Projection of a vector onto any axis. Projections of vectors onto coordinate axes

Definition 1. On a plane, a parallel projection of point A onto the l axis is a point - the point of intersection of the l axis with a straight line drawn through point A parallel to the vector that specifies the design direction.

Definition 2. The parallel projection of a vector onto the l axis (to the vector) is the coordinate of the vector relative to the basis axis l, where points and are parallel projections of points A and B onto the l axis, respectively (Fig. 1).

According to the definition we have

Definition 3. if and l axis basis Cartesian, that is, the projection of the vector onto the l axis called orthogonal (Fig. 2).

In space, definition 2 of the vector projection onto the axis remains in force, only the projection direction is specified by two non-collinear vectors (Fig. 3).

From the definition of the projection of a vector onto an axis it follows that each coordinate of a vector is a projection of this vector onto the axis defined by the corresponding basis vector. In this case, the design direction is specified by two other basis vectors if the design is carried out (considered) in space, or by another basis vector if the design is considered on a plane (Fig. 4).

Theorem 1. The orthogonal projection of a vector onto the l axis is equal to the product of the modulus of the vector and the cosine of the angle between the positive direction of the l axis and, i.e.

On the other side

From we find

Substituting AC into equality (2), we obtain

Since the numbers x and the same sign in both cases under consideration ((Fig. 5, a) ; (Fig. 5, b), then from equality (4) it follows

Comment. In what follows, we will consider only the orthogonal projection of the vector onto the axis and therefore the word “ort” (orthogonal) will be omitted from the notation.

Let us present a number of formulas that are used later in solving problems.

a) Projection of the vector onto the axis.

If, then the orthogonal projection onto the vector according to formula (5) has the form

c) Distance from a point to a plane.

Let b be a given plane with a normal vector, M be a given point,

d is the distance from point M to plane b (Fig. 6).

If N is an arbitrary point of the plane b, and and are projections of points M and N onto the axis, then

• G) The distance between intersecting lines.

Let a and b be given crossing lines, be a vector perpendicular to them, A and B be arbitrary points of lines a and b, respectively (Fig. 7), and and be projections of points A and B onto, then

e) Distance from a point to a line.

Let l- a given straight line with a direction vector, M - a given point,

N - its projection onto the line l, then - the required distance (Fig. 8).

If A is an arbitrary point on a line l, then in right triangle MNA, hypotenuse MA and legs can be found. Means,

f) The angle between a straight line and a plane.

Let be the direction vector of this line l, - normal vector of a given plane b, - projection of a straight line l to plane b (Fig. 9).

As is known, the angle μ between a straight line l and its projection onto plane b is called the angle between the line and the plane. We have

Let us give examples of solving metric problems using the vector-coordinate method.

A. The projection of point A onto the PQ axis (Fig. 4) is the base a of the perpendicular dropped from a given point to a given axis. The axis on which we project is called the projection axis.

b. Let two axes and a vector A B be given, shown in Fig. 5.

A vector whose beginning is the projection of the beginning and whose end is the projection of the end of this vector is called the projection of vector A B onto the PQ axis. It is written like this;

Sometimes the PQ indicator is not written at the bottom; this is done in cases where, besides PQ, there is no other OS on which to design.

With. Theorem I. The magnitudes of vectors lying on one axis are related as the magnitudes of their projections onto any axis.

Let the axes and vectors indicated in Fig. 6 be given. From the similarity of the triangles it is clear that the lengths of the vectors are related as the lengths of their projections, i.e.

Since the vectors in the drawing are directed in different directions, their magnitudes have different signs, therefore,

Obviously, the magnitudes of the projections also have different signs:

substituting (2) into (3) into (1), we get

Reversing the signs, we get

If the vectors are equally directed, then their projections will also be of the same direction; there will be no minus signs in formulas (2) and (3). Substituting (2) and (3) into equality (1), we immediately obtain equality (4). So, the theorem has been proven for all cases.

d. Theorem II. The magnitude of the projection of a vector onto any axis is equal to the magnitude of the vector multiplied by the cosine of the angle between the axis of projections and the axis of the vector. Let the axes be given as a vector as indicated in Fig. 7. Let's construct a vector with the same direction as its axis and plotted, for example, from the point of intersection of the axes. Let its length be equal to one. Then its magnitude

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 of a 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 the coordinate plane O x y, a 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 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 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 when 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 the scalar product of 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 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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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 are the coefficients of the only possible linear combination of basis vectors in the selected coordinate system, equal to this vector.

where are the coordinates of the vector.

Dot product 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

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 are reduced 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 with the direction of the Y axis acute angle 60°. 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.