Method of coordinates in space lesson. Method of coordinates in space, presentation for a geometry lesson (grade 11) on the topic. Examples of problems solved by the coordinate method
The coordinate method is a very effective and universal way of finding any angles or distances between stereometric objects in space. If your math tutor has highly qualified, then he should know this. Otherwise, I would advise changing the tutor for the “C” part. My preparation for the Unified State 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 the directions of straight lines a and b (parallel to them).
2) We determine the coordinates of the vectors using 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) 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 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 a given plane. Using these coordinates, we find the coordinates of the vectors and and require that the conditions and be met. By equating the scalar product of vectors to zero, we create a system of equations with three variables, from which we can find the coordinates of the normal.
Math tutor's note : It is not at all necessary to solve the system completely, because it is enough to select at least one normal. To do this, you can substitute any number (for example, one) instead of any of its unknown coordinates and solve the 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 whose value is one in the selected variable. Then substitute one for another variable (another coordinate) and solve new system. If you miss again, then your normal will have one at the last coordinate, and it itself will turn out to be parallel to some coordinate plane (in this case it is easy to find without a system).
Let's assume that we are given a straight line and a plane with the coordinates of the direction vector and normal
The angle between the straight line and the plane is calculated using the following formula:
Let and be any two normals to these 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 a 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 along with the coordinates of the found normal into the equation and find the coefficient.
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Slide 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 is chosen line measurements, then they say that a rectangular coordinate system is given in space
Straight lines with directions chosen on them are called coordinate axes, and their common point is the origin of coordinates. It is usually denoted by the letter O. The coordinate axes are designated as follows: Ox, Oy, O z - and have names: abscissa axis, ordinate axis, 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 rays. A ray whose direction coincides with the direction of the axis is called a positive semi-axis, and the other ray is called a negative semi-axis.
In a rectangular coordinate system, each point M in 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 expanded into coordinate vectors, i.e., represent it in the form where the expansion coefficients x, y, z are determined in a unique way.
The coefficients x, y and z in the expansion of a vector into coordinate vectors are called the coordinates of the vector in a given coordinate system.
Let's consider the rules that allow us to use the coordinates of these vectors 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.
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 and a number is equal to the product of the corresponding coordinate of the vector and this 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 (αх; αу; α z).
On the topic: methodological developments, presentations and notes
Didactic handout "Set of notes for students on the topic "Method of coordinates in space" for conducting lessons in the form of lectures. Geometry grades 10-11....
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 C2 Unified State Examination tasks.” Planned educational results: Students demonstrate: ...
Lesson test on geometry in 11th grade
Subject: " Method of coordinates in space."
Target: Test students’ theoretical knowledge, their skills and abilities to apply this knowledge in solving problems using vector and vector-coordinate methods.
Tasks:
1 .Create conditions for control (self-control, mutual control) for the acquisition of knowledge and skills.
2. Develop mathematical thinking, speech, attention.
3. Promote activity, mobility, communication skills, and general culture of students.
Form of conduct: work in groups.
Equipment and sources of information: screen, multimedia projector, knowledge table, test cards, tests.
During the classes
1.Mobilizing moment.
Lesson using CSR; students are distributed into 3 dynamic groups, in which students with an acceptable, optimal and advanced level. Each group selects a coordinator who leads the work of the entire group.
2 . Self-determination of students based on anticipation.
Task:goal setting according to the scheme: remember-learn-be able.
Entrance test - Fill in the blanks (in printouts)
Entrance test
Fill the gaps…
1.Three pairs of perpendicular straight lines are drawn through a point in space.
on each of them the direction and unit of measurement of the segments are selected,
then they say that it is given …………. in space.
2. Straight lines with directions chosen on them are called ……………..,
and their common point…………. .
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 are called
……………vectors 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 vector coordinate 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) 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…………………
Peer review of the entry test. Answers to test tasks on the screen.
Evaluation criteria:
1-2 errors – “5”
3-4 errors - “4”
5-6 errors - “3”
In other cases – “2”
3. Getting the job done. (by cards).
Each card contains two tasks: No. 1 - theoretical with proof, No. 2 includes tasks.
Explain the level of complexity 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. Discussing 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 No. 1
1.Derive formulas expressing the coordinates of the middle of a segment through the coordinates of its ends.
2.Task: 1) Given points A (-3; 1; 2) and B (1; -1; 2)
Find:
a) coordinates of the midpoint of segment AB
b) coordinates and length of vector AB
2) Given a cube ABCDA1 B1 C1 D1. Using the coordinate method, find the angle
between straight lines AB1 and A1 D.
CARD#2
Derive a formula for calculating the length of a vector from its coordinates.
Problem: 1) Given points M(-4; 7; 0),N(0; -1; 2). Find the distance from the origin to the middle of the segment MN.
→ → → → →
2) Vectors are given a And b. Find b(a+b), If a (-2; 3; 6), b = 6i-8k
CARD No. 3
Derive a formula to calculate the distance between points with given coordinates.
Problem: 1) Given points A(2;1;-8), B(1;-5;0), C(8;1;-4).
Prove that ∆ABC is isosceles and find the length midline triangle connecting the midpoints of the sides.
2) Calculate the angle between straight lines AB and CD, if A(1;1;0),
B(3;-1;2), D(0;1;0).
CARD#4
Derive formulas for the cosine of the angle between nonzero vectors with given coordinates.
Problem: 1) Given the coordinates of the three vertices of the parallelogram ABCD:
A(-6;-;4;0), B(6;-6;2), C(10;0;4). Find the coordinates of point D.
2) Find the angle between straight 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 scalar product vectorsa And b, If:
→ → → ^ →
a) | a| =4; | b| =√3 (ab)=30◦
b) a {2 ;-3; 1}, b = 3 i +2 k
2) Given points A(0;4;0), B(2;0;0), C(4;0;4) and D(2;4;4). Prove that ABCD is a rhombus.
4. Checking the work of dynamic groups using cards.
We listen to the performances of group representatives. The work of the groups is assessed by the teacher with the participation of students.
5. Reflection. Test grades.
Final test with multiple choice (in printouts).
1) Vectors are given a {2 ;-4 ;3} b(-3; ─ ; 1). Find the coordinates of the vector
→ 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 the coordinates of the vector
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 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.
Peer review of the test. Answer codes for test tasks on the screen: 1(b); 2(c);
3(a); 4(b); 5(c).
Evaluation criteria:
Everything is correct - “5”
1 error - “4”
2 errors - “3”
In other cases - “2”
Student Knowledge Table
Work on
cards
Final
test
Assessment for the test
Tasks
theory
practice
1 group
2nd group
3 group
Assessing students' preparation for the test.
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 a particular case and rewrite all the data using it. After that, all unknown quantities or proofs are carried out using this system. We discussed how to enter the coordinates of points in any coordinate system in another article - we will not dwell on this here.
Let us introduce the main statements that are used in the coordinate method.
Statement 1: The coordinates of the vector will be determined by the difference between the corresponding coordinates of the end of this vector and its beginning.
Statement 2: The coordinates of the middle of the segment will be determined 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 specified 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:
- Set the coordinate system most suitable for the task;
- The condition of the problem, the question of the problem are written down mathematically, and a drawing for this problem is constructed.
Write down all task data in the coordinates of the selected coordinate system.
- Make the necessary relations from the conditions of the problem, and also connect these relations with what needs to be found (proved in the problem).
- Translate the obtained result into the language of geometry.
Analyze what is given in the problem:
Examples of problems solved by the coordinate method
The main problems leading to the coordinate method can be identified as follows (we will not give their solutions here):
- Problems on finding the coordinates of a vector based on its end and beginning.
- Problems related to dividing a segment in any respect.
- Proof that three points lie on the same line or that four points lie on the same plane.
- Problems to find the distance between two given points.
- Problems on finding the volumes and areas of geometric figures.
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 problems using the coordinate method
Example 1
Find side a regular pyramid whose height is $3$ cm if the side of the base is $4$ cm.
May it be given to us regular pyramid$ABCDS$, whose height is $SO$. Let's introduce a coordinate system as in Figure 1.
Since point $A$ is the center of the coordinate system we have constructed, then
Since points $B$ and $D$ belong to the axes $Ox$ and $Oy$, respectively, then
$B=(4,0,0)$, $D=(0,4,0)$
Since point $C$ belongs to the $Oxy$ plane, then
Since the pyramid is regular, $O$ is the midpoint of 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 of $SO$
In order to use the coordinate method, you need to know the formulas well. There are three of them:
At first glance, it looks threatening, but with just a little practice, everything will work great.
Task. Find the cosine of the angle between the vectors a = (4; 3; 0) and b = (0; 12; 5).
Solution. Since the coordinates of the vectors are given to us, we substitute them into the first formula:
Task. Write an equation for a 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 of coordinates - the point (0; 0; 0) - then we put 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 correct numerical equality.
Let's 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 points N = (0; 1; 1) and K = (2; 1; 0) we obtain the following 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. Let's create and solve a system of equations:
We found that the equation of the plane has the form: − 0.25x − 0.5y − 0.5z + 1 = 0.
Task. The plane is given by the equation 7x − 2y + 4z + 1 = 0. Find the coordinates of the vector perpendicular to this plane.
Solution. Using the third formula, we get n = (7; − 2; 4) - that's all!
Calculation of vector coordinates
But what if there are no vectors in the problem - there are only points lying on straight lines, and you need 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, you need to subtract the coordinates of its beginning from the coordinates of its end.
This theorem works equally well both on a 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:
Task. There are three points in space, defined by their coordinates: A = (1; 6; 3), B = (3; − 1; 7) and C = (− 4; 3; − 2). Find the coordinates of the 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, we need 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 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 vector BC, you need 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 BC vector: many people make mistakes when working with negative numbers. This concerns the variable y: point B has coordinate y = − 1, and point C has coordinate y = 3. We get exactly 3 − (− 1) = 4, and not 3 − 1, as many people think. Don't make such stupid mistakes!
Calculation of 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.
First, let's look at the straight lines. Everything is simple here: on any line there are at least two distinct points and, conversely, any two distinct points define a unique line...
Did anyone understand what was written in the previous paragraph? I didn’t understand it myself, so I’ll explain it more simply: in problem C2, straight lines are always defined by a pair of points. If we introduce a coordinate system and consider a vector with the beginning and end at these points, we obtain the so-called direction vector for the line:
Why is this vector needed? The fact is that the angle between two straight lines is the angle between their direction vectors. Thus, we move from incomprehensible straight lines to specific vectors whose coordinates are easy to calculate. How easy is it? Take a look at the examples:
Task. In the cube ABCDA 1 B 1 C 1 D 1 the lines AC and BD 1 are drawn. 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. We introduce a coordinate system with the origin at point A and the x, y, z axes directed along the straight 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 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 look at the straight line BD 1. It also has two points: B = (1; 0; 0) and D 1 = (0; 1; 1). We obtain the direction vector BD 1 = (0 − 1; 1 − 0; 1 − 0) = (− 1; 1; 1).
Answer: AC = (1; 1; 0); BD 1 = (− 1; 1; 1)
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's 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 look at the straight line AB 1. Everything is simple here: we have points A = (0; 0; 0) and B 1 = (1; 0; 1). We obtain 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 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 of coordinates, 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 those vectors that are fine or feel good. By definition, a normal vector (normal) to a plane is a vector perpendicular to a given plane.
In other words, a normal is a vector perpendicular to any vector in a given plane. You've probably come across this definition - however, instead of vectors we were talking about straight lines. However, it was shown just above that in problem C2 you can operate with any convenient object - be it a straight line or a vector.
Let me remind you once again that every plane is defined in space by the equation Ax + By + Cz + D = 0, where A, B, C and D are some coefficients. Without losing 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 normal vector to this plane are equal to n = (A; B; C).
So, the plane can also be successfully replaced by a vector - the same normal. Every plane is defined in space by three points. We already discussed how to find the equation of the plane (and therefore the normal) at the very beginning of the article. However, this process causes problems for many, so I’ll give a couple more examples:
Task. In the cube ABCDA 1 B 1 C 1 D 1 a section A 1 BC 1 is drawn. Find the normal vector for the plane of this section if the origin of coordinates 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 = 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 following 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 we already know the coefficients A = − 1 and C = − 1, so it remains to find the coefficient B:
B = − 1 − A − C = − 1 + 1 + 1 = 1.
We obtain the equation of the plane: − A + B − C + 1 = 0. Therefore, the coordinates of the normal vector are equal to n = (− 1; 1; − 1).
Task. In the cube ABCDA 1 B 1 C 1 D 1 there is a section AA 1 C 1 C. Find the normal vector for the plane of this section if the origin of coordinates 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 = 0, and the equation of the plane looks like this: Ax + By + Cz = 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 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 obtain the equation:
A 1 + B 1 + C 0 = 0 ⇒ A + B = 0 ⇒ A = − B;
Let us set B = 1. Then A = − B = − 1, and the equation of the entire plane has the form: − A + B = 0. Therefore, the coordinates of the normal vector are equal to n = (− 1; 1; 0).
Generally speaking, in the above problems you need to create a system of equations and solve it. You will get 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 set B = 1 - without prejudice to the generality of the solution and the correctness of the answer.
Very often in Problem C2 you need to work with points that bisect a segment. The coordinates of such points are easily calculated if the coordinates of the ends of the segment are known.
So, let the segment be defined by its ends - points A = (x a; y a; z a) and B = (x b; y b; z b). Then the coordinates of the middle of the segment - let’s denote it by point H - can be found using the formula:
In other words, the coordinates of the middle of a segment are the arithmetic mean of the coordinates of its ends.
Task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in a coordinate system so that the x, y and z axes are directed along edges AB, AD and AA 1, respectively, and the origin coincides with point A. Point K is the middle of edge A 1 B 1 . Find the coordinates of this point.
Since 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:
Task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in a 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 at which they intersect diagonals of the square A 1 B 1 C 1 D 1 .
From the planimetry course we know 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 middle 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)