Determine the projection of the vector onto the axis. Vector projection. Coordinate axes. Projection of a point. Coordinates of a point on an axis
Algebraic projection of a vector on any axis is equal to the product of the length of the vector and the cosine of the angle between the axis and the vector:Pr a b = |b|cos(a,b) or
Where a b is the scalar product of vectors, |a| - modulus of vector a.
Instructions. To find the projection of the vector Пp a b in online mode it is necessary to indicate the coordinates of vectors a and b. In this case, the vector can be specified on the plane (two coordinates) and in space (three coordinates). The resulting solution is saved in a Word file. If vectors are specified through the coordinates of points, then you need to use this calculator.
Classification of vector projections
Types of projections by definition vector projection
Types of projections according to the coordinate system
Vector Projection Properties
- The geometric projection of a vector is a vector (has a direction).
- The algebraic projection of a vector is a number.
Vector projection theorems
Theorem 1. The projection of the sum of vectors onto any axis is equal to the projection of the summands of the vectors onto the same axis.
Theorem 2. The algebraic projection of a vector onto any axis is equal to the product of the length of the vector and the cosine of the angle between the axis and the vector:
Pr a b = |b|cos(a,b)
Types of vector projections
- projection onto the OX axis.
- projection onto the OY axis.
- projection onto a vector.
| Projection on the OX axis | Projection on the OY axis | Projection to vector |
If the direction of vector A’B’ coincides with the direction of the OX axis, then the projection of vector A’B’ has a positive sign.  | If the direction of vector A’B’ coincides with the direction of the OY axis, then the projection of vector A’B’ has a positive sign.  | If the direction of vector A’B’ coincides with the direction of vector NM, then the projection of vector A’B’ has a positive sign.  |
If the direction of the vector is opposite to the direction of the OX axis, then the projection of the vector A’B’ has a negative sign.  | If the direction of vector A’B’ is opposite to the direction of the OY axis, then the projection of vector A’B’ has a negative sign.  | If the direction of vector A’B’ is opposite to the direction of vector NM, then the projection of vector A’B’ has a negative sign.  |
If vector AB is parallel to the OX axis, then the projection of vector A’B’ is equal to the absolute value of vector AB.   | If vector AB is parallel to the OY axis, then the projection of vector A’B’ is equal to the absolute value of vector AB.   | If vector AB is parallel to vector NM, then the projection of vector A’B’ is equal to the absolute value of vector AB.   |
If the vector AB is perpendicular to the axis OX, then the projection A’B’ is equal to zero (null vector).   | If the vector AB is perpendicular to the OY axis, then the projection A’B’ is equal to zero (null vector).   | If the vector AB is perpendicular to the vector NM, then the projection A’B’ is equal to zero (null vector).   |
1. Question: Can the projection of a vector have a negative sign? Answer: Yes, the projection vector can be a negative value. In this case, the vector has the opposite direction (see how the OX axis and the AB vector are directed)
2. Question: Can the projection of a vector coincide with the absolute value of the vector? Answer: Yes, it can. In this case, the vectors are parallel (or lie on the same line).
3. Question: Can the projection of a vector be equal to zero (null vector). Answer: Yes, it can. In this case, the vector is perpendicular to the corresponding axis (vector).
Example 1. The vector (Fig. 1) forms an angle of 60° with the OX axis (it is specified by vector a). If OE is a unit of scale, then |b|=4, so 
.
Indeed, the length of the vector (geometric projection b) is equal to 2, and the direction coincides with the direction of the OX axis.
Example 2. The vector (Fig. 2) forms an angle (a,b) = 120 o with the OX axis (with vector a). Length |b| vector b is equal to 4, so pr a b=4·cos120 o = -2. 
Indeed, the length of the vector is 2, and the direction is opposite to the direction of the axis.
Now we are ready to introduce the all-important concept of projection of a vector onto an axis. It is constantly used in solving physical problems.
7.5.1 What is the projection of a vector onto an axis?
Let the vector ~a and the X axis be given. It is assumed that the X axis has a scale that allows you to measure the lengths of the segments and assign them the dimension of the vector ~a.
From the beginning and end of the vector ~a we lower perpendiculars to the X axis; let A and B be the bases of these perpendiculars (Fig. 7.26). We denote the length of the segment AB by jABj.
Rice. 7.26. Projection of a vector onto an axis
Definition. The projection ax of the vector ~a onto the X axis is equal to the length of the segment AB, taken with a plus sign if the angle " between the vector ~a and the X axis is acute, and taken with a minus sign if " is obtuse (or unfolded). If the angle is right, then ax = 0.
In short, we have the following formula:
Figure 7.27 illustrates all of these possibilities.
Here, as usual, a = j~aj modulus of the vector ~a.
Indeed, if"< 90 , то формула (7.10 ) даёт длину левого красного отрезка на рис.7.27 .
If " > 90, then, moving in the middle part of Fig. 7.27 to the angle adjacent to the angle ", we see that formula (7.10) gives the length of the middle red segment with a minus sign (due to the negativity of the cosine), which is exactly what we need need to.
Finally, if " = 90, then formula (7.10) gives ax = 0, since the cosine right angle equal to zero. This is exactly how it should be (right side of the picture).
Let us now assume that the X axis is also given an origin, so that it is a familiar coordinate axis. Then we have another formula for the projection ax, which also contains all three cases of Figure 7.27 in an “archived” form.
Corollary 2. Let x1 and x2 be the coordinates of the beginning and end of the vector ~a, respectively. Then the projection ax is calculated by the formula:
ax = x2 x1 : |
Indeed, let's look at Fig. 7.28. This is a case of positive projection. From the figure it is obvious that the difference x2 x1 is equal to the length of the red segment, and this length is in this case this is precisely the projection of ax.
Rice. 7.28. Projection of a vector onto an axis. To Corollary 2
What will happen in the remaining two cases (ax< 0 и ax = 0)? Убедитесь, пожалуйста, самостоятельно, что формула (7.11 ) и для них остаётся справедливой.
7.5.2 Properties of projecting a vector onto an axis
The operation of projecting a vector onto an axis is remarkably consistent with the operations of adding vectors and multiplying a scalar by a vector. Namely, whatever the X axis, the following two design properties hold.
1. The projection of the vector ~a + b onto the X axis is equal to ax + bx.
Brief verbal formulation: the projection of the sum of vectors is equal to the sum of their projections. This is true for the sum of any number of vectors, not just two.
Rice. 7.29. ~c = ~a + b) cx = ax | |||||||||
First of all, let us illustrate this statement in the figure. Let's place the beginning of the century-
torus b to the end of the vector ~a, and let ~c = ~a + b (Fig. 7.29).
This figure clearly shows that the projection cx is equal to the sum of the lengths of the red and green segments, that is, just ax + bx.
True, fig. 7.29 is made for the case ax > 0 and bx > 0. To prove our statement for all at once possible values projections ax and bx, we will carry out the following universal reasoning based on formula (7.11).
So, let the vectors ~a and b be arranged in an arbitrary manner. Let's combine the beginning again |
||
vector b with the end of the vector ~a and denote ~c = ~a + b. Let: |
||
coordinate of the beginning of the vector ~a and at the same time the beginning of the vector ~c; |
||
the coordinate of the end of the vector ~a and at the same time the beginning of the vector b; |
||
the coordinate of the end of vector b and at the same time the end of vector ~c. |
||
These designations are also present in Fig. 7.29.
By virtue of formula (7.11), we have: ax = x2 x1, bx = x3 x2, cx = x3 x1. Now it's easy to see that:
ax + bx = (x2 x1 ) + (x3 x2 ) = x3 x1 = cx :
Our first property of design is thus proven.
2. The projection of the vector ~a on the X axis is equal to a x.
Verbal formulation: the projection of the product of a scalar by a vector is equal to the product of a scalar by the projection of a vector.
Let's start again with an illustration. The left side of Figure 7.30 shows the vector ~a with a positive projection ax.
Rice. 7.30. The projection of the vector ~a is equal to ax
If you multiply the vector ~a by 2, then its length will double, the projection of the vector will also double (while maintaining the sign) and become equal to 2ax.
If you multiply the vector ~a by 2, then its length will again double, but the direction will change to the opposite. The projection will change sign and become equal to 2ax.
Thus, the essence of the second property is clear, and now a rigorous proof can be given.
So let ~ . We want to prove that x x . b = ~a b = a
Let's use formula (7.10) for this. We have:
ax = a cos "; bx = b cos ;
where is the angle between the vector and the axis, and the angle between the vector ~ and the axis. Except
Moreover, by virtue of the definition of multiplying a scalar by a vector:
Thus:
bx = j ja cos:
If, then j j ; in this case, the vector ~ is codirectional with the vector, and therefore.
> 0 = b ~a = "
bx = a cos " = ax :
If, then j j ; in this case, the vector ~ is opposite in the direction of the vector
ru ~a. It is not difficult to figure out that = " (for example, if " is acute, that is, adjacent to it is obtuse, and vice versa). We then have:
bx = ()a cos() = ()a(cos ") = a cos " = ax :
So, in all cases the desired relationship is obtained, and thus the second property of design is completely proven.
7.5.3 Design operation in physics
The proven properties of the design operation are very important to us. In mechanics, for example, we will use them at every step.
Thus, the solution of many problems in dynamics begins with writing Newton’s second law in vector form. Take, for example, a pendulum of mass m suspended on a thread. For a pendulum, Newton's second law will be:
Having written Newton's second law in vector form, we move on to projecting it onto
suitable axles. We take equality (7.12) and project onto the X axis: | |
max = mgx + Tx + fx : |
When moving from vector equality (7.12) to scalar equality (7.13), both projection properties are used! Namely, thanks to property 1, we wrote down the projection of a sum of vectors as the sum of their projections; By virtue of property 2, we were able to write down the projections of the vectors m~a and m~g in the form max and mgx.
Thus, both properties of the projection operation provide the transition from vector equalities to scalar ones, and this transition can be performed formally and without thinking: we discard the arrows in the vector notation and put projection indices in their place. This is exactly what the transition from equation (7.12) to equation (7.13) looks like.
A vector description of movement is useful, since in one drawing you can always depict many different vectors and get a visual “picture” of movement before your eyes. However, using a ruler and a protractor every time to perform operations with vectors is very labor-intensive. Therefore, these actions reduce to actions with positive and negative numbers– projections of vectors.
Projection of the vector onto the axis called a scalar quantity equal to the product of the modulus of the projected vector and the cosine of the angle between the directions of the vector and the selected coordinate axis.
The left drawing shows a displacement vector, the module of which is 50 km, and its direction forms obtuse angle 150° with the direction of the X axis. Using the definition, we find the projection of the displacement on the X axis:
sx = s cos(α) = 50 km cos(150°) = –43 km
Since the angle between the axes is 90°, it is easy to calculate that the direction of movement forms an acute angle of 60° with the direction of the Y axis. Using the definition, we find the projection of displacement on the Y axis:
sy = s cos(β) = 50 km cos(60°) = +25 km
As you can see, if the direction of the vector forms an acute angle with the direction of the axis, the projection is positive; if the direction of the vector forms an obtuse angle with the direction of the axis, the projection is negative.
The right drawing shows a velocity vector, the module of which is 5 m/s, and the direction forms an angle of 30° with the direction of the X axis. Let's find the projections:
υx = υ · cos(α) = 5 m/s · cos( 30°) = +4.3 m/s
υy = υ · cos(β) = 5 m/s · cos( 120°) = –2.5 m/s
It is much easier to find projections of vectors on axes if the projected vectors are parallel or perpendicular to the selected axes. Please note that for the case of parallelism, two options are possible: the vector is co-directional to the axis and the vector is opposite to the axis, and for the case of perpendicularity there is only one option.
The projection of a vector perpendicular to the axis is always zero (see sy and ay in the left drawing, and sx and υx in the right drawing). Indeed, for a vector perpendicular to the axis, the angle between it and the axis is 90°, so the cosine is zero, which means the projection is zero.
The projection of a vector codirectional with the axis is positive and equal to its absolute value, for example, sx = +s (see left drawing). Indeed, for a vector codirectional with the axis, the angle between it and the axis is zero, and its cosine is “+1”, that is, the projection is equal to the length of the vector: sx = x – xo = +s .
The projection of the vector opposite to the axis is negative and equal to its absolute value, taken with a minus sign, for example, sy = –s (see the right drawing). Indeed, for a vector opposite to the axis, the angle between it and the axis is 180°, and its cosine is “–1”, that is, the projection is equal to the length of the vector taken with a negative sign: sy = y – yo = –s .
The right-hand sides of both drawings show other cases where the vectors are parallel to one of the coordinate axes and perpendicular to the other. We invite you to make sure for yourself that in these cases, too, the rules formulated in the previous paragraphs are followed.
Answer:
Projection properties:
Vector Projection Properties
Property 1.
The projection of the sum of two vectors onto an axis is equal to the sum of the projections of vectors onto the same axis: 
This property allows you to replace the projection of a sum of vectors with the sum of their projections and vice versa.
Property 2. If a vector is multiplied by the number λ, then its projection onto the axis is also multiplied by this number:
Property 3.
The projection of the vector onto the l axis is equal to the product of the modulus of the vector and the cosine of the angle between the vector and the axis:
Orth axis. Decomposition of a vector in coordinate unit vectors. Vector coordinates. Coordinate properties
Answer:
Unit vectors of the axes.
A rectangular coordinate system (of any dimension) is also described by a set of unit vectors aligned with the coordinate axes. The number of unit vectors is equal to the dimension of the coordinate system and they are all perpendicular to each other.
In the three-dimensional case, the unit vectors are usually denoted
And Arrow symbols and may also be used.
In this case, in the case of a right coordinate system, the following formulas with vector products of unit vectors are valid:
Decomposition of a vector in coordinate unit vectors.
The unit vector of the coordinate axis is denoted by , axes by , axes by (Fig. 1)
For any vector that lies in the plane, the following expansion takes place:
If the vector 
located in space, then the expansion in unit vectors of the coordinate axes has the form:
Vector coordinates:
To calculate the coordinates of a vector, knowing the coordinates (x1; y1) of its beginning A and the coordinates (x2; y2) of its end B, you need to subtract the coordinates of the beginning from the coordinates of the end: (x2 – x1; y2 – y1).
Properties of coordinates.
Consider a coordinate line with the origin at point O and the unit vector i. Then for any vector a on this line: a = axi.
The number ax is called the coordinate of the vector a on the coordinate axis.
Property 1. When adding vectors on an axis, their coordinates are added.
Property 2. When a vector is multiplied by a number, its coordinate is multiplied by that number.
Dot product vectors. Properties.
Answer:
The scalar product of two non-zero vectors is the number
equal to the product of these vectors and the cosine of the angle between them.
Properties:
1. The scalar product has the commutative property: ab=ba
Scalar product of coordinate unit vectors. Determination of the scalar product of vectors specified by their coordinates.
Answer:
Dot product (×) of unit vectors
| (X) | I | J | K |
| I | |||
| J | |||
| K |
Determination of the scalar product of vectors specified by their coordinates.
The scalar product of two vectors and given by their coordinates can be calculated using the formula
The cross product of two vectors. Properties of a vector product.
Answer:
Three non-coplanar vectors form a right triple if from the end of the third the rotation from the first vector to the second is made counterclockwise. If clockwise, then left. If not, then in the opposite direction ( show how he showed with “handles”)
Cross product of a vector A to vector b called a vector from which:
1. Perpendicular to vectors A And b
2. Has length, numerically equal to the area parallelogram formed on a And b vectors
3. Vectors, a ,b, And c form a right-hand triple of vectors
Properties:
1. 
3. 
4. 
Vector product of coordinate unit vectors. Determination of the vector product of vectors specified by their coordinates.
Answer:
Vector product of coordinate unit vectors.
Determination of the vector product of vectors specified by their coordinates.
Let the vectors a = (x1; y1; z1) and b = (x2; y2; z2) be given by their coordinates in the rectangular Cartesian coordinate system O, i, j, k, and the triple i, j, k is right-handed.
Let's expand a and b into basis vectors:
a = x 1 i + y 1 j + z 1 k, b = x 2 i + y 2 j + z 2 k.
Using the properties of the vector product, we get
[A; b] = =
= x 1 x 2 + x 1 y 2 + x 1 z 2 +
+ y 1 x 2 + y 1 y 2 + y 1 z 2 +
+ z 1 x 2 + z 1 y 2 + z 1 z 2 . (1)
By the definition of a vector product we find
= 0, = k, = - j,
= - k, = 0, = i,
= j, = - i. = 0.
Taking these equalities into account, formula (1) can be written as follows:
[A; b] = x 1 y 2 k - x 1 z 2 j - y 1 x 2 k + y 1 z 2 i + z 1 x 2 j - z 1 y 2 i
[A; b] = (y 1 z 2 - z 1 y 2) i + (z 1 x 2 - x 1 z 2) j + (x 1 y 2 - y 1 x 2) k. (2)
Formula (2) gives an expression for the vector product of two vectors specified by their coordinates.
The resulting formula is cumbersome. Using the notation of determinants, you can write it in another form that is more convenient for memorization:
Usually formula (3) is written even shorter:
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 the important concept of 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 shows unequal vectors on the left and equal vectors \(\vec(a) \) and \(\vec(b) \) on the right. 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 a rectangular coordinate system Oxyz and an arbitrary vector \(\overrightarrow(AB)\) be given in space. 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 of a rectangular parallelepiped is equal to the sum of the 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 the vector \(\vec(c) \) to it, 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 of the operation 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 the vector \(\vec(a) \) by the number \( \lambda \) the vector \(\vec(a) \) is “stretched” by \(\lambda \) 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. Combinative property of 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 regarding 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 of 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 the unit vectors of the 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. 