Vectors can be represented graphically by directed segments. The length is chosen on a specific scale to indicate vector magnitude , and the direction of the segment represents vector direction . For example, if we assume that 1 cm represents 5 km/h, then a northeast wind with a speed of 15 km/h will be represented by a directional segment of length 3 cm, as shown in the figure.
Vector on a plane it is a directed segment. Two vectors equal if they have the same size And direction.
Consider a vector drawn from point A to point B. The point is called starting point vector, and point B is called end point. The symbolic notation for this vector is (read as “vector AB”). Vectors are also represented by bold letters such as U, V, and W. The four vectors in the figure on the left have the same length and direction. Therefore they represent equal winds; that is, 
In the context of vectors, we use = to indicate that they are equal.
Length, or magnitude is expressed as ||. In order to determine whether the vectors are equal, we find their magnitudes and directions.
Example 1 Vectors u, , w are shown in the figure below. Prove that u = = w. 
Solution First we find the length of each vector using the distance formula:
|u| = √ 2 + (4 - 3) 2 = √9 + 1 = √10,
|| = √ 2 + 2
= √9 + 1
= √10
,
|w| = √(4 - 1) 2 + [-1 - (-2)] 2 = √9 + 1 = √10 .
From here
|u| = | = |w|.
Vectors u, , and w, as can be seen from the figure, seem to have the same direction, but we will check their slope. If the lines on which they are located have the same slopes, then the vectors have the same direction. We calculate the slopes: 
Since u, , and w have equal magnitudes and the same direction,
u = = w.
Keep in mind that equal vectors only require the same magnitude and the same direction, not the same location. The topmost figure shows an example of vector equality.
Suppose a person takes 4 steps east and then 3 steps north. The person will then be 5 steps from the starting point in the direction shown on the left. A vector 4 units long with a direction to the right represents 4 steps east and a vector 3 units long with a direction up representing 3 steps north. Sum of these two vectors there is a vector of 5 steps of magnitude and in the direction shown. The amount is also called resulting two vectors.
In general, two nonzero vectors u and v can be added geometrically by placing the starting point of the vector v to the ending point of the vector u, and then finding a vector that has the same starting point as the vector u and the same ending point as the vector v as shown in the figure below. 
The sum is a vector represented by a directed segment from point A of vector u to end point C of vector v. Thus, if u = and v = , then
u + v = + =
We can also describe vector addition as placing the starting points of the vectors together, constructing a parallelogram, and finding the diagonal of the parallelogram. (in the figure below.) This addition is sometimes called as parallelogram rule
addition of vectors. Vector addition is commutative. As shown in the figure, both vectors u + v and v + u are represented by the same directional line segment. 
If two forces F 1 and F 2 act on one object, resulting force is the sum of F 1 + F 2 of these two separate forces. 
Example Two forces of 15 newtons and 25 newtons act on one object perpendicular to each other. Find their sum, or the resulting force, and the angle it makes with the greater force.
Solution Let's draw the condition of the problem, in this case a rectangle, using v or to represent the resultant. To find its value, we use the Pythagorean theorem:
|v| 2 = 15 2 + 25 2 Here |v| denotes the length or magnitude of v.
|v| = √15 2 + 25 2
|v| ≈ 29.2.
To find the direction, note that since OAB is a right angle,
tanθ = 15/25 = 0.6.
Using a calculator, we find θ, the angle that the larger force makes with the net force:
θ = tan - 1 (0.6) ≈ 31°
The resultant has a magnitude of 29.2 and an angle of 31° with greater force.
Pilots can adjust their flight direction if there is a crosswind. The wind and speed of an airplane can be represented as winds.
Example 3. Airplane speed and direction. The plane is moving along an azimuth of 100° at a speed of 190 km/h, while the wind speed is 48 km/h and its azimuth is 220°. Find the absolute speed of the plane and the direction of its movement, taking into account the wind.
Solution Let's make a drawing first. The wind is represented and the aircraft speed vector is . The resulting velocity vector is v, the sum of the two vectors. The angle θ between v and is called drift angle
.
Note that the COA value = 100° - 40° = 60°. Then the value of CBA is also equal to 60° (opposite angles of the parallelogram are equal). Since the sum of all the angles of a parallelogram is 360° and COB and OAB are the same magnitude, each must be 120°. By cosine rule
in OAB, we have
|v| 2 = 48 2 + 190 2 - 2.48.190.cos120°
|v| 2 = 47.524
|v| = 218
Then, |v| equals 218 km/h. According to rule of sines
, in the same triangle,
48
/sinθ = 218
/sin 120°,
or
sinθ = 48.sin120°/218 ≈ 0.1907
θ ≈ 11°
Then, θ = 11°, to the nearest integer angle. The absolute speed is 218 km/h, and the direction of its movement taking into account the wind: 100° - 11°, or 89°.
Given a vector w, we can find two other vectors u and v whose sum is w. The vectors u and v are called components w and the process of finding them is called decomposition , or the representation of a vector by its vector components.
When we expand a vector, we usually look for perpendicular components. Very often, however, one component will be parallel to the x-axis and the other will be parallel to the y-axis. Therefore, they are often called horizontal
And vertical
vector components. In the figure below, the vector w = is decomposed as the sum of u = and v =. 
The horizontal component of w is u and the vertical component is v.
Example 4 The vector w has a magnitude of 130 and a slope of 40° relative to the horizontal. Decompose the vector into horizontal and vertical components.
Solution First we will draw a picture with horizontal and vertical vectors u and v whose sum is w. 
From ABC, we find |u| and |v|, using the definitions of cosine and sine:
cos40° = |u|/130, or |u| = 130.cos40° ≈ 100,
sin40° = |v|/130, or |v| = 130.sin40° ≈ 84.
Then, the horizontal component of w is 100 to the right and the vertical component of w is 84 up.
The basis of space they call such a system of vectors in which all other vectors in space can be represented as a linear combination of vectors included in the basis.
In practice, this is all implemented quite simply. The basis, as a rule, is checked on a plane or in space, and for this you need to find the determinant of a second, third order matrix composed of vector coordinates. Below are schematically written conditions under which vectors form a basis
To expand vector b into basis vectors
e,e...,e[n] it is necessary to find the coefficients x, ..., x[n] for which the linear combination of vectors e,e...,e[n] is equal to the vector b:
x1*e+ ... + x[n]*e[n] = b.
To do this, the vector equation should be converted to a system of linear equations and solutions should be found. This is also quite simple to implement.
The found coefficients x, ..., x[n] are called coordinates of vector b in the basis e,e...,e[n].
Let's move on to the practical side of the topic.
Decomposition of a vector into basis vectors
Task 1. Check whether vectors a1, a2 form a basis on the plane
1) a1 (3; 5), a2 (4; 2)
Solution: We compose a determinant from the coordinates of the vectors and calculate it
Determinant is not zero, hence the vectors are linearly independent, which means they form a basis.
2) a1 (2;-3), a2 (5;-1)
Solution: We calculate the determinant made up of vectors 
The determinant is equal to 13 (not equal to zero) - from this it follows that the vectors a1, a2 are a basis on the plane.
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Let's look at typical examples from the MAUP program in the discipline "Higher Mathematics".
Task 2. Show that the vectors a1, a2, a3 form the basis of a three-dimensional vector space, and expand the vector b according to this basis (use Cramer’s method when solving a system of linear algebraic equations).
1) a1 (3; 1; 5), a2 (3; 2; 8), a3 (0; 1; 2), b (−3; 1; 2).
Solution: First, consider the system of vectors a1, a2, a3 and check the determinant of matrix A
built on non-zero vectors. The matrix contains one zero element, so it is more appropriate to calculate the determinant as a schedule in the first column or third row. 
As a result of the calculations, we found that the determinant is different from zero, therefore vectors a1, a2, a3 are linearly independent.
By definition, vectors form a basis in R3. Let's write down the schedule of vector b based on 
Vectors are equal when their corresponding coordinates are equal.
Therefore, from the vector equation we obtain a system of linear equations 
Let's solve SLAE Cramer's method. To do this, we write the system of equations in the form
The main determinant of a SLAE is always equal to the determinant composed of basis vectors
Therefore, in practice it is not counted twice. To find auxiliary determinants, we put a column of free terms in place of each column of the main determinant. Determinants are calculated using the triangle rule 
Let's substitute the found determinants into Cramer's formula 
So, the expansion of the vector b in terms of the basis has the form b=-4a1+3a2-a3. The coordinates of vector b in the basis a1, a2, a3 will be (-4,3, 1).
2)a1 (1; -5; 2), a2 (2; 3; 0), a3 (1; -1; 1), b (3; 5; 1).
Solution: We check the vectors for a basis - we compose a determinant from the coordinates of the vectors and calculate it 
The determinant is not equal to zero, therefore vectors form a basis in space. It remains to find the schedule of vector b through this basis. To do this, we write the vector equation 
and transform to a system of linear equations 
We write the matrix equation
Next, for Cramer’s formulas we find auxiliary determinants 
We apply Cramer's formulas 
So a given vector b has a schedule through two basis vectors b=-2a1+5a3, and its coordinates in the basis are equal to b(-2,0, 5).
Test assignments
Task 1 - 10. Vectors are given. Show that vectors form a basis of three-dimensional space and find the coordinates of the vector in this basis:
Given vectors ε 1 (3;1;6), ε 2 (-2;2;-3), ε 3 (-4;5;-1), X(3;0;1). Show that the vectors form the basis of three-dimensional space and find the coordinates of the vector X in this basis.
This task consists of two parts. First you need to check whether the vectors form a basis. Vectors form a basis if the determinant composed of the coordinates of these vectors is nonzero, otherwise the vectors are not basic and the vector X cannot be expanded over this basis.
Let's calculate the determinant of the matrix:
∆ = 3*(2*(-1) - 5*(-3)) - -2*(1*(-1) - 5*6) + -4*(1*(-3) - 2*6) = 37
The determinant of the matrix is ∆ =37
Since the determinant is nonzero, the vectors form a basis, therefore, the vector X can be expanded over this basis. Those. there are numbers α 1, α 2, α 3 such that the equality holds:
X = α 1 ε 1 + α 2 ε 2 + α 3 ε 3
Let us write this equality in coordinate form:
(3;0;1) = α(3;1;6) + α(-2;2;-3) + α(-4;5;-1)
Using the properties of vectors, we obtain the following equality:
(3;0;1) = (3α 1 ;1α 1 ;6α 1 ;) + (-2α 2 ;2α 2 ;-3α 2 ;) + (-4α 3 ;5α 3 ;-1α 3 ;)
(3;0;1) = (3α 1 -2α 2 -4α 3 ;1α 1 + 2α 2 + 5α 3 ;6α 1 -3α 2 -1α 3)
By the property of equality of vectors we have:
3α 1 -2α 2 -4α 3 = 3
1α 1 + 2α 2 + 5α 3 = 0
6α 1 -3α 2 -1α 3 = 1
We solve the resulting system of equations Gaussian method or Cramer's method.
X = ε 1 + 2ε 2 -ε 3
The solution was received and processed using the service:
Vector coordinates in the basis
Along with this problem they also solve:
Solving matrix equations
Cramer method
Gauss method
Inverse matrix using the Jordano-Gauss method
Inverse matrix via algebraic complements
Online matrix multiplication
Standard definition: “A vector is a directed segment.” This is usually the extent of a graduate’s knowledge about vectors. Who needs any “directional segments”?
But really, what are vectors and what are they for?
Weather forecast. “Wind northwest, speed 18 meters per second.” Agree, both the direction of the wind (where it blows from) and the magnitude (that is, the absolute value) of its speed matter.
Quantities that have no direction are called scalar. Mass, work, electric charge are not directed anywhere. They are characterized only by a numerical value - “how many kilograms” or “how many joules”.
Physical quantities that have not only an absolute value, but also a direction, are called vector quantities.
Speed, force, acceleration - vectors. For them, “how much” is important and “where” is important. For example, acceleration due to gravity 
directed towards the Earth's surface, and its magnitude is 9.8 m/s 2. Impulse, electric field strength, magnetic field induction are also vector quantities.
You remember that physical quantities are denoted by letters, Latin or Greek. The arrow above the letter indicates that the quantity is vector:
Here's another example.
A car moves from A to B. The end result is its movement from point A to point B, that is, movement by a vector.
Now it’s clear why a vector is a directed segment. Please note that the end of the vector is where the arrow is. Vector length is called the length of this segment. Indicated by: or
Until now, we have worked with scalar quantities, according to the rules of arithmetic and elementary algebra. Vectors are a new concept. This is another class of mathematical objects. They have their own rules.
Once upon a time we didn’t even know anything about numbers. My acquaintance with them began in elementary school. It turned out that numbers can be compared with each other, added, subtracted, multiplied and divided. We learned that there is a number one and a number zero.
Now we are introduced to vectors.
The concepts of “more” and “less” for vectors do not exist - after all, their directions can be different. Only vector lengths can be compared.
But there is a concept of equality for vectors.
Equal vectors that have the same length and the same direction are called. This means that the vector can be transferred parallel to itself to any point in the plane.
Single is a vector whose length is 1. Zero is a vector whose length is zero, that is, its beginning coincides with the end.
It is most convenient to work with vectors in a rectangular coordinate system - the same one in which we draw graphs of functions. Each point in the coordinate system corresponds to two numbers - its x and y coordinates, abscissa and ordinate.
The vector is also specified by two coordinates:
Here the coordinates of the vector are written in parentheses - in x and y.
They are found simply: the coordinate of the end of the vector minus the coordinate of its beginning.
If the vector coordinates are given, its length is found by the formula
Vector addition
There are two ways to add vectors.
1 . Parallelogram rule. To add the vectors and , we place the origins of both at the same point. We build up to a parallelogram and from the same point we draw a diagonal of the parallelogram. This will be the sum of the vectors and .
Remember the fable about the swan, crayfish and pike? They tried very hard, but they never moved the cart. After all, the vector sum of the forces they applied to the cart was equal to zero.
2. The second way to add vectors is the triangle rule. Let's take the same vectors and . We will add the beginning of the second to the end of the first vector. Now let's connect the beginning of the first and the end of the second. This is the sum of the vectors and .
Using the same rule, you can add several vectors. We arrange them one after another, and then connect the beginning of the first to the end of the last.
Imagine that you are going from point A to point B, from B to C, from C to D, then to E and to F. The end result of these actions is movement from A to F.
When adding vectors and we get:
Vector subtraction
The vector is directed opposite to the vector. The lengths of the vectors and are equal.
Now it’s clear what vector subtraction is. The vector difference and is the sum of the vector and the vector .
Multiplying a vector by a number
When a vector is multiplied by the number k, a vector is obtained whose length is k times different from the length . It is codirectional with the vector if k is greater than zero, and opposite if k is less than zero.
Dot product of vectors
Vectors can be multiplied not only by numbers, but also by each other.
The scalar product of vectors is the product of the lengths of the vectors and the cosine of the angle between them.
Please note that we multiplied two vectors, and the result was a scalar, that is, a number. For example, in physics, mechanical work is equal to the scalar product of two vectors - force and displacement:
If the vectors are perpendicular, their scalar product is zero.
And this is how the scalar product is expressed through the coordinates of the vectors and:
From the formula for the scalar product you can find the angle between the vectors:
This formula is especially convenient in stereometry. For example, in Problem 14 of the Profile Unified State Exam in Mathematics, you need to find the angle between intersecting lines or between a straight line and a plane. Often vector method problem 14 is solved several times faster than the classical one.
In the school mathematics curriculum, only the scalar product of vectors is taught.
It turns out that, in addition to the scalar product, there is also a vector product, when the result of multiplying two vectors is a vector. Who rents Unified State Exam in Physics, knows what the Lorentz force and the Ampere force are. The formulas for finding these forces include vector products.
Vectors are a very useful mathematical tool. You will see this in your first year.
