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 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
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.
Projection of a point
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 it 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 to for given vector presented a description of the movement, it should be placed in projection onto the 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 in this case the velocity 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).
Let us denote by a the angle between the vector and the projection axis and transfer the vector
so that its origin coincides with some point on the axis. If the directions of the vector component and the axis are the same, then angle a will be acute and, as can be seen from Fig. 24, a,
where a is the module of vector a. If the directions of the vector and the axis are opposite, then, taking into account the sign of the projection, we will have (see Fig. 24, b)
i.e. the previous expression (you must remember that in this case the angle a is obtuse and
Thus, the projection of the vector onto the axis is equal to the product of the modulus of the vector and the cosine of the angle between the vector and the axis:
In addition to this having exclusively important formulas for the projection of the vector onto the axis, you can give one more very simple formula. Let's set the origin on the axis and choose a scale that is common to the scale of the vectors. As is known, the coordinate of a point is a number that expresses, on a selected scale, the distance from the origin of the axis to the projection of a given point onto the axis, and this number is taken with a plus sign if the projection of the point is removed from the origin in the direction of the axis, and with a minus sign otherwise case. So, for example, the coordinate of point A (Fig. 23, b) will be a signed number expressing the length of the segment, and the coordinate of point B will be a signed number that determines the length of the segment (we do not dwell on this
in more detail, assuming that the reader is familiar with the concept of coordinates of a point from a course in elementary mathematics).
Let us denote by the coordinate of the beginning, and by the coordinate of the end of the vector on the x-axis. Then, as can be seen from Fig. 23, ah, we will have
The projection of the vector onto the x axis will be equal to
or, taking into account the previous equalities,
It is easy to see that this formula has general character and does not depend on the location of the vector relative to the axis and the origin. Indeed, consider the case depicted in Fig. 23, b. From the definition of the coordinates of the points and the projection of the vector we successively obtain
(the reader can easily check the validity of the formula and and at a different location of the vector relative to the axis and origin).
From (6.11) it follows that the projection of the vector onto the axis is equal to the difference between the coordinates of the end and beginning of the vector.
Calculating the projection of a vector onto an axis occurs quite often in the most various issues. Therefore, it is necessary to develop solid skills in calculating projections. You can indicate some techniques that facilitate the process of calculating projections.
1. The sign of the vector projection onto the axis, as a rule, can be determined directly from the drawing, and the projection modulus can be calculated using the formula
Where - acute angle between the vector and the axis of projections - if and if This technique, without introducing anything fundamentally new, is somewhat
facilitates the calculation of projection since it does not require trigonometric transformations.
2. If it is necessary to determine the projections of a vector onto two mutually perpendicular axes x and y (it is assumed that the vector lies in the plane of these axes) and is the acute angle between the vector and the x axis, then
(the sign of the projections is determined from the drawing).
Example. Find the projections on the x and y coordinate axes of the force shown in Fig. 25. From the drawing it is clear that both projections will be negative. Hence,
3. Sometimes the double design rule is applied, which is as follows. Let a vector be given and an axis lying in the plane. Let us drop perpendiculars from the end of the vector onto the plane and the straight line and then connect the bases of the perpendiculars with a straight line segment (Fig. 26). Let us denote the angle between the vector and the plane by the angle between and by and the angle between the vector and the axis of projections by a. Since the angle is right (by construction), then
Many physical quantities are completely determined by specifying a certain number. These are, for example, volume, mass, density, body temperature, etc. Such quantities are called scalar. Because of this, numbers are sometimes called scalars. But there are also quantities that are determined by specifying not only a number, but also a certain direction. For example, when a body moves, you should indicate not only the speed at which the body is moving, but also the direction of movement. In the same way, when studying the action of any force, it is necessary to indicate not only the value of this force, but also the direction of its action. Such quantities are called vector. To describe them, the concept of a vector was introduced, which turned out to be useful for mathematics.
Vector definition
Any ordered pair of points A to B in space defines directed segment, i.e. a segment along with the direction specified on it. If point A is the first, then it is called the beginning of the directed segment, and point B is its end. The direction of a segment is considered to be the direction from beginning to end.
Definition
A directed segment is called a vector.
We will denote a vector by the symbol \(\overrightarrow(AB) \), with the first letter indicating the beginning of the vector, and the second - its end.
A vector whose beginning and end coincide is called zero and is denoted by \(\vec(0)\) or simply 0.
The distance between the start and end of a vector is called its length and is denoted by \(|\overrightarrow(AB)| \) or \(|\vec(a)| \).
The vectors \(\vec(a) \) and \(\vec(b) \) are called collinear, if they lie on the same line or on parallel lines. Collinear vectors can have the same or opposite directions.
Now we can formulate important concept equality of two vectors.
Definition
Vectors \(\vec(a) \) and \(\vec(b) \) are said to be equal (\(\vec(a) = \vec(b) \)) if they are collinear, have the same direction and their lengths are equal .
In Fig. 1 depicted on the left are unequal, and on the right - equal vectors\(\vec(a) \) and \(\vec(b) \). From the definition of equality of vectors it follows that if a given vector is moved parallel to itself, then the result will be a vector equal to the given one. In this regard, the vectors in analytical geometry called free.
Projection of a vector onto an axis
Let the axis \(u\) and some vector \(\overrightarrow(AB)\) be given in space. Let us draw planes perpendicular to the \(u\) axis through points A and B. Let us denote by A" and B" the points of intersection of these planes with the axis (see Figure 2).
The projection of the vector \(\overrightarrow(AB) \) onto the axis \(u\) is the value A"B" of the directed segment A"B" on the axis \(u\). Let us recall that
\(A"B" = |\overrightarrow(A"B")| \) , if the direction \(\overrightarrow(A"B") \) coincides with the direction of the axis \(u\),
\(A"B" = -|\overrightarrow(A"B")| \) , if the direction \(\overrightarrow(A"B") \) is opposite to the direction of the axis \(u\),
The projection of the vector \(\overrightarrow(AB)\) onto the axis \(u\) is denoted as follows: \(Pr_u \overrightarrow(AB)\).
Theorem
The projection of the vector \(\overrightarrow(AB) \) onto the axis \(u\) is equal to the length of the vector \(\overrightarrow(AB) \) multiplied by the cosine of the angle between the vector \(\overrightarrow(AB) \) and the axis \( u\) , i.e.
Comment
Let \(\overrightarrow(A_1B_1)=\overrightarrow(A_2B_2) \) and some axis \(u\) be specified. Applying the formula of the theorem to each of these vectors, we obtain
Vector projections on coordinate axes
Let us be given in space rectangular system coordinates Oxyz and an arbitrary vector \(\overrightarrow(AB)\). Let, further, \(X = Pr_u \overrightarrow(AB), \;\; Y = Pr_u \overrightarrow(AB), \;\; Z = Pr_u \overrightarrow(AB) \). The projections of the X, Y, Z vector \(\overrightarrow(AB)\) on the coordinate axes are called coordinates. At the same time they write
\(\overrightarrow(AB) = (X;Y;Z) \)
Theorem
Whatever the two points A(x 1 ; y 1 ; z 1) and B(x 2 ; y 2 ; z 2), the coordinates of the vector \(\overrightarrow(AB) \) are determined by the following formulas:
X = x 2 -x 1 , Y = y 2 -y 1 , Z = z 2 -z 1
Comment
If the vector \(\overrightarrow(AB) \) leaves the origin, i.e. x 2 = x, y 2 = y, z 2 = z, then the coordinates X, Y, Z of the vector \(\overrightarrow(AB) \) are equal to the coordinates of its end:
X = x, Y = y, Z = z.
Direction cosines of a vector
Let an arbitrary vector \(\vec(a) = (X;Y;Z) \); we will assume that \(\vec(a) \) comes out from the origin and does not lie in any coordinate plane. Let us draw planes perpendicular to the axes through point A. Together with coordinate planes they form a rectangular parallelepiped, the diagonal of which is the segment OA (see figure).
From elementary geometry it is known that the square of the length of the diagonal rectangular parallelepiped equal to the sum squares of the lengths of its three dimensions. Hence,
\(|OA|^2 = |OA_x|^2 + |OA_y|^2 + |OA_z|^2 \)
But \(|OA| = |\vec(a)|, \;\; |OA_x| = |X|, \;\; |OA_y| = |Y|, \;\;|OA_z| = |Z| \); thus we get
\(|\vec(a)|^2 = X^2 + Y^2 + Z^2 \)
or
\(|\vec(a)| = \sqrt(X^2 + Y^2 + Z^2) \)
This formula expresses the length of an arbitrary vector through its coordinates.
Let us denote by \(\alpha, \; \beta, \; \gamma \) the angles between the vector \(\vec(a) \) and the coordinate axes. From the formulas for the projection of the vector onto the axis and the length of the vector we obtain
\(\cos \alpha = \frac(X)(\sqrt(X^2 + Y^2 + Z^2)) \)
\(\cos \beta = \frac(Y)(\sqrt(X^2 + Y^2 + Z^2)) \)
\(\cos \gamma = \frac(Z)(\sqrt(X^2 + Y^2 + Z^2)) \)
\(\cos \alpha, \;\; \cos \beta, \;\; \cos \gamma \) are called direction cosines of the vector \(\vec(a) \).
Squaring the left and right sides of each of the previous equalities and summing up the results obtained, we have
\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)
those. the sum of the squares of the direction cosines of any vector is equal to one.
Linear operations on vectors and their basic properties
Linear operations on vectors are the operations of adding and subtracting vectors and multiplying vectors by numbers.Addition of two vectors
Let two vectors \(\vec(a) \) and \(\vec(b) \) be given. The sum \(\vec(a) + \vec(b) \) is a vector that goes from the beginning of the vector \(\vec(a) \) to the end of the vector \(\vec(b) \) provided that the vector \(\vec(b) \) is attached to the end of the vector \(\vec(a) \) (see figure).Comment
The action of subtracting vectors is inverse to the action of addition, i.e. the difference \(\vec(b) - \vec(a) \) vectors \(\vec(b) \) and \(\vec(a) \) is a vector that, in sum with the vector \(\vec(a ) \) gives the vector \(\vec(b) \) (see figure).
Comment
By determining the sum of two vectors, you can find the sum of any number of given vectors. Let, for example, be given three vectors \(\vec(a),\;\; \vec(b), \;\; \vec(c) \). Adding \(\vec(a) \) and \(\vec(b) \), we obtain the vector \(\vec(a) + \vec(b) \). Now adding to it the vector \(\vec(c) \), we obtain the vector \(\vec(a) + \vec(b) + \vec(c) \) 
Product of a vector and a number
Let the vector \(\vec(a) \neq \vec(0) \) and the number \(\lambda \neq 0 \) be given. The product \(\lambda \vec(a) \) is a vector that is collinear to the vector \(\vec(a) \), has a length equal to \(|\lambda| |\vec(a)| \), and direction the same as the vector \(\vec(a) \) if \(\lambda > 0 \), and the opposite if \(\lambda Geometric meaning the operations of multiplying the vector \(\vec(a) \neq \vec(0) \) by the number \(\lambda \neq 0 \) can be expressed as follows: if \(|\lambda| >1 \), then when multiplying vector \(\vec(a) \) by the number \(\lambda \) the vector \(\vec(a) \) is “stretched” \(\lambda \) times, and if \(|\lambda| 1 \ ).If \(\lambda =0 \) or \(\vec(a) = \vec(0) \), then the product \(\lambda \vec(a) \) is considered equal to the zero vector.
Comment
Using the definition of multiplying a vector by a number, it is easy to prove that if the vectors \(\vec(a) \) and \(\vec(b) \) are collinear and \(\vec(a) \neq \vec(0) \), then there exists (and only one) number \(\lambda \) such that \(\vec(b) = \lambda \vec(a) \)
Basic properties of linear operations
1. Commutative property of addition
\(\vec(a) + \vec(b) = \vec(b) + \vec(a) \)
2. Matching property addition
\((\vec(a) + \vec(b))+ \vec(c) = \vec(a) + (\vec(b)+ \vec(c)) \)
3. Combinative property of multiplication
\(\lambda (\mu \vec(a)) = (\lambda \mu) \vec(a) \)
4. Distributive property relative to the sum of numbers
\((\lambda +\mu) \vec(a) = \lambda \vec(a) + \mu \vec(a) \)
5. Distributive property with respect to the sum of vectors
\(\lambda (\vec(a)+\vec(b)) = \lambda \vec(a) + \lambda \vec(b) \)
Comment
These properties linear operations are of fundamental importance, since they make it possible to perform ordinary algebraic operations on vectors. For example, due to properties 4 and 5, you can multiply a scalar polynomial by a vector polynomial “term by term”.
Vector projection theorems
Theorem
The projection of the sum of two vectors onto an axis is equal to the sum of their projections onto this axis, i.e.
\(Pr_u (\vec(a) + \vec(b)) = Pr_u \vec(a) + Pr_u \vec(b) \)
The theorem can be generalized to the case of any number of terms.
Theorem
When the vector \(\vec(a) \) is multiplied by the number \(\lambda \), its projection onto the axis is also multiplied by this number, i.e. \(Pr_u \lambda \vec(a) = \lambda Pr_u \vec(a) \)
Consequence
If \(\vec(a) = (x_1;y_1;z_1) \) and \(\vec(b) = (x_2;y_2;z_2) \), then
\(\vec(a) + \vec(b) = (x_1+x_2; \; y_1+y_2; \; z_1+z_2) \)
Consequence
If \(\vec(a) = (x;y;z) \), then \(\lambda \vec(a) = (\lambda x; \; \lambda y; \; \lambda z) \) for any number \(\lambda \)
From here it is easy to deduce condition of collinearity of two vectors in coordinates.
Indeed, the equality \(\vec(b) = \lambda \vec(a) \) is equivalent to the equalities \(x_2 = \lambda x_1, \; y_2 = \lambda y_1, \; z_2 = \lambda z_1 \) or
\(\frac(x_2)(x_1) = \frac(y_2)(y_1) = \frac(z_2)(z_1) \) i.e. the vectors \(\vec(a) \) and \(\vec(b) \) are collinear if and only if their coordinates are proportional.
Decomposition of a vector into a basis
Let the vectors \(\vec(i), \; \vec(j), \; \vec(k) \) be unit vectors coordinate axes, i.e. \(|\vec(i)| = |\vec(j)| = |\vec(k)| = 1 \), and each of them is equally directed with the corresponding coordinate axis (see figure). A triple of vectors \(\vec(i), \; \vec(j), \; \vec(k) \) is called basis.
The following theorem holds.
Theorem
Any vector \(\vec(a) \) can be uniquely expanded over the basis \(\vec(i), \; \vec(j), \; \vec(k)\; \), i.e. presented as
\(\vec(a) = \lambda \vec(i) + \mu \vec(j) + \nu \vec(k) \)
where \(\lambda, \;\; \mu, \;\; \nu \) are some numbers. 
