Method of coordinates in space lesson. Method of coordinates in space: formulas and tutor's comments
The position of any point in space can be uniquely determined using a rectangular coordinate system. This system includes three mutually perpendicular axes intersecting at one point O is the origin of coordinates. One of the axes is called x-axis(axis Oh), the other y-axis (OU), the third applicate axis (Oz). planes XOY, XOZ and YOZ are called coordinate planes. Any segment is taken as scale unit for all three axes . The positive directions on the axes are chosen so that the rotation by 90 0 that combines the positive beam OX with positive beam OY, seemed to go counter-clockwise when viewed from the beam oz. This coordinate system is called right.
Position of any point M in space can be defined by three coordinates as follows . ThroughMdraw planes parallel to planesXOY, XOZ and YOZ. At the intersection with the axes, we get points, for example, P, Q and R respectively. Numbers X (abscissa), at(ordinate), z (applique), measuring segmentsOP, OQandORon the chosen scale are calledrectangular coordinatespoints M. They are taken positive or negative depending on whether the corresponding segments lie on the positive or negative semiaxis. Each triple of numbers ( X; at; z) corresponds to one and only one point in space, and vice versa.
Distance between two points and is calculated by the formula: (1.6)
Coordinates (x; y; z) pointsM dividing in a given ratio line segment AB, (,) are determined by the formulas:
In particular, at (point M divides the segment AB in half), formulas are obtained for determining the coordinates of the midpoint of the segment:
Example 4: on axle OU find a point equidistant from two points and .
Decision: Dot M lying on the axis OU, has coordinates . According to the task |AM| = |VM|. Let's find distances |AM| and |VM|, using formula (1.6):
We get the equation: .
Hence we find that 4 at= 16, i.e. y= 4. The desired point is M(0; 4; 0).
Example 5: Line segment AB divided into 3 equal parts. Find the coordinates of the division points, if the points are known and .
Decision:
Denote the points of division of the segment AB in the following order: FROM and D. According to the task |AC| = |CD| = |DB|. Therefore, the point FROM divides the segment AB in a relationship . Using formulas (1.7), we find the coordinates of the point C:
By formulas (1.8) we find the coordinates of the point D- the middle of the segment SW:
That is, point D has coordinates: .
Example 6: At points , ,, masses are concentrated accordingly m 1 , m 2 , m 3 , m four . Find the coordinates of the center of gravity of the system of these masses.
Decision:
As is known from the course of physics, the center of gravity of masses m 1 and m 2 placed at points AND and AT, divides the segment AB into parts inversely proportional to the masses concentrated at the ends of the segment (). Based on this, we first find the center of gravity of the system of two masses m 1 and m 2 placed at points AND 1 and AND 2 :
,
,
.
Center of gravity of a three-mass system m 1 and m 2 and m 3 () we find similarly:
,
,.
We finally find the center of gravity of the system of three massesm 1 , m 2 , m 3 andm 4 :
,
,
.
Questions to control:
Describe a rectangular coordinate system in the plane and all its components.
How are the coordinates of an arbitrary point on a plane determined?
Write a formula to find pdistance between two points on plane .
How to findcoordinates of a point dividing a segment in a given ratio?
Write the formulas for the coordinates of the midpoint of the segment.
Write a formula that calculates the area of a triangle if the coordinates of its vertices are known .
Describe the polar coordinate system.
What is the polar radius? To what extent is it measured?
What is a polar angle? Limits of its measurement?
As find the rectangular coordinates of a point for which the polar coordinates are known?
As find the polar coordinates of a point for which the rectangular coordinates are known?
How to find distance between points in polar coordinate system?
Describe a rectangular coordinate system in space and all its components.
How to determine the coordinates of a point in space?
Write down the formula for finding the distance between two points in space.
Write down formulas for finding the coordinates of a point dividing a segment in a given ratio for a three-dimensional coordinate system.
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Slides captions:
Rectangular coordinate system in space. Vector coordinates.
Rectangular coordinate system
If three pairwise perpendicular lines are drawn through a point in space, a direction is chosen on each of them, and a unit of measurement of segments is chosen, then they say that a rectangular coordinate system is set in space
Straight lines, with directions chosen on them, are called coordinate axes, and their common point is called the origin of coordinates. It is usually denoted by the letter O. The coordinate axes are denoted as follows: Ox, Oy, O z - and have names: the abscissa axis, the y-axis, the applicate axis.
The entire coordinate system is denoted Oxy z . The planes passing through the coordinate axes Ox and Oy, Oy and O z , O z and Ox, respectively, are called coordinate planes and are denoted Oxy, Oy z , O z x.
Point O divides each of the coordinate axes into two beams. The ray whose direction coincides with the direction of the axis is called the positive semi-axis, and the other ray is the negative semi-axis.
In a rectangular coordinate system, each point M of space is associated with a triple of numbers, which are called its coordinates.
The figure shows six points A (9; 5; 10), B (4; -3; 6), C (9; 0; 0), D (4; 0; 5), E (0; 3; 0) , F(0; 0; -3).
Vector coordinates
Any vector can be decomposed into coordinate vectors, i.e., represented in the form where the expansion coefficients x, y, z are uniquely determined.
The coefficients x, y and z in the expansion of a vector in terms of coordinate vectors are called the coordinates of the vector in the given coordinate system.
Consider the rules that allow us to find the coordinates of their sum and difference, as well as the coordinates of the product of a given vector by a given number, using the coordinates of these vectors.
10 . Each coordinate of the sum of two or more vectors is equal to the sum of the corresponding coordinates of these vectors. In other words, if a (x 1, y 1, z 1) and b (x 2, y 2, z 2 ) are given vectors, then the vector a + b has coordinates (x 1 + x 2, y 1 + y 2 , z 1 + z 2 ).
20 . Each coordinate of the difference of two vectors is equal to the difference of the corresponding coordinates of these vectors. In other words, if a (x 1, y 1, z 1) and b (x 2 y 2; z 2) are given vectors, then the vector a - b has coordinates (x 1 - x 2, y 1 - y 2, z 1 - z 2 ).
thirty . Each coordinate of the product of a vector by a number is equal to the product of the corresponding coordinate of the vector by that number. In other words, if a (x; y; x) is a given vector, α is a given number, then the vector α a has coordinates (αx; αy; α z).
On the topic: methodological developments, presentations and notes
Didactic handout "A set of notes for students on the topic "Method of coordinates in space" for conducting lessons in the form of lectures. Geometry grade 10-11....
The purpose of the lesson: To test the knowledge, skills and abilities of students on the topic "Using the method of coordinates in space to solve tasks C2 USE." Planned educational results: Students demonstrate: ...
Lesson test in geometry in grade 11
Theme: " Method of coordinates in space”.
Target: Check the theoretical knowledge of students, their skills and abilities to apply this knowledge in solving problems in vector, vector-coordinate ways.
Tasks:
1 .Create conditions for control (self-control, mutual control) of the assimilation of knowledge and skills.
2. Develop mathematical thinking, speech, attention.
3. To promote activity, mobility, ability to communicate, the general culture of students.
Conduct form: work in groups.
Equipment and sources of information: screen, multimedia projector, spreadsheet, credit cards, tests.
During the classes
1. Mobilizing moment.
Lesson using CSR; students are divided into 3 dynamic groups, in which students with an acceptable, optimal and advanced level. Each group has a coordinator who manages the work of the entire group.
2 . Self-determination of students on the basis of anticipation.
A task:goal-setting according to the scheme: remember-learn-be able.
Entrance test - Fill in the blanks (on printouts)
entrance test
Fill the gaps…
1.Three pairwise perpendicular lines are drawn through a point in space
we, on each of them, the direction and unit of measurement of the segments are selected,
then they say that it is set …………. in space.
2. Straight lines with directions chosen on them are called ……………..,
and their common point is …………. .
3. In a rectangular coordinate system, each point M of space is associated with a triple of numbers that call it ………………..
4. The coordinates of a point in space are called ………………..
5. A vector whose length is equal to one is called …………..
6. Vectors iykare called………….
7. Odds xyz in decomposition a= xi + yj + zk called
……………vector a .
8. Each coordinate of the sum of two or more vectors is equal to ……………..
9. Each coordinate of the difference of two vectors is equal to ……………….
10. Each coordinate of the product of a vector and a number is equal to………………..
11.Each coordinate of the vector is equal to…………….
12. Each coordinate of the middle of the segment is equal to……………….
13. Vector length a { xyz) is calculated by the formula ……………………
14. Distance between points M 1(x 1 ; y 1; z 1) and M 2 (x 2; y 2 ; z2) is calculated by the formula …………………
15. The scalar product of two vectors is called……………..
16. The scalar product of non-zero vectors is equal to zero………………..
17. Dot product of vectorsa{ x 1; y 1; z 1} b { x 2 ; y 2 ; z 2) in expressed by the formula…………………
Mutual verification of the entrance test. Answers to the tasks of the test on the screen.
Evaluation criteria:
1-2 mistakes - "5"
3-4 errors - "4"
5-6 errors - "3"
In other cases - "2"
3. Doing work. (for cards).
Each card contains two tasks: No. 1 - theoretical with proof, No. 2 includes tasks.
Explain the level of difficulty of the tasks included in the work. The group performs one task, but having 2 parts. The group coordinator manages the work of the entire group. Discussion of the same information with several partners increases responsibility not only for one's own successes, but also for the results of collective work, which has a positive effect on the microclimate in the team.
CARD #1
1. Derive formulas expressing the coordinates of the middle of the segment in terms of the coordinates of its ends.
2. Task: 1) Points A (-3; 1; 2) and B (1; -1; 2) are given
Find:
a) the coordinates of the midpoint of the segment AB
b) coordinates and length of the vector AB
2) The cube ABCDA1 B1 C1 D1 is given. Using the coordinate method, find the angle
between lines AB1 and A1 D.
CARD#2
Derive a formula for calculating the length of a vector from its coordinates.
Task: 1) Given points M(-4; 7; 0),N(0; -1; 2). Find the distance from the origin of coordinates to the middle of the segment MN.
→ → → → →
2) Vector data a and b. Find b(a+b), if a(-2;3;6),b=6i-8k
CARD #3
Derive a formula for calculating the distance between points with given coordinates.
Task: 1) Points A(2;1;-8), B(1;-5;0), C(8;1;-4) are given.
Prove that ∆ABC is isosceles and find the length of the midline of the triangle connecting the midpoints of the sides.
2) Calculate the angle between straight lines AB and SD if A(1;1;0),
B(3;-1;2), D(0;1;0).
CARD#4
Derive formulas for the cosine of the angle between non-zero vectors with given coordinates.
Task: 1) The coordinates of three vertices of the parallelogram ABCD are given:
A(-6;-;4;0), B(6;-6;2), C(10;0;4). Find the coordinates of point D.
2) Find the angle between the lines AB and CD, if A (1; 1; 2), B (0; 1; 1), C (2; -2; 2), D (2; -3; 1).
CARD#5
Tell us how to calculate the angle between two lines in space using the direction vectors of these lines. →→
Task: 1) Find the scalar product of vectorsa and b, if:
→ → → ^ →
a) | a| =4; | b| =√3 (ab)=30◦
b) a {2 ;-3; 1}, b = 3 i +2 k
2) Points A(0;4;0), B(2;0;0), C(4;0;4) and D(2;4;4) are given. Prove that ABCD is a rhombus.
4. Checking the work of dynamic groups on cards.
We listen to the speeches of the representatives of the groups. The work of the groups is evaluated by the teacher with the participation of students.
5. Reflection. Grades for credit.
Final test with a choice of answers (in printouts).
1) Vectors are given a {2 ;-4 ;3} b(-3; ─ ; 1). Find vector coordinates
→ 2
c = a+ b
a) (-5; 3 −; 4); b) (-1; -3.5; 4) c) (5; -4 −; 2) d) (-1; 3.5; -4)
2) Vectors are given a(4; -3; 5) and b(-3; 1; 2). Find vector coordinates
C=2 a – 3 b
a) (7;-2;3); b) (11; -7; 8); c) (17; -9; 4); d) (-1; -3; 4).
→ → → → → →
3) Calculate the scalar product of vectorsm and n, if m = a + 2 b- c
→ → → → →^ → → → → →
n= 2 a - b if | a|=2 , | b |=3, (ab)=60°, c ┴ a , c ┴ b.
a)-1; b) -27; in 1; d) 35.
4) Vector length a { xyz) is equal to 5. Find the coordinates of the vector a ifx=2, z=-√5
a) 16; b) 4 or -4; at 9; d) 3 or -3.
5) Find the area ∆ABC if A(1;-1;3); B(3;-1;1) and C(-1;1;-3).
a) 4√3; b) √3; c) 2√3; d) √8.
Cross-validation test. Response codes to test tasks on the screen: 1(b); 2(c);
3(a); 4(b); 5(c).
Evaluation criteria:
Everything is correct - "5"
1 mistake - "4"
2 errors - "3"
In other cases - "2"
Student knowledge table
Work on
cards
final
test
Credit score
Tasks
theory
practice
1 group
2 group
3 group
Evaluation of students' preparation for the test.
In order to use the coordinate method, you need to know the formulas well. There are three of them:
At first glance, it looks menacing, but just a little practice - and everything will work great.
A task. Find the cosine of the angle between the vectors a = (4; 3; 0) and b = (0; 12; 5).
Decision. Since we are given the coordinates of the vectors, we substitute them into the first formula:
A task. Write an equation for the plane passing through the points M = (2; 0; 1), N = (0; 1; 1) and K = (2; 1; 0), if it is known that it does not pass through the origin.
Decision. The general equation of the plane: Ax + By + Cz + D = 0, but since the desired plane does not pass through the origin - the point (0; 0; 0) - then we set D = 1. Since this plane passes through the points M, N and K, then the coordinates of these points should turn the equation into a true numerical equality.
Let us substitute the coordinates of the point M = (2; 0; 1) instead of x, y and z. We have:
A 2 + B 0 + C 1 + 1 = 0 ⇒ 2A + C + 1 = 0;
Similarly, for the points N = (0; 1; 1) and K = (2; 1; 0) we obtain the equations:
A 0 + B 1 + C 1 + 1 = 0 ⇒ B + C + 1 = 0;
A 2 + B 1 + C 0 + 1 = 0 ⇒ 2A + B + 1 = 0;
So we have three equations and three unknowns. We compose and solve the system of equations:
We got that the equation of the plane has the form: − 0.25x − 0.5y − 0.5z + 1 = 0.
A task. The plane is given by the equation 7x − 2y + 4z + 1 = 0. Find the coordinates of the vector perpendicular to the given plane.
Decision. Using the third formula, we get n = (7; − 2; 4) - that's all!
Calculation of coordinates of vectors
But what if there are no vectors in the problem - there are only points lying on straight lines, and it is required to calculate the angle between these straight lines? It's simple: knowing the coordinates of the points - the beginning and end of the vector - you can calculate the coordinates of the vector itself.
To find the coordinates of a vector, it is necessary to subtract the coordinates of the beginning from the coordinates of its end.
This theorem works equally on the plane and in space. The expression “subtract coordinates” means that the x coordinate of another point is subtracted from the x coordinate of one point, then the same must be done with the y and z coordinates. Here are some examples:
A task. There are three points in space, given by their coordinates: A = (1; 6; 3), B = (3; − 1; 7) and C = (− 4; 3; − 2). Find the coordinates of vectors AB, AC and BC.
Consider the vector AB: its beginning is at point A, and its end is at point B. Therefore, to find its coordinates, it is necessary to subtract the coordinates of point A from the coordinates of point B:
AB = (3 - 1; - 1 - 6; 7 - 3) = (2; - 7; 4).
Similarly, the beginning of the vector AC is still the same point A, but the end is point C. Therefore, we have:
AC = (− 4 − 1; 3 − 6; − 2 − 3) = (− 5; − 3; − 5).
Finally, to find the coordinates of the vector BC, it is necessary to subtract the coordinates of point B from the coordinates of point C:
BC = (− 4 − 3; 3 − (− 1); − 2 − 7) = (− 7; 4; − 9).
Answer: AB = (2; − 7; 4); AC = (−5;−3;−5); BC = (−7; 4; − 9)
Pay attention to the calculation of the coordinates of the last vector BC: a lot of people make mistakes when they work with negative numbers. This applies to the variable y: point B has the coordinate y = − 1, and point C has y = 3. We get exactly 3 − (− 1) = 4, and not 3 − 1, as many people think. Don't make such stupid mistakes!
Computing Direction Vectors for Straight Lines
If you carefully read problem C2, you will be surprised to find that there are no vectors there. There are only straight lines and planes.
Let's start with straight lines. Everything is simple here: on any line there are at least two different points and, conversely, any two different points define a single line...
Does anyone understand what is written in the previous paragraph? I didn’t understand it myself, so I’ll explain it more simply: in problem C2, lines are always given by a pair of points. If we introduce a coordinate system and consider a vector with the beginning and end at these points, we get the so-called directing vector for a straight line:
Why is this vector needed? The point is that the angle between two straight lines is the angle between their direction vectors. Thus, we are moving from incomprehensible straight lines to specific vectors, the coordinates of which are easily calculated. How easy? Take a look at the examples:
A task. Lines AC and BD 1 are drawn in the cube ABCDA 1 B 1 C 1 D 1 . Find the coordinates of the direction vectors of these lines.
Since the length of the edges of the cube is not specified in the condition, we set AB = 1. Let us introduce a coordinate system with the origin at point A and axes x, y, z directed along the lines AB, AD and AA 1, respectively. The unit segment is equal to AB = 1.
Now let's find the coordinates of the direction vector for the straight line AC. We need two points: A = (0; 0; 0) and C = (1; 1; 0). From here we get the coordinates of the vector AC = (1 - 0; 1 - 0; 0 - 0) = (1; 1; 0) - this is the direction vector.
Now let's deal with the straight line BD 1 . It also has two points: B = (1; 0; 0) and D 1 = (0; 1; 1). We get the direction vector BD 1 = (0 − 1; 1 − 0; 1 − 0) = (− 1; 1; 1).
Answer: AC = (1; 1; 0); BD 1 = (− 1; 1; 1)
A task. In a regular triangular prism ABCA 1 B 1 C 1 , all edges of which are equal to 1, straight lines AB 1 and AC 1 are drawn. Find the coordinates of the direction vectors of these lines.
Let us introduce a coordinate system: the origin is at point A, the x-axis coincides with AB, the z-axis coincides with AA 1 , the y-axis forms the OXY plane with the x-axis, which coincides with the ABC plane.
First, let's deal with the straight line AB 1 . Everything is simple here: we have points A = (0; 0; 0) and B 1 = (1; 0; 1). We get the direction vector AB 1 = (1 − 0; 0 − 0; 1 − 0) = (1; 0; 1).
Now let's find the direction vector for AC 1 . Everything is the same - the only difference is that the point C 1 has irrational coordinates. So, A = (0; 0; 0), so we have:
Answer: AB 1 = (1; 0; 1);
A small but very important note about the last example. If the beginning of the vector coincides with the origin, the calculations are greatly simplified: the coordinates of the vector are simply equal to the coordinates of the end. Unfortunately, this is only true for vectors. For example, when working with planes, the presence of the origin of coordinates on them only complicates the calculations.
Calculation of normal vectors for planes
Normal vectors are not vectors that are doing well, or that feel good. By definition, a normal vector (normal) to a plane is a vector perpendicular to the given plane.
In other words, a normal is a vector perpendicular to any vector in a given plane. Surely you have come across such a definition - however, instead of vectors, it was about straight lines. However, just above it was shown that in the C2 problem one can operate with any convenient object - even a straight line, even a vector.
Let me remind you once again that any plane is defined in space by the equation Ax + By + Cz + D = 0, where A, B, C and D are some coefficients. Without diminishing the generality of the solution, we can assume D = 1 if the plane does not pass through the origin, or D = 0 if it does. In any case, the coordinates of the normal vector to this plane are n = (A; B; C).
So, the plane can also be successfully replaced by a vector - the same normal. Any plane is defined in space by three points. How to find the equation of the plane (and hence the normal), we have already discussed at the very beginning of the article. However, this process causes problems for many, so I will give a couple more examples:
A task. The section A 1 BC 1 is drawn in the cube ABCDA 1 B 1 C 1 D 1 . Find the normal vector for the plane of this section if the origin is at point A and the x, y, and z axes coincide with the edges AB, AD, and AA 1, respectively.
Since the plane does not pass through the origin, its equation looks like this: Ax + By + Cz + 1 = 0, i.e. coefficient D \u003d 1. Since this plane passes through points A 1, B and C 1, the coordinates of these points turn the equation of the plane into the correct numerical equality.
A 0 + B 0 + C 1 + 1 = 0 ⇒ C + 1 = 0 ⇒ C = − 1;
Similarly, for points B = (1; 0; 0) and C 1 = (1; 1; 1) we obtain the equations:
A 1 + B 0 + C 0 + 1 = 0 ⇒ A + 1 = 0 ⇒ A = − 1;
A 1 + B 1 + C 1 + 1 = 0 ⇒ A + B + C + 1 = 0;
But the coefficients A = − 1 and C = − 1 are already known to us, so it remains to find the coefficient B:
B = − 1 − A − C = − 1 + 1 + 1 = 1.
We get the equation of the plane: - A + B - C + 1 = 0, Therefore, the coordinates of the normal vector are n = (- 1; 1; - 1).
A task. A section AA 1 C 1 C is drawn in the cube ABCDA 1 B 1 C 1 D 1. Find the normal vector for the plane of this section if the origin is at point A, and the x, y and z axes coincide with the edges AB, AD and AA 1 respectively.
In this case, the plane passes through the origin, so the coefficient D \u003d 0, and the equation of the plane looks like this: Ax + By + Cz \u003d 0. Since the plane passes through points A 1 and C, the coordinates of these points turn the equation of the plane into the correct numerical equality.
Let us substitute the coordinates of the point A 1 = (0; 0; 1) instead of x, y and z. We have:
A 0 + B 0 + C 1 = 0 ⇒ C = 0;
Similarly, for the point C = (1; 1; 0) we get the equation:
A 1 + B 1 + C 0 = 0 ⇒ A + B = 0 ⇒ A = − B;
Let B = 1. Then A = − B = − 1, and the equation of the entire plane is: − A + B = 0. Therefore, the coordinates of the normal vector are n = (− 1; 1; 0).
Generally speaking, in the above problems it is necessary to compose a system of equations and solve it. There will be three equations and three variables, but in the second case one of them will be free, i.e. take arbitrary values. That is why we have the right to put B = 1 - without prejudice to the generality of the solution and the correctness of the answer.
Very often in problem C2 it is required to work with points that divide the segment in half. The coordinates of such points are easily calculated if the coordinates of the ends of the segment are known.
So, let the segment be given by its ends - points A \u003d (x a; y a; z a) and B \u003d (x b; y b; z b). Then the coordinates of the middle of the segment - we denote it by the point H - can be found by the formula:
In other words, the coordinates of the middle of a segment are the arithmetic mean of the coordinates of its ends.
A task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in the coordinate system so that the x, y and z axes are directed along the edges AB, AD and AA 1, respectively, and the origin coincides with point A. Point K is the midpoint of edge A 1 B 1 . Find the coordinates of this point.
Since the point K is the middle of the segment A 1 B 1 , its coordinates are equal to the arithmetic mean of the coordinates of the ends. Let's write down the coordinates of the ends: A 1 = (0; 0; 1) and B 1 = (1; 0; 1). Now let's find the coordinates of point K:
A task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in the coordinate system so that the x, y and z axes are directed along the edges AB, AD and AA 1 respectively, and the origin coincides with point A. Find the coordinates of the point L where they intersect diagonals of the square A 1 B 1 C 1 D 1 .
From the course of planimetry it is known that the point of intersection of the diagonals of a square is equidistant from all its vertices. In particular, A 1 L = C 1 L, i.e. point L is the midpoint of the segment A 1 C 1 . But A 1 = (0; 0; 1), C 1 = (1; 1; 1), so we have:
Answer: L = (0.5; 0.5; 1)
