

Thus, if , then . Strictly speaking, the vector product itself is zero vector, but in practice this is often neglected and written that it is simply equal to zero.

special case is the cross product of a vector and itself:

Using the cross product, you can check the collinearity of three-dimensional vectors, and this task among others, we will also analyze.

For solutions practical examples may be required trigonometric table to find the values ​​of the sines from it.

Well, let's start a 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

Decision: No, this is not a typo, I intentionally made the initial data in the condition items the same. Because the design of the solutions will be different!

a) According to the condition, it is required to find length vector (vector product). According to the corresponding formula:

Answer:

Since it was asked about the length, then in the answer we indicate the dimension - units.

b) According to the condition, it is required to find square parallelogram built on vectors . The area of ​​this parallelogram is numerically equal to the length of the cross product:

Answer:

Please note that in the answer about the vector product there is no talk at all, we were asked about figure area, respectively, the dimension is square units.

We always look at WHAT is required to be found by the condition, and, based on this, we formulate clear answer. It may seem like literalism, but there are enough literalists among the teachers, and the task with good chances will be returned for revision. Although this is not a particularly strained nitpick - if the answer is incorrect, then one gets the impression that the person does not understand simple things and / or did not understand the essence of the task. This moment must always be kept under control, solving any problem on higher mathematics and in other subjects as well.

Where did the big letter "en" go? In principle, it could be additionally stuck to the solution, but in order to shorten the record, I did not. I hope everyone understands that and is the designation of the same thing.

Popular Example for independent decision:

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. Solution and answer at the end of the lesson.

In practice, the task is really very common, triangles can generally be tortured.

To solve other problems, we need:

Properties of the cross 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 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) - combination or associative vector product laws. The constants are easily taken out of the limits of the vector product. Really, what are they doing there?

4) - distribution or distribution vector product laws. There are no problems with opening brackets either.

As a demonstration, consider a short example:

Example 3

Find if

Decision: By condition, it is again required to find the length of the vector product. Let's paint our miniature:

(1) According to the associative laws, we take out the constants beyond the limits of the vector product.

(2) We take the constant out of the module, while the module “eats” the minus sign. The length cannot be negative.

(3) What follows is clear.

Answer:

It's time to throw wood on the fire:

Example 4

Calculate the area of ​​a triangle built on vectors if

Decision: Find the area of ​​a triangle using the formula . The snag is that the vectors "ce" and "te" are themselves represented as sums of vectors. The algorithm here is standard and is somewhat reminiscent of examples No. 3 and 4 of the lesson. Dot product of vectors. Let's break it down into three steps for clarity:

1) At the first step, we express the vector product through the vector product, in fact, express the vector in terms of the vector. No word on length yet!

(1) We substitute expressions of vectors .

(2) Using distributive laws, we open the brackets according to the rule of multiplication of polynomials.

(3) Using the associative laws, we take out all the constants beyond the vector products. With little experience, actions 2 and 3 can be performed simultaneously.

(4) The first and last terms are equal to zero (zero vector) due to the pleasant property . In the second term, we use the anticommutativity property of the vector product:

(5) We present similar terms.

As a result, the vector turned out to be expressed through a vector, which was what was required to be achieved:

2) At the second step, we find the length of the vector product we need. This action reminiscent of Example 3:

3) Find the area of ​​the required triangle:

Steps 2-3 of the solution could be arranged in one line.

Answer:

The considered problem is quite common in control work, here's an example for a do-it-yourself solution:

Example 5

Find if

Quick Solution and the 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

, given in the orthonormal basis , is expressed by the formula:

The formula is really simple: we write the coordinate vectors in the top line of the determinant, we “pack” the coordinates of the vectors into the second and third lines, and we put in strict order- first, the coordinates of the vector "ve", then the coordinates of the vector "double-ve". If the vectors need to be multiplied in a different order, then the lines should also be swapped:

Example 10

Check if the following space vectors are collinear:
a)
b)

Decision: Validation based on one of the assertions this lesson: if the vectors are collinear, then their vector product is zero (zero vector): .

a) Find the vector product:

So 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 rest on the definition, geometric meaning and a couple of working formulas.

The mixed product of vectors is product of three vectors:

This is how they lined up like a train and wait, they can’t wait until they are calculated.

First again the definition and picture:

Definition: Mixed product non-coplanar vectors , taken in this order, is called volume of the parallelepiped, 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 by a dotted line:

Let's dive into the definition:

2) Vectors taken in a certain order, that is, the permutation of vectors in the product, as you might guess, does not go without consequences.

3) Before commenting on the geometric meaning, I note obvious fact: the mixed product of vectors is a NUMBER: . In educational literature, the design may be somewhat different, I used to designate a mixed product through, and the result of calculations with 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 the given parallelepiped.

Note : The drawing is schematic.

4) Let's not bother again with the concept of the 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, the mixed product can be negative: .

The formula for calculating the volume of a parallelepiped built on vectors follows directly from the definition.

Definition An ordered set of (x 1 , x 2 , ... , x n) n real numbers called n-dimensional vector, and the numbers x i (i = ) - components or coordinates,

Example. If, for example, a certain automobile plant has to produce 50 cars, 100 trucks, 10 buses, 50 sets of spare parts for cars and 150 sets for trucks and buses, then the production program of this plant can be written as a vector (50, 100, 10, 50, 150) with five components.

Notation. Vectors are denoted in bold lower case or letters with a bar or arrow at the top, for example, a or. The two vectors are called equal if they have the same number component and their corresponding components are equal.

Vector components cannot be interchanged, e.g. (3, 2, 5, 0, 1) and (2, 3, 5, 0, 1) different vectors.
Operations on vectors. work x= (x 1 , x 2 , ... ,x n) to a real numberλ called vectorλ x= (λ x 1 , λ x 2 , ... , λ x n).

sumx= (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).

The space of vectors. N -dimensional vector space R n is defined as the set of all n-dimensional vectors for which operations of multiplication by real numbers and addition.

Economic illustration. An economic illustration of an n-dimensional vector space: space of goods (goods). Under commodity we will understand some good or service that went on sale at a certain time in certain place. Assume that there is a finite number of goods available n; 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 bought. Then all possible sets of goods are vectors of the space of goods C = ( x= (x 1 , x 2 , ... , x n) x i ≥ 0, i = ).

Linear independence. System e 1 , e 2 , ... , e m n-dimensional vectors is called linearly dependent if there are such numbersλ 1 , λ 2 , ... , λ m , of which at least one is nonzero, which satisfies the equalityλ1 e 1 + λ2 e 2+...+λm e m = 0; otherwise this system vectors is called linearly independent, that is, this equality is possible only in the case when all . geometric sense linear dependence 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. For two vectors to be linearly dependent, it is necessary and sufficient that they be collinear (parallel).

Theorem 3 . For three vectors to be linearly dependent, it is necessary and sufficient that they be coplanar (lying in the same plane).

Left and right triples of vectors. A triple of non-coplanar vectors a, b, c called right, if the observer from them common beginning bypassing the ends of vectors a, b, c in that order seems to proceed clockwise. Otherwise a, b, c -left triple. All right (or left) triples of vectors are called equally oriented.

Basis and coordinates. Troika e 1, e 2 , e 3 non-coplanar vectors in R 3 called basis, and the vectors themselves e 1, e 2 , e 3 - basic. Any vector a can be expanded in a unique way in terms of basis vectors, that is, it can be 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 basis e 1, e 2 , e 3 and are denoted 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 i, j, k.

We will assume that in space R 3 the right system of Cartesian rectangular coordinates (0, i, j, k}.

vector product. vector art a per vector b called vector c, which is determined by the following three conditions:

1. Vector length c numerically equal to the area of ​​the parallelogram built on the 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 that order, form a right triple.

For vector 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 orts: [ ij]=k, [jk] = i, [ki]=j.

If the vectors a and b given 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 cross product of two vectors a and b scalar multiplied by the third vector c, then such a product of three vectors is called mixed product and is denoted by the symbol a bc.

If the vectors a, b and c in basis i, j, k set 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, according to absolute value equal to the volume of the 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 the 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 the right orthonormal basis. Unit vector codirectional to vector a, denoted by the symbol a about. Symbol r=OM denoted by the radius vector of the point M, the symbols a, AB or|a|, | AB |the modules of vectors are denoted a and AB.

Example 1.2. Find the angle between vectors a= 2m+4n and b= m-n, where m and n- unit vectors and angle between m and n equal to 120 o.

Decision. 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, so a = . b= ; b 2 =
= (m-n)(m-n) = m 2 -2mn+n 2 = 1-2(-0.5)+1 = 3, so b = . Finally we have: cosφ \u003d -1/2, φ \u003d 120 o.

Example 1.3.Knowing vectors AB(-3,-2.6) and BC(-2,4,4), calculate the height AD of triangle ABC.

Decision. Denoting the area of ​​triangle ABC by S, we get:
S = 1/2 B.C. AD. Then AD=2S/BC, BC== = 6,
S = 1/2| AB ×AC|. AC=AB+BC, so the vector AC has coordinates
. .

Example 1.4 . Given two vectors 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.

Decision.Let us denote the coordinates of the vector c with respect to the given right orthonormal basis in terms of x, y, z.

Insofar as c ⊥ a, c ⊥b, then ca= 0, cb= 0. By the condition 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 get z = -4/3 x, y = -5/6 x. Substituting y and z into the third equation, we will have: x 2 = 36/125, whence
x=± . Using 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, whence it follows that x>0. So x = , y = - , z = - .

7.1. Definition of cross product

Three non-coplanar vectors a , b and c , taken in the indicated order, form a right triple 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 one if clockwise (see Fig. . sixteen).

The vector product of a 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. It has a length numerically equal to the area of ​​the parallelogram built on the vectors a andb as on the sides (see fig. 17), i.e.

3. The vectors a , b and c form a right triple.

The vector product is denoted a x b or [a,b]. From the definition of a vector product, the following relations between the orts i follow directly, j and k(see fig. 18):

i x j \u003d k, j x k \u003d i, k x i \u003d j.
Let us prove, for example, that i xj \u003d 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. Cross product properties

1. When the factors are rearranged, the vector product changes sign, i.e. and xb \u003d (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, and xb and a, b, b x a of opposite orientation). That is axb = -(bxa).

2. The vector product has associative property with respect to a scalar factor, i.e. l ​​(a xb) \u003d (l a) x b \u003d a x (l b).

Let l >0. The vector l (a xb) is perpendicular to the vectors a and b. Vector ( l a) x b is also perpendicular to the vectors a and b(vectors a, l but lie in the same plane). So the vectors l(a xb) and ( l a) x b collinear. It is obvious that their directions coincide. They have the same length:

So l(a xb)= l a xb. It is proved similarly 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., and ||b<=>and xb \u003d 0.

In particular, i *i =j *j =k *k =0 .

4. The vector product has a distribution property:

(a+b) xs = a xs + b xs .

Accept without proof.

7.3. Cross product expression in terms of coordinates

We will use the vector cross product table 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 match, the third vector is taken with a minus sign.

Let two vectors a =a x i +a y j+az k and b=bx i+by j+bz 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 shorter:

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 the cross product

Establishing collinearity of vectors

Finding the area of ​​a parallelogram and a triangle

According to the definition of the cross product of vectors a and b |a xb | =| a | * |b |sin g , i.e. S par = |a x b |. And, therefore, D S \u003d 1/2 | a x b |.

Determining the moment of force about a point

Let a force be applied at point A F =AB let it go O- some point in space (see Fig. 20).

It is known from physics that torque F relative to the point O called vector M , which passes through the point O and:

1) perpendicular to the plane passing through the points O, A, B;

2) numerically equal to the product of the force and the shoulder

3) forms a right triple with vectors OA and A B .

Therefore, M \u003d OA x F.

Finding the linear speed of rotation

Speed v point M of a rigid body rotating at an angular velocity w around a fixed axis, is determined by the Euler formula v \u003d w x r, where r \u003d OM, where O is some fixed point of the axis (see Fig. 21).

Definition. The vector product of a vector a (multiplier) by a vector (multiplier) that is not collinear with it is the third vector c (product), which is constructed as follows:

1) its modulus 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 (out of two possible ones) so that the vectors c form a right-handed system (§ 110).

Designation: or

Addendum to the definition. If the vectors are collinear, then considering the figure as a (conditionally) 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 convention does not contradict items 2 and 3 of the definition.

Remark 1. In the term "vector product", the first word indicates that the result of an action is a vector (as opposed to a scalar product; cf. § 104, remark 1).

Example 1. Find the vector product where the main vectors of the right coordinate system (Fig. 156).

1. Since the lengths of the main vectors are equal to the scale unit, the area of ​​the parallelogram (square) is numerically equal to one. Hence, the modulus of the vector product is equal to one.

2. Since the perpendicular to the plane is the axis, the desired vector product is a vector collinear to the vector k; and since both of them have modulus 1, the required cross product is either k or -k.

3. Of these two possible vectors, the first must be chosen, since the vectors k form a right system (and the vectors form a left one).

Example 2. Find the cross product

Decision. As in example 1, we conclude that the vector is either k or -k. But now we need to choose -k, since the vectors form the right system (and the vectors form the left). So,

Example 3 The vectors have lengths of 80 and 50 cm, respectively, and form an angle of 30°. Taking a meter as a unit of length, find the length of the vector product a

Decision. 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 cross product of the same vectors, taking a centimeter as a unit of length.

Decision. Since the area of ​​the parallelogram built on vectors is equal to the length of the vector product is 2000 cm, i.e.

Comparison of examples 3 and 4 shows that the length of the vector depends not only on the lengths of the factors, but also on the choice of the length unit.

The physical meaning of the vector product. Of the many physical quantities represented by the vector product, we will consider only the moment of force.

Let A be the point of application of the force. The moment of force relative to the point O is called the vector product. Since the module of this vector product is numerically equal to the area of ​​the parallelogram (Fig. 157), the module of the moment is equal to the product of the base by the height, i.e., the force multiplied by the distance from the point O to the straight line along which the force acts.

In mechanics, it is proved that for the equilibrium of a rigid body, it is necessary that not only the sum of the vectors representing the forces applied to the body, but also the sum of the moments of forces should be equal to zero. In the case when all forces are parallel to the same plane, the addition of the vectors representing the moments can be replaced by the addition and subtraction of their moduli. But for arbitrary directions of forces, such a replacement is impossible. In accordance with this, the cross product is defined precisely as a vector, and not as a number.