Find the projection of vector a on the coordinate axes. Linear operations on vectors. Vector projection theorems

Solving problems on the equilibrium of converging forces by constructing closed force polygons involves cumbersome constructions. A universal method for solving such problems is to move on to determining the projections of given forces onto the coordinate axes and operating with these projections. An axis is a straight line that is assigned a specific direction.

The projection of a vector onto an axis is a scalar quantity, which is determined by the segment of the axis cut off by the perpendiculars dropped onto it from the beginning and end of the vector.

A vector projection is considered positive if the direction from the beginning of the projection to its end coincides with the positive direction of the axis. A vector projection is considered negative if the direction from the beginning of the projection to its end is opposite to the positive direction of the axis.

Thus, the projection of the force onto the coordinate axis is equal to the product of the force modulus and the cosine of the angle between the force vector and the positive direction of the axis.

Let's consider a number of cases of projecting forces onto an axis:

Force vector F(Fig. 15) is with the positive direction of the x axis acute angle.

To find the projection, from the beginning and end of the force vector we lower perpendiculars to the axis oh; we get

1. F x = F cos α

The projection of the vector in this case is positive

Strength F(Fig. 16) is with the positive direction of the axis X obtuse angle α.

Then F x = F cos α, but since α = 180 0 - φ,

F x = F cos α = F cos180 0 - φ =- F cos φ.

Projection of force F per axis oh in this case it is negative.

Strength F(Fig. 17) perpendicular to the axis oh.

Projection of force F onto the axis X equal to zero

F x = F cos 90° = 0.

Force located on the plane howe(Fig. 18), can be projected on two coordinate axes Oh And oh.

Strength F can be broken down into components: F x and F y. Vector module F x is equal to the projection of the vector F per axis ox, and the vector modulus F y is equal to the projection of the vector F per axis oh.

From Δ OAV: F x = F cos α, F x = F sin α.

From Δ OAS: F x = F cos φ, F x = F sin φ.

The magnitude of the force can be found using the Pythagorean theorem:

The projection of a vector sum or resultant onto any axis is equal to the algebraic sum of the projections of the summands of the vectors onto the same axis.

Consider the converging forces F 1 , F 2 , F 3, and F 4, (Fig. 19, a). The geometric sum, or resultant, of these forces F determined by the closing side of the force polygon

Let us drop from the vertices of the force polygon to the axis x perpendiculars.

Considering the obtained projections of forces directly from the completed construction, we have

F= F 1x+ F 2x+ F 3x+ F 4x

where n is the number of vector terms. Their projections enter the above equation with the corresponding sign.

In a plane, the geometric sum of forces can be projected onto two coordinate axes, and in space, respectively, onto three.

First, let's remember what it is coordinate axis, projection of a point onto an axis And coordinates of a point on the axis.

Coordinate axis- This is a straight line that is given some direction. You can think of it as a vector with an infinitely large modulus.

Coordinate axis denoted by some letter: X, Y, Z, s, t... Usually a point is selected (arbitrarily) on the axis, which is called the origin and, as a rule, denoted by the letter O. From this point the distances to other points of interest to us are measured.

Projection of a point onto an axis- this is the base of the perpendicular lowered from this point to this axis (Fig. 8). That is, the projection of a point onto the axis is a point.

Point coordinate on axis- this is the number absolute value which is equal to the length of the axis segment (on the selected scale) enclosed between the origin of the axis and the projection of the point onto this axis. This number is taken with a plus sign if the projection of the point is located in the direction of the axis from its origin and with a minus sign if in the opposite direction.

Scalar projection of a vector onto an axis- This number, the absolute value of which is equal to the length of the axis segment (on the selected scale) enclosed between the projections of the start point and the end point of the vector. Important! Usually instead of the expression scalar projection of a vector onto an axis they simply say - projection of the vector onto the axis, that is, the word scalar lowered. Vector projection is denoted by the same letter as the projected vector (in normal, non-bold writing), with a lower (as a rule) index of the name of the axis on which this vector is projected. For example, if a vector is projected onto the X axis A, then its projection is denoted by a x. When projecting the same vector onto another axis, say, the Y axis, its projection will be denoted a y (Fig. 9).

To calculate projection of the vector onto the axis(for example, the X axis), it is necessary to subtract the coordinate of the starting point from the coordinate of its end point, that is

a x = x k − x n.

We must remember: the scalar projection of a vector onto an axis (or, simply, the projection of a vector onto an axis) is a number (not a vector)! Moreover, the projection can be positive if the value x k is greater than the value x n, negative if the value x k less than the value x n and equal to zero if x k is equal to x n (Fig. 10).

The projection of a vector onto an axis can also be found by knowing the modulus of the vector and the angle it makes with this axis.

From Figure 11 it is clear that a x = a Cos α

That is, 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 axis direction and vector direction. If the angle is acute, then Cos α > 0 and a x > 0, and if it is obtuse, then the cosine of the obtuse angle is negative, and the projection of the vector onto the axis will also be negative.

Angles measured from the axis counterclockwise are considered positive, and angles measured along the axis are negative. However, since cosine is an even function, that is, Cos α = Cos (− α), when calculating projections, angles can be counted both clockwise and counterclockwise.

When solving problems, the following properties of projections will often be used: if

A = b + c +…+ d, then a x = b x + c x +…+ d x (similar to other axes),

a= m b, then a x = mb x (similarly for other axes).

The formula a x = a Cos α will be very often occur when solving problems, so you definitely need to know it. You need to know the rule for determining projection by heart!

Remember!

To find the projection of a vector onto an axis, the modulus of this vector must be multiplied by the cosine of the angle between the direction of the axis and the direction of the vector.

Once again - by heart!

Projection vector onto an axis is a vector that is obtained by multiplying the scalar projection of a vector onto this axis and the unit vector of this axis. For example, if a x – scalar projection vector A to the X axis, then a x i- its vector projection onto this axis.

Let's denote vector projection the same as the vector itself, but with the index of the axis on which the vector is projected. So, the vector projection of the vector A on the X axis we denote A x( fat a letter denoting a vector and a subscript of the axis name) or (a non-bold letter denoting a vector, but with an arrow at the top (!) and a subscript of the axis name).

Scalar projection vector per axis is called number, the absolute value of which is equal to the length of the axis segment (on the selected scale) enclosed between the projections of the start point and the end point of the vector. Usually instead of the expression scalar projection they simply say - projection. The projection is denoted by the same letter as the projected vector (in normal, non-bold writing), with a lower index (as a rule) of the name of the axis on which this vector is projected. For example, if a vector is projected onto the X axis A, then its projection is denoted by a x. When projecting the same vector onto another axis, if the axis is Y, its projection will be denoted as y.

To calculate the projection vector on an axis (for example, the X axis), it is necessary to subtract the coordinate of the starting point from the coordinate of its end point, that is
a x = x k − x n.
The projection of a vector onto an axis is a number. Moreover, the projection can be positive if the value x k is greater than the value x n,

negative if the value x k is less than the value x n

and equal to zero if x k equals x n.

The projection of a vector onto an axis can also be found by knowing the modulus of the vector and the angle it makes with this axis.

From the figure it is clear that a x = a Cos α

that is, 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 direction of the axis and vector direction. If the angle is acute, then
Cos α > 0 and a x > 0, and, if obtuse, then the cosine of the obtuse angle is negative, and the projection of the vector onto the axis will also be negative.

Angles measured from the axis counterclockwise are considered positive, and angles measured along the axis are negative. However, since cosine is an even function, that is, Cos α = Cos (− α), when calculating projections, angles can be counted both clockwise and counterclockwise.

To find the projection of a vector onto an axis, the modulus of this vector must be multiplied by the cosine of the angle between the direction of the axis and the direction of the vector.

Vector coordinates— coefficients of the only possible linear combination of basis vectors in the selected coordinate system, equal to the given vector.

where are the coordinates of the vector.

Dot product of vectors

Scalar product of vectors[- in finite-dimensional vector space is defined as the sum of the products of identical components being multiplied vectors.

For example, S.p.v. a = (a 1 , ..., a n) And b = (b 1 , ..., b n):

(a , b ) = a 1 b 1 + a 2 b 2 + ... + a n b n

In physics for grade 9 (I.K.Kikoin, A.K.Kikoin, 1999),
task №5
to the chapter " CHAPTER 1. GENERAL INFORMATION ABOUT TRAFFIC».

1. What is called the projection of a vector onto the coordinate axis?

1. Projection of vector a onto coordinate axis call the length of the segment between the projections of the beginning and end of the vector a (perpendiculars dropped from these points to the axis) onto this coordinate axis.

2. How is the displacement vector of a body related to its coordinates?

2. The projections of the displacement vector s on the coordinate axes are equal to the change in the corresponding body coordinates.

3. If the coordinate of a point increases over time, then what sign does the projection of the displacement vector onto the coordinate axis have? What if it decreases?

3. If the coordinate of a point increases over time, then the projection of the displacement vector onto the coordinate axis will be positive, because in this case we will go from the projection of the beginning to the projection of the end of the vector in the direction of the axis itself.

If the coordinate of a point decreases over time, then the projection of the displacement vector onto the coordinate axis will be negative, because in this case we will go from the projection of the beginning to the projection of the end of the vector against the guide of the axis itself.

4. If the displacement vector is parallel to the X axis, then what is the modulus of the projection of the vector onto this axis? And what about the modulus of the projection of the same vector onto the Y axis?

4. If the displacement vector is parallel to the X axis, then the modulus of the vector’s projection onto this axis is equal to the modulus of the vector itself, and its projection onto the Y axis is zero.

5. Determine the signs of the projections onto the X axis of the displacement vectors shown in Figure 22. How do the coordinates of the body change during these displacements?

5. In all the following cases, the Y coordinate of the body does not change, and the X coordinate of the body will change as follows:

a) s 1;

the projection of the vector s 1 onto the X axis is negative and is equal in absolute value to the length of the vector s 1 . With such a movement, the X coordinate of the body will decrease by the length of the vector s 1.

b) s 2 ;

the projection of the vector s 2 onto the X axis is positive and equal in magnitude to the length of the vector s 1 . With such a movement, the X coordinate of the body will increase by the length of the vector s 2.

c) s 3 ;

the projection of the vector s 3 onto the X axis is negative and equal in magnitude to the length of the vector s 3 . With such a movement, the X coordinate of the body will decrease by the length of the vector s 3.

d)s 4;

the projection of the vector s 4 onto the X axis is positive and equal in magnitude to the length of the vector s 4 . With such a movement, the X coordinate of the body will increase by the length of the vector s 4.

e)s 5;

the projection of the vector s 5 on the X axis is negative and equal in magnitude to the length of the vector s 5 . With such a movement, the X coordinate of the body will decrease by the length of the vector s 5.

6. If the distance traveled is large, then can the displacement module be small?

6. Maybe. This is due to the fact that displacement (displacement vector) is a vector quantity, i.e. is a directed straight line segment connecting the initial position of the body with its subsequent positions. And the final position of the body (regardless of the distance traveled) can be as close as desired to the initial position of the body. If the final and initial positions of the body coincide, the displacement module will be equal to zero.

7. Why is the vector of movement of a body more important in mechanics than the path it has traveled?

7. The main task of mechanics is to determine the position of the body at any time. Knowing the vector of movement of the body, we can determine the coordinates of the body, i.e. the position of the body at any moment in time, and knowing only the distance traveled, we cannot determine the coordinates of the body, because we have no information about the direction of movement, but can only judge the length of the distance traveled at the moment time.

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.

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 are talking 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 designated 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 where 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 the 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 °. This means that we substitute the numerical 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 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 → onto 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 . The length b → is found 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 this 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.

If you notice an error in the text, please highlight it and press Ctrl+Enter