Let two vectors and be given in space. Let's postpone from an arbitrary point O vectors and . Angle between vectors is called the smallest of the angles. Designated 
.
Consider the axis l and plot a unit vector on it (i.e., a vector whose length is equal to one).
At an angle between the vector and the axis l understand the angle between the vectors and .
So let l is some axis and is a vector.
Let us denote by A 1 And B 1 projections onto the axis l respectively points A And B. Let's assume that A 1 has a coordinate x 1, A B 1– coordinate x 2 on the axis l.
Then projection vector per axis l called difference x 1 – x 2 between the coordinates of the projections of the end and beginning of the vector onto this axis.
Projection of the vector onto the axis l we will denote .
It is clear that if the angle between the vector and the axis l spicy then x 2> x 1, and projection x 2 – x 1> 0; if this angle is obtuse, then x 2< x 1 and projection x 2 – x 1< 0. Наконец, если вектор перпендикулярен оси l, That x 2= x 1 And x 2– x 1=0.
Thus, the projection of the vector onto the axis l is the length of the segment A 1 B 1, taken with a certain sign. Therefore, the projection of the vector onto the axis is a number or a scalar.
The projection of one vector onto another is determined similarly. In this case, the projections of the ends of this vector onto the line on which the 2nd vector lies are found.
Let's look at some basic properties of projections.
LINEARLY DEPENDENT AND LINEARLY INDEPENDENT VECTOR SYSTEMS
Let's consider several vectors.
Linear combination of these vectors is any vector of the form , where are some numbers. The numbers are called linear combination coefficients. They also say that in this case it is linearly expressed through these vectors, i.e. is obtained from them using linear actions.
For example, if three vectors are given, then the following vectors can be considered as their linear combination: 
If a vector is represented as a linear combination of some vectors, then it is said to be laid out along these vectors.
The vectors are called linearly dependent, if there are numbers, not all equal to zero, such that 
. It is clear that given vectors will be linearly dependent if any of these vectors is linearly expressed in terms of the others.
Otherwise, i.e. when the ratio performed only when 
, these vectors are called linearly independent.
Theorem 1. Any two vectors are linearly dependent if and only if they are collinear.
Proof:
The following theorem can be proven similarly.
Theorem 2. Three vectors are linearly dependent if and only if they are coplanar.
Proof.
BASIS
Basis is the collection of non-zeros linearly independent vectors. We will denote the elements of the basis by .
In the previous paragraph, we saw that two non-collinear vectors on a plane are linearly independent. Therefore, according to Theorem 1 from the previous paragraph, a basis on a plane is any two non-collinear vectors on this plane.
Similarly, any three non-coplanar vectors are linearly independent in space. Consequently, we call three non-coplanar vectors a basis in space.
The following statement is true.
Theorem. Let a basis be given in space. Then any vector can be represented as a linear combination 
, Where x, y, z- some numbers. This is the only decomposition.
Proof.
Thus, the basis allows each vector to be uniquely associated with a triple of numbers - the coefficients of the expansion of this vector into the basis vectors: . The converse is also true, for every three numbers x, y, z using the basis, you can compare the vector if you make a linear combination 
.
If the basis and , then the numbers x, y, z are called coordinates vector in a given basis. Vector coordinates are denoted by .
CARTESIAN COORDINATE SYSTEM
Let a point be given in space O and three non-coplanar vectors.
Cartesian system coordinates in space (on the plane) is the collection of a point and a basis, i.e. a set of a point and three non-coplanar vectors (2 non-collinear vectors) emanating from this point.
Dot O called the origin; straight lines passing through the origin of coordinates in the direction of the basis vectors are called coordinate axes - the abscissa, ordinate and applicate axis. Planes passing through the coordinate axes are called coordinate planes.
Let us consider in the selected coordinate system arbitrary point M. Let us introduce the concept of point coordinates M. Vector connecting the origin to a point M. called radius vector points M.
A vector in the selected basis can be associated with a triple of numbers – its coordinates: 
.
Coordinates of the radius vector of the point M. are called coordinates of point M. in the coordinate system under consideration. M(x,y,z). The first coordinate is called the abscissa, the second is the ordinate, and the third is the applicate.
Similarly defined Cartesian coordinates on a plane. Here the point has only two coordinates - abscissa and ordinate.
It is easy to see that when given system coordinates, each point has certain coordinates. On the other hand, for each triple of numbers there is a single point that has these numbers as coordinates.
If the vectors taken as a basis in the selected coordinate system have unit length and are pairwise perpendicular, then the coordinate system is called Cartesian rectangular.
It is easy to show that .
The direction cosines of a vector completely determine its direction, but say nothing about its length.
and on an axis or some other vector there are the concepts of its geometric projection and numerical (or algebraic) projection. The result of a geometric projection will be a vector, and the result of an algebraic projection will be a non-negative real number. But before we move on to these concepts, let us remember necessary information.
Preliminary information
The main concept is the concept of a vector itself. In order to introduce the definition of a geometric vector, let us remember what a segment is. Let us introduce the following definition.
Definition 1
A segment is a part of a straight line that has two boundaries in the form of points.
A segment can have 2 directions. To denote the direction, we will call one of the boundaries of the segment its beginning, and the other boundary its end. The direction is indicated from its beginning to the end of the segment.
Definition 2
We will call a vector or a directed segment a segment for which it is known which of the boundaries of the segment is considered the beginning and which is its end.
Designation: In two letters: $\overline(AB)$ – (where $A$ is its beginning, and $B$ is its end).
In one small letter: $\overline(a)$ (Fig. 1).
Let us introduce a few more concepts related to the concept of a vector.
Definition 3
We will call two non-zero vectors collinear if they lie on the same line or on lines parallel to each other (Fig. 2).
Definition 4
We will call two non-zero vectors codirectional if they satisfy two conditions:
- These vectors are collinear.
- If they are directed in one direction (Fig. 3).
Notation: $\overline(a)\overline(b)$
Definition 5
We will call two non-zero vectors oppositely directed if they satisfy two conditions:
- These vectors are collinear.
- If they are directed to different sides(Fig. 4).
Notation: $\overline(a)↓\overline(d)$
Definition 6
The length of the vector $\overline(a)$ will be the length of the segment $a$.
Notation: $|\overline(a)|$
Let's move on to determining the equality of two vectors
Definition 7
We will call two vectors equal if they satisfy two conditions:
- They are co-directional;
- Their lengths are equal (Fig. 5).
Geometric projection
As we said earlier, the result of a geometric projection will be a vector.
Definition 8
The geometric projection of the vector $\overline(AB)$ onto an axis is a vector that is obtained as follows: The origin point of the vector $A$ is projected onto this axis. We obtain point $A"$ - the beginning of the desired vector. The end point of vector $B$ is projected onto this axis. We obtain point $B"$ - the end of the desired vector. The vector $\overline(A"B")$ will be the desired vector.
Let's consider the problem:
Example 1
Build geometric projection$\overline(AB)$ to the $l$ axis, shown in Figure 6.
Let us draw a perpendicular from point $A$ to the axis $l$, we obtain point $A"$ on it. Next, we draw a perpendicular from point $B$ to the axis $l$, we obtain point $B"$ on it (Fig. 7).
and on an axis or some other vector there are the concepts of its geometric projection and numerical (or algebraic) projection. The result of a geometric projection will be a vector, and the result of an algebraic projection will be a non-negative real number. But before moving on to these concepts, let’s remember the necessary information.
Preliminary information
The main concept is the concept of a vector itself. In order to introduce the definition of a geometric vector, let us remember what a segment is. Let us introduce the following definition.
Definition 1
A segment is a part of a straight line that has two boundaries in the form of points.
A segment can have 2 directions. To denote the direction, we will call one of the boundaries of the segment its beginning, and the other boundary its end. The direction is indicated from its beginning to the end of the segment.
Definition 2
We will call a vector or a directed segment a segment for which it is known which of the boundaries of the segment is considered the beginning and which is its end.
Designation: In two letters: $\overline(AB)$ – (where $A$ is its beginning, and $B$ is its end).
In one small letter: $\overline(a)$ (Fig. 1).
Let us introduce a few more concepts related to the concept of a vector.
Definition 3
We will call two non-zero vectors collinear if they lie on the same line or on lines parallel to each other (Fig. 2).
Definition 4
We will call two non-zero vectors codirectional if they satisfy two conditions:
- These vectors are collinear.
- If they are directed in one direction (Fig. 3).
Notation: $\overline(a)\overline(b)$
Definition 5
We will call two non-zero vectors oppositely directed if they satisfy two conditions:
- These vectors are collinear.
- If they are directed in different directions (Fig. 4).
Notation: $\overline(a)↓\overline(d)$
Definition 6
The length of the vector $\overline(a)$ will be the length of the segment $a$.
Notation: $|\overline(a)|$
Let's move on to determining the equality of two vectors
Definition 7
We will call two vectors equal if they satisfy two conditions:
- They are co-directional;
- Their lengths are equal (Fig. 5).
Geometric projection
As we said earlier, the result of a geometric projection will be a vector.
Definition 8
The geometric projection of the vector $\overline(AB)$ onto an axis is a vector that is obtained as follows: The origin point of the vector $A$ is projected onto this axis. We obtain point $A"$ - the beginning of the desired vector. The end point of vector $B$ is projected onto this axis. We obtain point $B"$ - the end of the desired vector. The vector $\overline(A"B")$ will be the desired vector.
Let's consider the problem:
Example 1
Construct a geometric projection $\overline(AB)$ onto the $l$ axis shown in Figure 6.
Let us draw a perpendicular from point $A$ to the axis $l$, we obtain point $A"$ on it. Next, we draw a perpendicular from point $B$ to the axis $l$, we obtain point $B"$ on it (Fig. 7).
Pictures in drawings geometric bodies are constructed using the projection method. But for this one image is not enough; at least two projections are needed. With their help, points in space are determined. Therefore, you need to know how to find the projection of a point.
Point projection
To do this you will need to consider the space dihedral angle, with a point (A) located inside. Here the horizontal P1 and vertical P2 projection planes are used. Point (A) is projected orthogonally onto the projection planes. As for the perpendicular projecting rays, they are combined into a projecting plane, perpendicular to the planes projections. Thus, when combining the horizontal P1 and frontal P2 planes by rotating along the P2 / P1 axis, we obtain a flat drawing.
Then a line with projection points located on it is shown perpendicular to the axis. So it turns out complex drawing. Thanks to the constructed segments on it and vertical line connection, you can easily determine the position of a point relative to the projection planes.
To make it easier to understand how to find the projection, you need to consider right triangle. Its short side is the leg, and its long side is the hypotenuse. If you project a leg onto the hypotenuse, it will be divided into two segments. To determine their value, you need to calculate a set of initial data. Let's look at given triangle, methods for calculating the main projections.
As a rule, in this problem they indicate the length of the leg N and the length of the hypotenuse D, whose projection is required to be found. To do this, we will find out how to find the projection of the leg.
Let's consider a method for finding the length of the leg (A). Considering that the geometric mean of the projection of the leg and the length of the hypotenuse is equal to the value of the leg we are looking for: N = √(D*Nd).
How to find the projection length
The root of the product can be found by squaring the length of the desired leg (N), and then dividing by the length of the hypotenuse: Nd = (N / √ D)² = N² / D. When specifying the values of only legs D and N in the source data, the length projections should be found using the Pythagorean theorem.
Let's find the length of the hypotenuse D. To do this, you need to use the values of the legs √ (N² + T²), and then substitute the resulting value into the following formula for finding the projection: Nd = N² / √ (N² + T²).
When the source data contains data on the length of the projection of the leg RD, as well as data on the value of the hypotenuse D, the length of the projection of the second leg ND should be calculated using a simple subtraction formula: ND = D – RD.
Projection of speed
Let's look at how to find the projection of velocity. In order for a given vector to represent a description of movement, it should be placed in projection onto coordinate axes. There is one coordinate axis (ray), two coordinate axes (plane) and three coordinate axes (space). When finding a projection, it is necessary to lower perpendiculars from the ends of the vector onto the axis.
In order to understand the meaning of projection, you need to know how to find the projection of a vector.
Vector projection
When the body moves perpendicular to the axis, the projection will be represented as a point, and have the value equal to zero. If the movement is carried out parallel to the coordinate axis, then the projection will coincide with the vector module. In the case when the body moves in such a way that the velocity vector is directed at an angle φ relative to the (x) axis, the projection onto this axis will be a segment: V(x) = V cos(φ), where V is the model of the velocity vector. When the directions of the velocity vector and the coordinate axis coincide, then the projection is positive, and vice versa.
Let's take the following coordinate equation: x = x(t), y = y(t), z = z(t). In this case, the speed function will be projected onto three axes and will have the following form: V(x) = dx / dt = x"(t), V(y) = dy / dt = y"(t), V(z) = dz / dt = z"(t). It follows that to find the speed it is necessary to take derivatives. The speed vector itself is expressed by an equation of the following form: V = V(x) i + V(y) j + V(z) k . Here i, j, k are unit vectors coordinate axes x, y, z respectively. Thus, the velocity module is calculated by the following formula: V = √ (V(x) ^ 2 + V(y) ^ 2 + V(z) ^ 2).
The axis is the direction. This means that projection onto an axis or onto a directed line is considered the same. Projection can be algebraic or geometric. In geometric terms, the projection of a vector onto an axis is understood as a vector, and in algebraic terms, it is a number. That is, the concepts of projection of a vector onto an axis and numerical projection of a vector onto an axis are used.
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If we have an L axis and a non-zero vector A B →, then we can construct a vector A 1 B 1 ⇀, denoting the projections of its points A 1 and B 1.
A 1 B → 1 will be the projection of the vector A B → onto L.
Definition 1
Projection of the vector onto the axis is a vector whose beginning and end are projections of the beginning and end beyond given vector. n p L A B → → it is customary to denote the projection A B → onto L. To construct a projection onto L, perpendiculars are dropped onto L.
Example 1
An example of a vector projection onto an axis.
On coordinate plane About x y the point M 1 (x 1 , y 1) is specified. It is necessary to construct projections on O x and O y to image the radius vector of point M 1. We get the coordinates of the vectors (x 1, 0) and (0, y 1).
If we're talking about about the projection of a → onto a non-zero b → or the projection of a → onto the direction b → , then we mean the projection of a → onto the axis with which the direction b → coincides. The projection of a → onto the line defined by b → is denoted n p b → a → → . It is known that when the angle between a → and b → , n p b → a → → and b → can be considered codirectional. In the case when the angle is obtuse, n p b → a → → and b → are in opposite directions. In a situation of perpendicularity a → and b →, and a → is zero, the projection of a → in the direction b → is a zero vector.
The numerical characteristic of the projection of a vector onto an axis is the numerical projection of a vector onto a given axis.
Definition 2
Numerical projection of the vector onto the axis is a number that is equal to the product of the length of a given vector and the cosine of the angle between the given vector and the vector that determines the direction of the axis.
The numerical projection of A B → onto L is denoted n p L A B → , and a → onto b → - n p b → a → .
Based on the formula, we obtain n p b → a → = a → · cos a → , b → ^ , from where a → is the length of the vector a → , a ⇀ , b → ^ is the angle between the vectors a → and b → .
We obtain the formula for calculating the numerical projection: n p b → a → = a → · cos a → , b → ^ . It is applicable for known lengths a → and b → and the angle between them. The formula is applicable for known coordinates a → and b →, but there is a simplified form.
Example 2
Find out the numerical projection of a → onto a straight line in the direction b → with a length a → equal to 8 and an angle between them of 60 degrees. By condition we have a ⇀ = 8, a ⇀, b → ^ = 60 °. So, let's substitute numeric values into the formula n p b ⇀ a → = a → · cos a → , b → ^ = 8 · cos 60 ° = 8 · 1 2 = 4 .
Answer: 4.
With known cos (a → , b → ^) = a ⇀ , b → a → · b → , we have a → , b → as dot product a → and b → . Following from the formula n p b → a → = a → · cos a ⇀ , b → ^ , we can find the numerical projection a → directed along the vector b → and get n p b → a → = a → , b → b → . The formula is equivalent to the definition given at the beginning of the paragraph.
Definition 3
The numerical projection of the vector a → onto an axis coinciding in direction with b → is the ratio of the scalar product of the vectors a → and b → to the length b → . The formula n p b → a → = a → , b → b → is applicable to find the numerical projection of a → onto a line coinciding in direction with b → , with known a → and b → coordinates.
Example 3
Given b → = (- 3 , 4) . Find the numerical projection a → = (1, 7) onto L.
Solution
On the coordinate plane n p b → a → = a → , b → b → has the form n p b → a → = a → , b → b → = a x b x + a y b y b x 2 + b y 2 , with a → = (a x , a y ) and b → = b x , b y . To find the numerical projection of the vector a → onto the L axis, you need: n p L a → = n p b → a → = a → , b → b → = a x · b x + a y · b y b x 2 + b y 2 = 1 · (- 3) + 7 · 4 (- 3) 2 + 4 2 = 5.
Answer: 5.
Example 4
Find the projection of a → on L, coinciding with the direction b →, where there are a → = - 2, 3, 1 and b → = (3, - 2, 6). Three-dimensional space is specified.
Solution
Given a → = a x , a y , a z and b → = b x , b y , b z , we calculate the scalar product: a ⇀ , b → = a x · b x + a y · b y + a z · b z . We find the length b → using the formula b → = b x 2 + b y 2 + b z 2 . It follows that the formula for determining the numerical projection a → will be: n p b → a ⇀ = a → , b → b → = a x · b x + a y · b y + a z · b z b x 2 + b y 2 + b z 2 .
Substitute the numerical values: n p L a → = n p b → a → = (- 2) 3 + 3 (- 2) + 1 6 3 2 + (- 2) 2 + 6 2 = - 6 49 = - 6 7 .
Answer: - 6 7.
Let's look at the connection between a → on L and the length of the projection a → on L. Let's draw an axis L, adding a → and b → from a point on L, after which we draw a perpendicular line from the end a → to L and draw a projection onto L. There are 5 variations of the image:
First the case with a → = n p b → a → → means a → = n p b → a → → , hence n p b → a → = a → · cos (a , → b → ^) = a → · cos 0 ° = a → = n p b → a → → .
Second the case implies the use of n p b → a → ⇀ = a → · cos a → , b → , which means n p b → a → = a → · cos (a → , b →) ^ = n p b → a → → .
Third the case explains that when n p b → a → → = 0 → we obtain n p b ⇀ a → = a → · cos (a → , b → ^) = a → · cos 90 ° = 0 , then n p b → a → → = 0 and n p b → a → = 0 = n p b → a → → .
Fourth the case shows n p b → a → → = a → · cos (180 ° - a → , b → ^) = - a → · cos (a → , b → ^) , follows n p b → a → = a → · cos (a → , b → ^) = - n p b → a → → .
Fifth the case shows a → = n p b → a → → , which means a → = n p b → a → → , hence we have n p b → a → = a → · cos a → , b → ^ = a → · cos 180° = - a → = - n p b → a → .
Definition 4
The numerical projection of the vector a → onto the L axis, which is directed in the same way as b →, has the following value:
- the length of the projection of the vector a → onto L, provided that the angle between a → and b → is less than 90 degrees or equal to 0: n p b → a → = n p b → a → → with the condition 0 ≤ (a → , b →) ^< 90 ° ;
- zero provided that a → and b → are perpendicular: n p b → a → = 0, when (a → , b → ^) = 90 °;
- projection length a → onto L, multiplied by -1, when there is an obtuse or rotated angle of the vectors a → and b →: n p b → a → = - n p b → a → → with the condition of 90 °< a → , b → ^ ≤ 180 ° .
Example 5
Given the length of the projection a → onto L, equal to 2. Find the numerical projection a → provided that the angle is 5 π 6 radians.
Solution
From the condition it is clear that given angle is obtuse: π 2< 5 π 6 < π . Тогда можем найти числовую проекцию a → на L: n p L a → = - n p L a → → = - 2 .
Answer: - 2.
Example 6
Given a plane O x y z with a vector length a → equal to 6 3, b → (- 2, 1, 2) with an angle of 30 degrees. Find the coordinates of the projection a → onto the L axis.
Solution
First, we calculate the numerical projection of the vector a →: n p L a → = n p b → a → = a → · cos (a → , b →) ^ = 6 3 · cos 30 ° = 6 3 · 3 2 = 9 .
By condition, the angle is acute, then the numerical projection a → = the length of the projection of the vector a →: n p L a → = n p L a → → = 9. This case shows that the vectors n p L a → → and b → are co-directed, which means there is a number t for which the equality is true: n p L a → → = t · b → . From here we see that n p L a → → = t · b → , which means we can find the value of the parameter t: t = n p L a → → b → = 9 (- 2) 2 + 1 2 + 2 2 = 9 9 = 3 .
Then n p L a → → = 3 · b → with the coordinates of the projection of vector a → onto the L axis equal to b → = (- 2 , 1 , 2) , where it is necessary to multiply the values by 3. We have n p L a → → = (- 6 , 3 , 6) . Answer: (- 6, 3, 6).
It is necessary to repeat the previously learned information about the condition of collinearity of vectors.
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