The essence of the coordinate method for solving geometric problems

The essence of solving problems using the coordinate method is to introduce a coordinate system that is convenient for us in one case or another and rewrite all the data using it. After that, all unknown quantities or proofs are held using this system. How to enter the coordinates of points in any coordinate system was discussed by us in another article - we will not dwell on this here.

Let us introduce the main assertions that are used in the coordinate method.

Statement 1: The vector coordinates will be determined by the difference between the corresponding coordinates of the end of this vector and its beginning.

Statement 2: The midpoint coordinates of the segment will be defined as half the sum of the corresponding coordinates of its boundaries.

Statement 3: The length of any vector $\overline(δ)$ with given coordinates $(δ_1,δ_2,δ_3)$ will be determined by the formula

$|\overline(δ)|=\sqrt(δ_1^2+δ_2^2+δ_3^2)$

Statement 4: The distance between any two points given by the coordinates $(δ_1,δ_2,δ_3)$ and $(β_1,β_2,β_3)$ will be determined by the formula

$d=\sqrt((δ_1-β_1)^2+(δ_2-β_2)^2+(δ_3-β_3)^2)$

Scheme for solving geometric problems using the coordinate method

To solve geometric problems using the coordinate method, it is best to use this scheme:

Analyze what is given in the problem:

• Set the most appropriate coordinate system for the task;
• Mathematically, the condition of the problem, the question of the problem are written down, a drawing is built for this problem.

1. Write down all data of the problem in the coordinates of the selected coordinate system.

2. Compose the necessary relations from the condition of the problem, and also connect these relations with what needs to be found (proved in the problem).
3. The result obtained is translated into the language of geometry.

Examples of problems solved by the coordinate method

The following tasks can be singled out as the main tasks leading to the coordinate method (their solutions will not be given here):

1. Tasks for finding the coordinates of a vector at its end and beginning.
2. Tasks related to the division of a segment in any respect.
3. Proof that three points lie on the same line or that four points lie on the same plane.
4. Tasks to find the distance between two given points.
5. Problems for finding volumes and areas of geometric shapes.

The results of solving the first and fourth problems are presented by us as the main statements above and are quite often used to solve other problems using the coordinate method.

Examples of tasks for applying the coordinate method

Example 1

Find the side of a regular pyramid whose height is $3$ cm if the side of the base is $4$ cm.

Let us be given a regular pyramid $ABCDS$, whose height is $SO$. Let's introduce a coordinate system, as in Figure 1.

Since the point $A$ is the center of the coordinate system we have constructed, then

Since the points $B$ and $D$ belong to the axes $Ox$ and $Oy$, respectively, then

$B=(4,0,0)$, $D=(0,4,0)$

Since the point $C$ belongs to the plane $Oxy$, then

Since the pyramid is regular, then $O$ is the midpoint of the segment $$. According to statement 2, we get:

$O=(\frac(0+4)(2),\frac(0+4)(2),\frac(0+0)(2))=(2,2,0)$

Since the height $SO$

Lesson test in geometry in grade 11

Topic: " 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.

To use the preview of presentations, create a Google account (account) and sign in: https://accounts.google.com

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.

ten . 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 ).

twenty . 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: ...

The coordinate method is a very efficient and versatile way to find any angles or distances between stereometric objects in space. If your math tutor is highly qualified, then he should know this. Otherwise, I would advise for the "C" part to change the tutor. My preparation for the exam in mathematics C1-C6 usually includes an analysis of the basic algorithms and formulas described below.

Angle between lines a and b

The angle between lines in space is the angle between any intersecting lines parallel to them. This angle is equal to the angle between the direction vectors of these lines (or complements it to 180 degrees).

What algorithm does the math tutor use to find the angle?

1) Choose any vectors and having directions of lines a and b (parallel to them).
2) We determine the coordinates of the vectors and by the corresponding coordinates of their beginnings and ends (the coordinates of the beginning must be subtracted from the coordinates of the end of the vector).
3) We substitute the found coordinates into the formula:
. To find the angle itself, you need to find the arc cosine of the result.

Normal to plane

A normal to a plane is any vector perpendicular to that plane.
How to find the normal? To find the coordinates of the normal, it is enough to know the coordinates of any three points M, N and K lying in the given plane. Using these coordinates, we find the coordinates of the vectors and and require that the conditions and be satisfied. Equating the scalar product of vectors to zero, we compose a system of equations with three variables, from which we can find the coordinates of the normal.

Math tutor's note : It is not necessary to solve the system completely, because it is enough to choose at least one normal. To do this, you can substitute any number (for example, one) instead of any of its unknown coordinates and solve a system of two equations with the remaining two unknowns. If it has no solutions, then this means that in the family of normals there is no one that has a unit for the selected variable. Then substitute one for another variable (another coordinate) and solve a new system. If you miss again, then your normal will have a unit on the last coordinate, and it will turn out to be parallel to some coordinate plane (in this case, it is easy to find it without a system).

Let's say that we are given a line and a plane with the coordinates of the direction vector and the normal
The angle between a straight line and a plane is calculated using the following formula:

Let and be any two normals to the given planes. Then the cosine of the angle between the planes is equal to the modulus of the cosine of the angle between the normals:

Equation of a plane in space

Points satisfying the equality form a plane with the normal . The coefficient is responsible for the amount of deviation (parallel shift) between two planes with the same given normal. In order to write the equation of a plane, you must first find its normal (as described above), and then substitute the coordinates of any point on the plane, together with the coordinates of the found normal, into the equation and find the coefficient .

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).

Solution. 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.

Solution. 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.

Solution. 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 one . 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)