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Back forward

Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

Goals:

• generalization and systematization of knowledge and skills of students;
• development of skills to analyze, compare, draw conclusions.

Equipment:

• multimedia projector;
• a computer;
• task sheets

STUDY PROCESS

I. Organizational moment

II. The stage of updating knowledge(slide 2)

We repeat how the distance from a point to a plane is determined

III. Lecture(slides 3-15)

In the lesson, we will look at various ways to find the distance from a point to a plane.

First method: step-by-step computational

Distance from point M to plane α:
is equal to the distance to the plane α from an arbitrary point P lying on the line a, which passes through the point M and is parallel to the plane α;
– is equal to the distance to the plane α from an arbitrary point P lying on the plane β, which passes through the point M and is parallel to the plane α.

We will solve the following tasks:

№1. In the cube A ... D 1 find the distance from the point C 1 to the plane AB 1 C.

It remains to calculate the value of the length of the segment O 1 N.

№2. In a regular hexagonal prism A ... F 1, all edges of which are equal to 1, find the distance from point A to the plane DEA 1.

Next method: volume method.

If the volume of the pyramid ABCM is V, then the distance from the point M to the plane α containing ∆ABC is calculated by the formula ρ(M; α) = ρ(M; ABC) =
When solving problems, we use the equality of the volumes of one figure, expressed in two different ways.

Let's solve the following problem:

№3. The edge AD of the pyramid DABC is perpendicular to the plane of the base ABC. Find the distance from A to the plane passing through the midpoints of the edges AB, AC and AD, if.

When solving problems coordinate method the distance from the point M to the plane α can be calculated by the formula ρ(M; α) = , where M(x 0; y 0; z 0), and the plane is given by the equation ax + by + cz + d = 0

Let's solve the following problem:

№4. In the unit cube A…D 1 find the distance from point A 1 to plane BDC 1 .

Let us introduce a coordinate system with the origin at point A, the y axis will pass along the edge AB, the x axis - along the edge AD, the z axis - along the edge AA 1. Then the coordinates of the points B (0; 1; 0) D (1; 0; 0;) C 1 (1; 1; 1)
Let us compose the equation of the plane passing through the points B, D, C 1 .

Then – dx – dy + dz + d = 0 x + y – z – 1= 0. Therefore, ρ =

The following method, which can be used in solving problems of this type - method of reference tasks.

The application of this method consists in the application of well-known reference problems, which are formulated as theorems.

Let's solve the following problem:

№5. In a unit cube A ... D 1 find the distance from the point D 1 to the plane AB 1 C.

Consider Application vector method.

№6. In a unit cube A ... D 1 find the distance from point A 1 to the plane BDC 1.

So, we have considered various methods that can be used in solving this type of problem. The choice of one or another method depends on the specific task and your preferences.

IV. Group work

Try to solve the problem in different ways.

№1. The edge of the cube А…D 1 is equal to . Find the distance from vertex C to plane BDC 1 .

№2. In a regular tetrahedron ABCD with an edge, find the distance from point A to plane BDC

№3. In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance from A to the plane BCA 1.

№4. In a regular quadrangular pyramid SABCD, all edges of which are equal to 1, find the distance from A to the plane SCD.

V. Lesson summary, homework, reflection

This article talks about determining the distance from a point to a plane. let's analyze the coordinate method, which will allow us to find the distance from a given point in three-dimensional space. To consolidate, consider examples of several tasks.

Yandex.RTB R-A-339285-1

The distance from a point to a plane is found by means of a known distance from a point to a point, where one of them is given, and the other is a projection onto a given plane.

When a point M 1 with a plane χ is given in space, then a straight line perpendicular to the plane can be drawn through the point. H 1 is a common point of their intersection. From here we get that the segment M 1 H 1 is a perpendicular, which was drawn from the point M 1 to the plane χ, where the point H 1 is the base of the perpendicular.

Definition 1

They call the distance from a given point to the base of the perpendicular, which was drawn from a given point to a given plane.

The definition can be written in different formulations.

Definition 2

Distance from point to plane called the length of the perpendicular, which was drawn from a given point to a given plane.

The distance from the point M 1 to the plane χ is defined as follows: the distance from the point M 1 to the plane χ will be the smallest from a given point to any point in the plane. If the point H 2 is located in the χ plane and is not equal to the point H 2, then we get a right triangle of the form M 2 H 1 H 2 , which is rectangular, where there is a leg M 2 H 1, M 2 H 2 - hypotenuse. Hence, this implies that M 1 H 1< M 1 H 2 . Тогда отрезок М 2 H 1 is considered inclined, which is drawn from the point M 1 to the plane χ. We have that the perpendicular drawn from a given point to a plane is less than the inclined one drawn from a point to a given plane. Consider this case in the figure below.

Distance from a point to a plane - theory, examples, solutions

There are a number of geometric problems whose solutions must contain the distance from a point to a plane. Ways to detect this may be different. To resolve, use the Pythagorean theorem or the similarity of triangles. When, according to the condition, it is necessary to calculate the distance from a point to a plane, given in a rectangular coordinate system of three-dimensional space, they solve using the coordinate method. This paragraph deals with this method.

According to the condition of the problem, we have that a point in three-dimensional space with coordinates M 1 (x 1, y 1, z 1) with the plane χ is given, it is necessary to determine the distance from M 1 to the plane χ. Several solutions are used to solve.

First way

This method is based on finding the distance from a point to a plane using the coordinates of the point H 1, which are the base of the perpendicular from the point M 1 to the plane χ. Next, you need to calculate the distance between M 1 and H 1.

To solve the problem in the second way, the normal equation of a given plane is used.

Second way

By condition, we have that H 1 is the base of the perpendicular, which was lowered from the point M 1 to the plane χ. Then we determine the coordinates (x 2, y 2, z 2) of the point H 1. The desired distance from M 1 to the χ plane is found by the formula M 1 H 1 \u003d (x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2, where M 1 (x 1, y 1 , z 1) and H 1 (x 2 , y 2 , z 2) . To solve, you need to know the coordinates of the point H 1.

We have that H 1 is the point of intersection of the plane χ with the line a, which passes through the point M 1 located perpendicular to the plane χ. It follows that it is necessary to formulate the equation of a straight line passing through a given point perpendicular to a given plane. It is then that we can determine the coordinates of the point H 1 . It is necessary to calculate the coordinates of the point of intersection of the line and the plane.

Algorithm for finding the distance from a point with coordinates M 1 (x 1, y 1, z 1) to the χ plane:

Definition 3

• compose the equation of a straight line a passing through the point M 1 and at the same time
• perpendicular to the χ plane;
• find and calculate the coordinates (x 2, y 2, z 2) of the point H 1, which are points
• intersection of the line a with the plane χ ;
• calculate the distance from M 1 to χ using the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + z 2 - z 1 2.

Third way

In a given rectangular coordinate system O x y z there is a plane χ, then we obtain a normal equation of the plane of the form cos α · x + cos β · y + cos γ · z - p = 0 . From here we get that the distance M 1 H 1 with the point M 1 (x 1 , y 1 , z 1) drawn to the plane χ, calculated by the formula M 1 H 1 = cos α x + cos β y + cos γ z-p. This formula is valid, since it is established thanks to the theorem.

Theorem

If a point M 1 (x 1 , y 1 , z 1) is given in three-dimensional space, having a normal equation of the χ plane of the form cos α x + cos β y + cos γ z - p = 0, then calculating the distance from the point to plane M 1 H 1 is derived from the formula M 1 H 1 = cos α · x + cos β · y + cos γ · z - p, since x = x 1 , y = y 1 , z = z 1 .

Proof

The proof of the theorem is reduced to finding the distance from a point to a line. From here we get that the distance from M 1 to the χ plane is the modulus of the difference between the numerical projection of the radius vector M 1 with the distance from the origin to the χ plane. Then we get the expression M 1 H 1 = n p n → O M → - p . The normal vector of the plane χ has the form n → = cos α , cos β , cos γ , and its length is equal to one, n p n → O M → is the numerical projection of the vector O M → = (x 1 , y 1 , z 1) in the direction determined by the vector n → .

Let's apply the formula for calculating scalar vectors. Then we obtain an expression for finding a vector of the form n → , O M → = n → n p n → O M → = 1 n p n → O M → = n p n → O M → , since n → = cos α , cos β , cos γ z and O M → = (x 1 , y 1 , z 1) . The coordinate form of the notation will take the form n →, O M → = cos α x 1 + cos β y 1 + cos γ z 1, then M 1 H 1 = n p n → O M → - p = cos α x 1 + cos β · y 1 + cos γ · z 1 - p . The theorem has been proven.

From here we get that the distance from the point M 1 (x 1, y 1, z 1) to the plane χ is calculated by substituting into the left side of the normal equation of the plane cos α x + cos β y + cos γ z - p = 0 instead of x, y, z coordinates x 1 , y 1 and z1 relating to the point M 1 , taking the absolute value of the obtained value.

Consider examples of finding the distance from a point with coordinates to a given plane.

Example 1

Calculate the distance from the point with coordinates M 1 (5 , - 3 , 10) to the plane 2 x - y + 5 z - 3 = 0 .

Solution

Let's solve the problem in two ways.

The first method will start by calculating the direction vector of the line a . By condition, we have that the given equation 2 x - y + 5 z - 3 = 0 is a general plane equation, and n → = (2 , - 1 , 5) is the normal vector of the given plane. It is used as a directing vector for the straight line a, which is perpendicular to the given plane. You should write the canonical equation of a straight line in space passing through M 1 (5, - 3, 10) with a direction vector with coordinates 2, - 1, 5.

The equation will look like x - 5 2 = y - (- 3) - 1 = z - 10 5 ⇔ x - 5 2 = y + 3 - 1 = z - 10 5 .

Intersection points should be defined. To do this, gently combine the equations into a system for the transition from the canonical to the equations of two intersecting lines. Let's take this point as H 1 . We get that

x - 5 2 = y + 3 - 1 = z - 10 5 ⇔ - 1 (x - 5) = 2 (y + 3) 5 (x - 5) = 2 (z - 10) 5 ( y + 3) = - 1 (z - 10) ⇔ ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 5 y + z + 5 = 0 ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0

Then you need to enable the system

x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 2 x - y + 5 z - 3 = 0 ⇔ x + 2 y = 1 5 x - 2 z = 5 2 x - y + 5 z = 3

Let us turn to the rule for solving the system according to Gauss:

1 2 0 - 1 5 0 - 2 5 2 - 1 5 3 ~ 1 2 0 - 1 0 - 10 - 2 10 0 - 5 5 5 ~ 1 2 0 - 1 0 - 10 - 2 10 0 0 6 0 ⇒ ⇒ z = 0 6 = 0 , y = - 1 10 10 + 2 z = - 1 , x = - 1 - 2 y = 1

We get that H 1 (1, - 1, 0) .

We calculate the distance from a given point to a plane. We take points M 1 (5, - 3, 10) and H 1 (1, - 1, 0) and get

M 1 H 1 \u003d (1 - 5) 2 + (- 1 - (- 3)) 2 + (0 - 10) 2 \u003d 2 30

The second solution is to first bring the given equation 2 x - y + 5 z - 3 = 0 to normal form. We determine the normalizing factor and get 1 2 2 + (- 1) 2 + 5 2 = 1 30 . From here we derive the equation of the plane 2 30 · x - 1 30 · y + 5 30 · z - 3 30 = 0 . The left side of the equation is calculated by substituting x \u003d 5, y \u003d - 3, z \u003d 10, and you need to take the distance from M 1 (5, - 3, 10) to 2 x - y + 5 z - 3 = 0 modulo. We get the expression:

M 1 H 1 \u003d 2 30 5 - 1 30 - 3 + 5 30 10 - 3 30 \u003d 60 30 \u003d 2 30

Answer: 2 30 .

When the χ plane is given by one of the methods of the plane definition methods section, then you first need to obtain the equation of the χ plane and calculate the desired distance using any method.

Example 2

Points with coordinates M 1 (5 , - 3 , 10) , A (0 , 2 , 1) , B (2 , 6 , 1) , C (4 , 0 , - 1) are set in three-dimensional space. Calculate the distance from M 1 to the plane A B C.

Solution

First you need to write down the equation of the plane passing through the given three points with coordinates M 1 (5, - 3, 10) , A (0 , 2 , 1) , B (2 , 6 , 1) , C (4 , 0 , - one) .

x - 0 y - 2 z - 1 2 - 0 6 - 2 1 - 1 4 - 0 0 - 2 - 1 - 1 = 0 ⇔ x y - 2 z - 1 2 4 0 4 - 2 - 2 = 0 ⇔ ⇔ - 8x + 4y - 20z + 12 = 0 ⇔ 2x - y + 5z - 3 = 0

It follows that the problem has a solution similar to the previous one. Hence, the distance from the point M 1 to the plane A B C is 2 30 .

Answer: 2 30 .

Finding the distance from a given point on a plane or to a plane to which they are parallel is more convenient by applying the formula M 1 H 1 = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p . From here we get that the normal equations of the planes are obtained in several steps.

Example 3

Find the distance from a given point with coordinates M 1 (- 3 , 2 , - 7) to the coordinate plane O x y z and the plane given by the equation 2 y - 5 = 0 .

Solution

The coordinate plane O y z corresponds to an equation of the form x = 0. For the O y z plane, it is normal. Therefore, it is necessary to substitute the values ​​x \u003d - 3 into the left side of the expression and take the absolute value of the distance from the point with coordinates M 1 (- 3, 2, - 7) to the plane. We get the value equal to - 3 = 3 .

After the transformation, the normal equation of the plane 2 y - 5 = 0 will take the form y - 5 2 = 0 . Then you can find the required distance from the point with coordinates M 1 (- 3 , 2 , - 7) to the plane 2 y - 5 = 0 . Substituting and calculating, we get 2 - 5 2 = 5 2 - 2.

Answer: The desired distance from M 1 (- 3 , 2 , - 7) to O y z has a value of 3 , and to 2 y - 5 = 0 has a value of 5 2 - 2 .

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First level

Coordinates and vectors. Comprehensive Guide (2019)

In this article, you and I will begin a discussion of one "magic wand" that will allow you to reduce many problems in geometry to simple arithmetic. This “wand” can make your life much easier, especially when you feel insecure in building spatial figures, sections, etc. All this requires a certain imagination and practical skills. The method, which we will begin to consider here, will allow you to abstract almost completely from all kinds of geometric constructions and reasoning. The method is called "coordinate method". In this article, we will consider the following questions:

1. Coordinate plane
2. Points and vectors on the plane
3. Building a vector from two points​
4. Vector length (distance between two points)​
5. Midpoint coordinates
6. Dot product of vectors​
7. Angle between two vectors

I think you already guessed why the coordinate method is called that? It is true that it got such a name, since it does not operate with geometric objects, but with their numerical characteristics (coordinates). And the transformation itself, which makes it possible to move from geometry to algebra, consists in introducing a coordinate system. If the original figure was flat, then the coordinates are two-dimensional, and if the figure is three-dimensional, then the coordinates are three-dimensional. In this article, we will consider only the two-dimensional case. And the main purpose of the article is to teach you how to use some basic techniques of the coordinate method (they sometimes turn out to be useful when solving problems in planimetry in part B of the Unified State Examination). The following two sections on this topic are devoted to the discussion of methods for solving problems C2 (the problem of stereometry).

Where would it be logical to start discussing the coordinate method? Probably with the concept of a coordinate system. Remember when you first met her. It seems to me that in the 7th grade, when you learned about the existence of a linear function, for example. Let me remind you that you built it point by point. Do you remember? You chose an arbitrary number, substituted it into the formula and calculated in this way. For example, if, then, if, then, etc. What did you get as a result? And you received points with coordinates: and. Then you drew a “cross” (coordinate system), chose a scale on it (how many cells you will have as a single segment) and marked the points you received on it, which you then connected with a straight line, the resulting line is the graph of the function.

There are a few things that need to be explained to you in a little more detail:

1. You choose a single segment for reasons of convenience, so that everything fits nicely and compactly in the picture

2. It is assumed that the axis goes from left to right, and the axis goes from bottom to top

3. They intersect at a right angle, and the point of their intersection is called the origin. It is marked with a letter.

4. In the record of the coordinate of a point, for example, on the left in brackets is the coordinate of the point along the axis, and on the right, along the axis. In particular, simply means that the point

5. In order to set any point on the coordinate axis, you need to specify its coordinates (2 numbers)

6. For any point lying on the axis,

7. For any point lying on the axis,

8. The axis is called the x-axis

9. The axis is called the y-axis

Now let's take the next step with you: mark two points. Connect these two points with a line. And let's put the arrow as if we were drawing a segment from point to point: that is, we will make our segment directed!

Remember what another name for a directed segment is? That's right, it's called a vector!

Thus, if we connect a dot to a dot, and the beginning will be point A, and the end will be point B, then we get a vector. You also did this construction in the 8th grade, remember?

It turns out that vectors, like points, can be denoted by two numbers: these numbers are called the coordinates of the vector. Question: do you think it is enough for us to know the coordinates of the beginning and end of the vector to find its coordinates? It turns out that yes! And it's very easy to do:

Thus, since in the vector the point is the beginning, and the end, the vector has the following coordinates:

For example, if, then the coordinates of the vector

Now let's do the opposite, find the coordinates of the vector. What do we need to change for this? Yes, you need to swap the beginning and end: now the beginning of the vector will be at a point, and the end at a point. Then:

Look closely, what is the difference between vectors and? Their only difference is the signs in the coordinates. They are opposite. This fact is written like this:

Sometimes, if it is not specifically stated which point is the beginning of the vector, and which is the end, then the vectors are denoted not by two capital letters, but by one lower case, for example:, etc.

Now a little practice and find the coordinates of the following vectors:

Examination:

Now solve the problem a little more difficult:

A vector torus with on-cha-scrap at a point has co-or-di-on-you. Find-di-te abs-cis-su points.

All the same is quite prosaic: Let be the coordinates of the point. Then

I compiled the system by determining what the coordinates of a vector are. Then the point has coordinates. We are interested in the abscissa. Then

Answer:

What else can you do with vectors? Yes, almost everything is the same as with ordinary numbers (except that you cannot divide, but you can multiply in two ways, one of which we will discuss here a little later)

1. Vectors can be stacked with each other
2. Vectors can be subtracted from each other
3. Vectors can be multiplied (or divided) by an arbitrary non-zero number
4. Vectors can be multiplied with each other

All these operations have a quite visual geometric representation. For example, the triangle (or parallelogram) rule for addition and subtraction:

A vector stretches or shrinks or changes direction when multiplied or divided by a number:

However, here we will be interested in the question of what happens to the coordinates.

1. When adding (subtracting) two vectors, we add (subtract) their coordinates element by element. That is:

2. When multiplying (dividing) a vector by a number, all its coordinates are multiplied (divided) by this number:

For example:

· Find-di-the sum of ko-or-di-nat century-to-ra.

Let's first find the coordinates of each of the vectors. Both of them have the same origin - the origin point. Their ends are different. Then, . Now we calculate the coordinates of the vector Then the sum of the coordinates of the resulting vector is equal to.

Answer:

Now solve the following problem yourself:

· Find the sum of the coordinates of the vector

We check:

Let's now consider the following problem: we have two points on the coordinate plane. How to find the distance between them? Let the first point be, and the second. Let's denote the distance between them as . Let's make the following drawing for clarity:

What I've done? I, firstly, connected the points and, and also drew a line parallel to the axis from the point, and drew a line parallel to the axis from the point. Did they intersect at a point, forming a wonderful figure? Why is she wonderful? Yes, you and I almost know everything about a right triangle. Well, the Pythagorean theorem, for sure. The desired segment is the hypotenuse of this triangle, and the segments are the legs. What are the coordinates of the point? Yes, they are easy to find from the picture: Since the segments are parallel to the axes and, respectively, their lengths are easy to find: if we denote the lengths of the segments, respectively, through, then

Now let's use the Pythagorean theorem. We know the lengths of the legs, we will find the hypotenuse:

Thus, the distance between two points is the root sum of the squared differences from the coordinates. Or - the distance between two points is the length of the segment connecting them. It is easy to see that the distance between the points does not depend on the direction. Then:

From this we draw three conclusions:

Let's practice a bit on calculating the distance between two points:

For example, if, then the distance between and is

Or let's go differently: find the coordinates of the vector

And find the length of the vector:

As you can see, it's the same!

Now practice a little on your own:

Task: find the distance between the given points:

We check:

Here are a couple more problems for the same formula, though they sound a little different:

1. Find-di-te the square of the length of the eyelid-to-ra.

2. Nai-di-te square of eyelid length-to-ra

I'm guessing you can handle them easily? We check:

1. And this is for attentiveness) We have already found the coordinates of the vectors before: . Then the vector has coordinates. The square of its length will be:

2. Find the coordinates of the vector

Then the square of its length is

Nothing complicated, right? Simple arithmetic, nothing more.

The following puzzles cannot be unambiguously classified, they are rather for general erudition and the ability to draw simple pictures.

1. Find-di-those sine of the angle on-clo-on-from-cut, connect-one-n-th-th point, with the abscissa axis.

and

How are we going to do it here? You need to find the sine of the angle between and the axis. And where can we look for the sine? That's right, in a right triangle. So what do we need to do? Build this triangle!

Since the coordinates of the point and, then the segment is equal, and the segment. We need to find the sine of the angle. Let me remind you that the sine is the ratio of the opposite leg to the hypotenuse, then

What are we left to do? Find the hypotenuse. You can do it in two ways: by the Pythagorean theorem (the legs are known!) or by the formula for the distance between two points (actually the same as the first method!). I will go the second way:

Answer:

The next task will seem even easier to you. She - on the coordinates of the point.

Task 2. From the point, the per-pen-di-ku-lar is lowered onto the abs-ciss axis. Nai-di-te abs-cis-su os-no-va-niya per-pen-di-ku-la-ra.

Let's make a drawing:

The base of the perpendicular is the point at which it intersects the x-axis (axis) for me this is a point. The figure shows that it has coordinates: . We are interested in the abscissa - that is, the "X" component. She is equal.

Answer: .

Task 3. Under the conditions of the previous problem, find the sum of the distances from the point to the coordinate axes.

The task is generally elementary if you know what the distance from a point to the axes is. You know? I hope, but still I remind you:

So, in my drawing, located a little higher, I have already depicted one such perpendicular? What axis is it? to the axis. And what is its length then? She is equal. Now draw a perpendicular to the axis yourself and find its length. It will be equal, right? Then their sum is equal.

Answer: .

Task 4. In the conditions of problem 2, find the ordinate of the point symmetrical to the point about the x-axis.

I think you intuitively understand what symmetry is? Very many objects have it: many buildings, tables, planes, many geometric shapes: a ball, a cylinder, a square, a rhombus, etc. Roughly speaking, symmetry can be understood as follows: a figure consists of two (or more) identical halves. This symmetry is called axial. What then is an axis? This is exactly the line along which the figure can, relatively speaking, be “cut” into identical halves (in this picture, the axis of symmetry is straight):

Now let's get back to our task. We know that we are looking for a point that is symmetric about the axis. Then this axis is the axis of symmetry. So, we need to mark a point so that the axis cuts the segment into two equal parts. Try to mark such a point yourself. Now compare with my solution:

Did you do the same? Good! At the found point, we are interested in the ordinate. She is equal

Answer:

Now tell me, after thinking for a second, what will be the abscissa of the point symmetrical to point A about the y-axis? What is your answer? Correct answer: .

In general, the rule can be written like this:

A point symmetrical to a point about the x-axis has the coordinates:

A point symmetrical to a point about the y-axis has coordinates:

Well, now it's really scary. a task: Find the coordinates of a point that is symmetrical to a point, relative to the origin. You first think for yourself, and then look at my drawing!

Answer:

Now parallelogram problem:

Task 5: The points are ver-shi-na-mi-pa-ral-le-lo-gram-ma. Find-dee-te or-dee-on-tu points.

You can solve this problem in two ways: logic and the coordinate method. I will first apply the coordinate method, and then I will tell you how you can decide differently.

It is quite clear that the abscissa of the point is equal. (it lies on the perpendicular drawn from the point to the x-axis). We need to find the ordinate. Let's take advantage of the fact that our figure is a parallelogram, which means that. Find the length of the segment using the formula for the distance between two points:

We lower the perpendicular connecting the point with the axis. The point of intersection is denoted by a letter.

The length of the segment is equal. (find the problem yourself, where we discussed this moment), then we will find the length of the segment using the Pythagorean theorem:

The length of the segment is exactly the same as its ordinate.

Answer: .

Another solution (I'll just provide a picture that illustrates it)

Solution progress:

1. Spend

2. Find point coordinates and length

3. Prove that.

Another one cut length problem:

The points are-la-yut-xia top-shi-on-mi tri-angle-no-ka. Find the length of his midline, par-ral-lel-noy.

Do you remember what the middle line of a triangle is? Then for you this task is elementary. If you do not remember, then I will remind you: the middle line of a triangle is a line that connects the midpoints of opposite sides. It is parallel to the base and equal to half of it.

The base is a segment. We had to look for its length earlier, it is equal. Then the length of the midline is half as long and equal.

Answer: .

Comment: This problem can be solved in another way, which we will turn to a little later.

In the meantime, here are a few tasks for you, practice on them, they are quite simple, but they help to “get your hand in” using the coordinate method!

1. The points appear-la-yut-xia top-shi-on-mi tra-pe-tion. Find the length of its midline.

2. Points and yav-la-yut-xia ver-shi-na-mi pa-ral-le-lo-gram-ma. Find-dee-te or-dee-on-tu points.

3. Find the length from the cut, connect the second point and

4. Find-di-te the area for-the-red-shen-noy fi-gu-ry on the ko-or-di-nat-noy plane.

5. A circle centered at na-cha-le ko-or-di-nat passes through a point. Find-de-te her ra-di-mustache.

6. Nai-di-te ra-di-us circle-no-sti, describe-san-noy near the right-angle-no-ka, the tops-shi-ny of something-ro-go have co-or -di-na-you co-from-reply-but

Solutions:

1. It is known that the midline of a trapezoid is equal to half the sum of its bases. The base is equal, but the base. Then

Answer:

2. The easiest way to solve this problem is to notice that (parallelogram rule). Calculate the coordinates of the vectors and is not difficult: . When adding vectors, the coordinates are added. Then has coordinates. The point has the same coordinates, since the beginning of the vector is a point with coordinates. We are interested in the ordinate. She is equal.

Answer:

3. We act immediately according to the formula for the distance between two points:

Answer:

4. Look at the picture and say, between which two figures is the shaded area “squeezed”? It is sandwiched between two squares. Then the area of ​​the desired figure is equal to the area of ​​the large square minus the area of ​​the small one. The side of the small square is a segment connecting the points and its length is

Then the area of ​​the small square is

We do the same with a large square: its side is a segment connecting the points and its length is equal to

Then the area of ​​the large square is

The area of ​​the desired figure is found by the formula:

Answer:

5. If the circle has the origin as its center and passes through a point, then its radius will be exactly equal to the length of the segment (make a drawing and you will understand why this is obvious). Find the length of this segment:

Answer:

6. It is known that the radius of a circle circumscribed about a rectangle is equal to half of its diagonal. Let's find the length of any of the two diagonals (after all, in a rectangle they are equal!)

Answer:

Well, did you manage everything? It wasn't that hard to figure it out, was it? There is only one rule here - to be able to make a visual picture and simply “read” all the data from it.

We have very little left. There are literally two more points that I would like to discuss.

Let's try to solve this simple problem. Let two points and be given. Find the coordinates of the middle of the segment. The solution to this problem is as follows: let the point be the desired middle, then it has coordinates:

That is: coordinates of the middle of the segment = arithmetic mean of the corresponding coordinates of the ends of the segment.

This rule is very simple and usually does not cause difficulties for students. Let's see in what problems and how it is used:

1. Find-di-te or-di-na-tu se-re-di-us from-cut, connect-nya-yu-th-th point and

2. The points are yav-la-yut-xia ver-shi-na-mi-che-you-reh-coal-no-ka. Find-di-te or-di-na-tu points of re-re-se-che-niya of his dia-go-on-lei.

3. Find-di-te abs-cis-su of the center of the circle, describe-san-noy near the rectangle-no-ka, the tops-shi-we have something-ro-go co-or-di-na-you co-from-vet-stvenno-but.

Solutions:

1. The first task is just a classic. We act immediately by determining the midpoint of the segment. She has coordinates. The ordinate is equal.

Answer:

2. It is easy to see that the given quadrilateral is a parallelogram (even a rhombus!). You can prove it yourself by calculating the lengths of the sides and comparing them with each other. What do I know about a parallelogram? Its diagonals are bisected by the intersection point! Aha! So the point of intersection of the diagonals is what? This is the middle of any of the diagonals! I will choose, in particular, the diagonal. Then the point has coordinates. The ordinate of the point is equal to.

Answer:

3. What is the center of the circle circumscribed about the rectangle? It coincides with the point of intersection of its diagonals. What do you know about the diagonals of a rectangle? They are equal and the intersection point is divided in half. The task has been reduced to the previous one. Take, for example, the diagonal. Then if is the center of the circumscribed circle, then is the middle. I'm looking for coordinates: The abscissa is equal.

Answer:

Now practice a little on your own, I will only give the answers to each problem so that you can check yourself.

1. Nai-di-te ra-di-us circle-no-sti, describe-san-noy near the triangle-no-ka, the tops of someone-ro-go have ko-or-di -no misters

2. Find-di-te or-di-na-tu the center of the circle, describe the san-noy near the triangle-no-ka, the tops-shi-we have something-ro-go coordinates

3. What kind of ra-di-y-sa should there be a circle with a center at a point so that it touches the abs-ciss axis?

4. Find-di-te or-di-on-that point of re-re-se-che-ing of the axis and from-cut, connect-nya-yu-th-th point and

Answers:

Did everything work out? I really hope for it! Now - the last push. Now be especially careful. The material that I'm going to explain now is not only relevant to the simple coordinate method problems in Part B, but is also ubiquitous in Problem C2.

Which of my promises have I not yet kept? Remember what operations on vectors I promised to introduce and which ones I eventually introduced? Am I sure I haven't forgotten anything? Forgot! I forgot to explain what multiplication of vectors means.

There are two ways to multiply a vector by a vector. Depending on the chosen method, we will get objects of a different nature:

The vector product is quite tricky. How to do it and why it is needed, we will discuss with you in the next article. And in this we will focus on the scalar product.

There are already two ways that allow us to calculate it:

As you guessed, the result should be the same! So let's look at the first way first:

Dot product through coordinates

Find: - common notation for dot product

The formula for the calculation is as follows:

That is, the dot product = the sum of the products of the coordinates of the vectors!

Example:

Find-dee-te

Solution:

Find the coordinates of each of the vectors:

We calculate the scalar product by the formula:

Answer:

You see, absolutely nothing complicated!

Well, now try it yourself:

Find-di-te scalar-noe pro-from-ve-de-nie century-to-ditch and

Did you manage? Maybe he noticed a little trick? Let's check:

Vector coordinates, as in the previous task! Answer: .

In addition to the coordinate, there is another way to calculate the scalar product, namely, through the lengths of the vectors and the cosine of the angle between them:

Denotes the angle between the vectors and.

That is, the scalar product is equal to the product of the lengths of the vectors and the cosine of the angle between them.

Why do we need this second formula, if we have the first one, which is much simpler, at least there are no cosines in it. And we need it so that from the first and second formulas we can deduce how to find the angle between vectors!

Let Then remember the formula for the length of a vector!

Then if I plug this data into the dot product formula, I get:

But on the other side:

So what have we got? We now have a formula to calculate the angle between two vectors! Sometimes, for brevity, it is also written like this:

That is, the algorithm for calculating the angle between vectors is as follows:

1. We calculate the scalar product through the coordinates
2. Find the lengths of vectors and multiply them
3. Divide the result of point 1 by the result of point 2

Let's practice with examples:

1. Find the angle between the eyelids-to-ra-mi and. Give your answer in degrees.

2. Under the conditions of the previous problem, find the cosine between the vectors

Let's do this: I'll help you solve the first problem, and try to do the second one yourself! I agree? Then let's start!

1. These vectors are our old friends. We have already considered their scalar product and it was equal. Their coordinates are: , . Then we find their lengths:

Then we are looking for the cosine between the vectors:

What is the cosine of the angle? This is the corner.

Answer:

Well, now solve the second problem yourself, and then compare! I'll just give a very short solution:

2. has coordinates, has coordinates.

Let be the angle between the vectors and, then

Answer:

It should be noted that the tasks directly on the vectors and the method of coordinates in part B of the examination paper are quite rare. However, the vast majority of C2 problems can be easily solved by introducing a coordinate system. So you can consider this article as a foundation, on the basis of which we will make quite tricky constructions that we will need to solve complex problems.

COORDINATES AND VECTORS. INTERMEDIATE LEVEL

You and I continue to study the method of coordinates. In the last part, we derived a number of important formulas that allow:

1. Find vector coordinates
2. Find the length of a vector (alternatively: the distance between two points)
3. Add, subtract vectors. Multiply them by a real number
4. Find the midpoint of a segment
5. Calculate dot product of vectors
6. Find the angle between vectors

Of course, the entire coordinate method does not fit into these 6 points. It underlies such a science as analytical geometry, which you will get acquainted with at the university. I just want to build a foundation that will allow you to solve problems in a single state. exam. We figured out the tasks of part B in Now it's time to move to a qualitatively new level! This article will be devoted to a method for solving those C2 problems in which it would be reasonable to switch to the coordinate method. This reasonableness is determined by what needs to be found in the problem, and what figure is given. So, I would use the coordinate method if the questions are:

1. Find the angle between two planes
2. Find the angle between a line and a plane
3. Find the angle between two lines
4. Find the distance from a point to a plane
5. Find the distance from a point to a line
6. Find the distance from a straight line to a plane
7. Find the distance between two lines

If the figure given in the condition of the problem is a body of revolution (ball, cylinder, cone ...)

Suitable figures for the coordinate method are:

1. cuboid
2. Pyramid (triangular, quadrangular, hexagonal)

Also in my experience it is inappropriate to use the coordinate method for:

1. Finding the areas of sections
2. Calculations of volumes of bodies

However, it should be immediately noted that three “unfavorable” situations for the coordinate method are quite rare in practice. In most tasks, it can become your savior, especially if you are not very strong in three-dimensional constructions (which are sometimes quite intricate).

What are all the figures I have listed above? They are no longer flat, such as a square, triangle, circle, but voluminous! Accordingly, we need to consider not a two-dimensional, but a three-dimensional coordinate system. It is built quite easily: just in addition to the abscissa and ordinates, we will introduce another axis, the applicate axis. The figure schematically shows their relative position:

All of them are mutually perpendicular, intersect at one point, which we will call the origin. The abscissa axis, as before, will be denoted, the ordinate axis - , and the introduced applicate axis - .

If earlier each point on the plane was characterized by two numbers - the abscissa and the ordinate, then each point in space is already described by three numbers - the abscissa, the ordinate, the applicate. For example:

Accordingly, the abscissa of the point is equal, the ordinate is , and the applicate is .

Sometimes the abscissa of a point is also called the projection of the point onto the abscissa axis, the ordinate is the projection of the point onto the y-axis, and the applicate is the projection of the point onto the applicate axis. Accordingly, if a point is given then, a point with coordinates:

called the projection of a point onto a plane

called the projection of a point onto a plane

A natural question arises: are all the formulas derived for the two-dimensional case valid in space? The answer is yes, they are just and have the same appearance. For a small detail. I think you already guessed which one. In all formulas, we will have to add one more term responsible for the applicate axis. Namely.

1. If two points are given: , then:

• Vector coordinates:
• Distance between two points (or vector length)
• The middle of the segment has coordinates

2. If two vectors are given: and, then:

• Their dot product is:
• The cosine of the angle between the vectors is:

However, space is not so simple. As you understand, the addition of one more coordinate introduces a significant variety in the spectrum of figures "living" in this space. And for further narration, I need to introduce some, roughly speaking, "generalization" of the straight line. This "generalization" will be a plane. What do you know about plane? Try to answer the question, what is a plane? It's very difficult to say. However, we all intuitively imagine what it looks like:

Roughly speaking, this is a kind of endless “leaf” thrust into space. "Infinity" should be understood that the plane extends in all directions, that is, its area is equal to infinity. However, this explanation "on the fingers" does not give the slightest idea about the structure of the plane. And we will be interested in it.

Let's remember one of the basic axioms of geometry:

• A straight line passes through two different points on a plane, moreover, only one:

Or its analog in space:

Of course, you remember how to derive the equation of a straight line from two given points, this is not at all difficult: if the first point has coordinates: and the second, then the equation of the straight line will be as follows:

You went through this in 7th grade. In space, the equation of a straight line looks like this: let us have two points with coordinates: , then the equation of a straight line passing through them has the form:

For example, a line passes through points:

How should this be understood? This should be understood as follows: a point lies on a line if its coordinates satisfy the following system:

We will not be very interested in the equation of a straight line, but we need to pay attention to the very important concept of the directing vector of a straight line. - any non-zero vector lying on a given line or parallel to it.

For example, both vectors are direction vectors of a straight line. Let be a point lying on a straight line, and be its directing vector. Then the equation of a straight line can be written in the following form:

Once again, I will not be very interested in the equation of a straight line, but I really need you to remember what a direction vector is! Again: it is ANY non-zero vector lying on a line, or parallel to it.

Withdraw three-point equation of a plane is no longer so trivial, and is usually not covered in a high school course. But in vain! This technique is vital when we resort to the coordinate method to solve complex problems. However, I assume that you are full of desire to learn something new? Moreover, you will be able to impress your teacher at the university when it turns out that you already know how to use the technique that is usually studied in the course of analytic geometry. So let's get started.

The equation of a plane is not too different from the equation of a straight line on a plane, namely, it has the form:

some numbers (not all equal to zero), but variables, for example: etc. As you can see, the equation of a plane is not very different from the equation of a straight line (linear function). However, remember what we argued with you? We said that if we have three points that do not lie on one straight line, then the equation of the plane is uniquely restored from them. But how? I'll try to explain to you.

Since the plane equation is:

And the points belong to this plane, then when substituting the coordinates of each point into the equation of the plane, we should get the correct identity:

Thus, there is a need to solve three equations already with unknowns! Dilemma! However, we can always assume that (for this we need to divide by). Thus, we get three equations with three unknowns:

However, we will not solve such a system, but write out the cryptic expression that follows from it:

Equation of a plane passing through three given points

\[\left| (\begin(array)(*(20)(c))(x - (x_0))&((x_1) - (x_0))&((x_2) - (x_0))\\(y - (y_0) )&((y_1) - (y_0))&((y_2) - (y_0))\\(z - (z_0))&((z_1) - (z_0))&((z_2) - (z_0)) \end(array)) \right| = 0\]

Stop! What else is this? Some very unusual module! However, the object that you see in front of you has nothing to do with the module. This object is called a third-order determinant. From now on, when you deal with the method of coordinates on a plane, you will often come across these very determinants. What is a third order determinant? Oddly enough, it's just a number. It remains to understand what specific number we will compare with the determinant.

Let's first write the third-order determinant in a more general form:

Where are some numbers. Moreover, by the first index we mean the row number, and by the index - the column number. For example, it means that the given number is at the intersection of the second row and the third column. Let's pose the following question: how exactly are we going to calculate such a determinant? That is, what specific number will we compare it with? For the determinant of precisely the third order, there is a heuristic (visual) triangle rule, it looks like this:

1. The product of the elements of the main diagonal (from upper left to lower right) the product of the elements that form the first triangle "perpendicular" to the main diagonal the product of the elements that form the second triangle "perpendicular" to the main diagonal
2. The product of the elements of the secondary diagonal (from the upper right to the lower left) the product of the elements that form the first triangle "perpendicular" to the secondary diagonal the product of the elements that form the second triangle "perpendicular" to the secondary diagonal
3. Then the determinant is equal to the difference between the values ​​obtained at the step and

If we write all this in numbers, then we get the following expression:

However, you don’t need to memorize the calculation method in this form, it’s enough to just keep the triangles in your head and the very idea of ​​\u200b\u200bwhat is added to what and what is then subtracted from what).

Let's illustrate the triangle method with an example:

1. Calculate the determinant:

Let's figure out what we add and what we subtract:

Terms that come with a "plus":

This is the main diagonal: the product of the elements is

The first triangle, "perpendicular to the main diagonal: the product of the elements is

The second triangle, "perpendicular to the main diagonal: the product of the elements is

We add three numbers:

Terms that come with a "minus"

This is a side diagonal: the product of the elements is

The first triangle, "perpendicular to the secondary diagonal: the product of the elements is

The second triangle, "perpendicular to the secondary diagonal: the product of the elements is

We add three numbers:

All that remains to be done is to subtract from the sum of the plus terms the sum of the minus terms:

In this way,

As you can see, there is nothing complicated and supernatural in the calculation of third-order determinants. It is simply important to remember about triangles and not to make arithmetic mistakes. Now try to calculate yourself:

We check:

1. The first triangle perpendicular to the main diagonal:
2. The second triangle perpendicular to the main diagonal:
3. The sum of the plus terms:
4. First triangle perpendicular to the side diagonal:
5. The second triangle, perpendicular to the side diagonal:
6. The sum of terms with a minus:
7. Sum of plus terms minus sum of minus terms:

Here's a couple more determinants for you, calculate their values ​​yourself and compare with the answers:

Answers:

Well, did everything match? Great, then you can move on! If there are difficulties, then my advice is this: on the Internet there are a bunch of programs for calculating the determinant online. All you need is to come up with your own determinant, calculate it yourself, and then compare it with what the program calculates. And so on until the results start to match. I'm sure this moment will not be long in coming!

Now let's return to the determinant that I wrote out when I talked about the equation of a plane passing through three given points:

All you have to do is calculate its value directly (using the triangle method) and set the result equal to zero. Naturally, since they are variables, you will get some expression that depends on them. It is this expression that will be the equation of a plane passing through three given points that do not lie on one straight line!

Let's illustrate this with a simple example:

1. Construct the equation of the plane passing through the points

We compose a determinant for these three points:

Simplifying:

Now we calculate it directly according to the rule of triangles:

\[(\left| (\begin(array)(*(20)(c))(x + 3)&2&6\\(y - 2)&0&1\\(z + 1)&5&0\end(array)) \ right| = \left((x + 3) \right) \cdot 0 \cdot 0 + 2 \cdot 1 \cdot \left((z + 1) \right) + \left((y - 2) \right) \cdot 5 \cdot 6 - )\]

Thus, the equation of the plane passing through the points is:

Now try to solve one problem yourself, and then we will discuss it:

2. Find the equation of the plane passing through the points

Well, let's discuss the solution now:

We make a determinant:

And calculate its value:

Then the equation of the plane has the form:

Or, reducing by, we get:

Now two tasks for self-control:

1. Construct the equation of a plane passing through three points:

Answers:

Did everything match? Again, if there are certain difficulties, then my advice is this: you take three points from your head (with a high degree of probability they will not lie on one straight line), build a plane on them. And then check yourself online. For example, on the site:

However, with the help of determinants, we will construct not only the equation of the plane. Remember, I told you that for vectors, not only the dot product is defined. There is also a vector, as well as a mixed product. And if the scalar product of two vectors will be a number, then the vector product of two vectors will be a vector, and this vector will be perpendicular to the given ones:

Moreover, its modulus will be equal to the area of ​​the parallelogram built on the vectors and. We will need this vector to calculate the distance from a point to a line. How can we calculate the cross product of vectors and if their coordinates are given? The determinant of the third order again comes to our aid. However, before I move on to the algorithm for calculating the cross product, I have to make a small lyrical digression.

This digression concerns the basis vectors.

Schematically they are shown in the figure:

Why do you think they are called basic? The fact is that :

Or in the picture:

The validity of this formula is obvious, because:

vector product

Now I can start introducing the cross product:

The vector product of two vectors is a vector that is calculated according to the following rule:

Now let's give some examples of calculating the cross product:

Example 1: Find the cross product of vectors:

Solution: I make a determinant:

And I calculate it:

Now, from writing through basis vectors, I will return to the usual vector notation:

In this way:

Now try.

Ready? We check:

And traditionally two tasks to control:

1. Find the cross product of the following vectors:
2. Find the cross product of the following vectors:

Answers:

Mixed product of three vectors

The last construction I need is the mixed product of three vectors. It, like a scalar, is a number. There are two ways to calculate it. - through the determinant, - through the mixed product.

Namely, let's say we have three vectors:

Then the mixed product of three vectors, denoted by can be calculated as:

1. - that is, the mixed product is the scalar product of a vector and the vector product of two other vectors

For example, the mixed product of three vectors is:

Try to calculate it yourself using the vector product and make sure that the results match!

And again - two examples for an independent decision:

Answers:

Choice of coordinate system

Well, now we have all the necessary foundation of knowledge to solve complex stereometric problems in geometry. However, before proceeding directly to the examples and algorithms for solving them, I believe that it will be useful to dwell on the following question: how exactly choose a coordinate system for a particular figure. After all, it is the choice of the relative position of the coordinate system and the figure in space that will ultimately determine how cumbersome the calculations will be.

I remind you that in this section we are considering the following shapes:

1. cuboid
2. Straight prism (triangular, hexagonal…)
3. Pyramid (triangular, quadrangular)
4. Tetrahedron (same as triangular pyramid)

For a cuboid or cube, I recommend the following construction:

That is, I will place the figure “in the corner”. The cube and the box are very good figures. For them, you can always easily find the coordinates of its vertices. For example, if (as shown in the picture)

then the vertex coordinates are:

Of course, you don’t need to remember this, but remembering how best to position a cube or a rectangular box is desirable.

straight prism

Prism is a more harmful figure. You can arrange it in space in different ways. However, I think the following is the best option:

Triangular prism:

That is, we put one of the sides of the triangle entirely on the axis, and one of the vertices coincides with the origin.

Hexagonal prism:

That is, one of the vertices coincides with the origin, and one of the sides lies on the axis.

Quadrangular and hexagonal pyramid:

A situation similar to a cube: we combine two sides of the base with the coordinate axes, we combine one of the vertices with the origin. The only small difficulty will be to calculate the coordinates of the point.

For a hexagonal pyramid - the same as for a hexagonal prism. The main task will again be in finding the coordinates of the vertex.

Tetrahedron (triangular pyramid)

The situation is very similar to the one I gave for the triangular prism: one vertex coincides with the origin, one side lies on the coordinate axis.

Well, now you and I are finally close to starting to solve problems. From what I said at the very beginning of the article, you could draw the following conclusion: most C2 problems fall into 2 categories: problems for the angle and problems for the distance. First, we will consider problems for finding an angle. They, in turn, are divided into the following categories (as the complexity increases):

Problems for finding corners

1. Finding the angle between two lines
2. Finding the angle between two planes

Let's consider these problems sequentially: let's start by finding the angle between two straight lines. Come on, remember, have you and I solved similar examples before? You remember, because we already had something similar ... We were looking for an angle between two vectors. I remind you, if two vectors are given: and, then the angle between them is found from the relation:

Now we have a goal - finding the angle between two straight lines. Let's turn to the "flat picture":

How many angles do we get when two lines intersect? Already things. True, only two of them are not equal, while others are vertical to them (and therefore coincide with them). So what angle should we consider the angle between two straight lines: or? Here the rule is: the angle between two straight lines is always no more than degrees. That is, from two angles, we will always choose the angle with the smallest degree measure. That is, in this picture, the angle between the two lines is equal. In order not to bother with finding the smallest of the two angles every time, cunning mathematicians suggested using the module. Thus, the angle between two straight lines is determined by the formula:

You, as an attentive reader, should have had a question: where, in fact, do we get these very numbers that we need to calculate the cosine of an angle? Answer: we will take them from the direction vectors of the lines! Thus, the algorithm for finding the angle between two lines is as follows:

1. We apply formula 1.

Or in more detail:

1. We are looking for the coordinates of the direction vector of the first straight line
2. We are looking for the coordinates of the direction vector of the second line
3. Calculate the modulus of their scalar product
4. We are looking for the length of the first vector
5. We are looking for the length of the second vector
6. Multiply the results of point 4 by the results of point 5
7. We divide the result of point 3 by the result of point 6. We get the cosine of the angle between the lines
8. If this result allows us to calculate the angle exactly, we look for it
9. Otherwise, we write through the arccosine

Well, now is the time to move on to the tasks: I will demonstrate the solution of the first two in detail, I will present the solution of another one in brief, and I will only give answers to the last two tasks, you must do all the calculations for them yourself.

Tasks:

1. In the right tet-ra-ed-re, find-di-te the angle between you-so-that tet-ra-ed-ra and the me-di-a-noy bo-ko-how side.

2. In the right-forward six-coal-pi-ra-mi-de, the hundred-ro-na-os-no-va-niya are somehow equal, and the side ribs are equal, find the angle between the straight lines and.

3. The lengths of all edges of the right-handed four-you-rech-coal-noy pi-ra-mi-dy are equal to each other. Find the angle between the straight lines and if from-re-zok - you-so-that given pi-ra-mi-dy, the point is se-re-di-on her bo-ko- th rib

4. On the edge of the cube from-me-che-to a point so that Find-di-te the angle between the straight lines and

5. Point - se-re-di-on the edges of the cube Nai-di-te the angle between the straight lines and.

It is no coincidence that I placed the tasks in this order. While you have not yet had time to begin to navigate the coordinate method, I myself will analyze the most “problematic” figures, and I will leave you to deal with the simplest cube! Gradually you have to learn how to work with all the figures, I will increase the complexity of the tasks from topic to topic.

Let's start solving problems:

1. Draw a tetrahedron, place it in the coordinate system as I suggested earlier. Since the tetrahedron is regular, then all its faces (including the base) are regular triangles. Since we are not given the length of the side, I can take it equal. I think you understand that the angle will not really depend on how much our tetrahedron will be "stretched" ?. I will also draw the height and median in the tetrahedron. Along the way, I will draw its base (it will also come in handy for us).

I need to find the angle between and. What do we know? We only know the coordinate of the point. So, we need to find more coordinates of the points. Now we think: a point is a point of intersection of heights (or bisectors or medians) of a triangle. A dot is an elevated point. The point is the midpoint of the segment. Then finally we need to find: the coordinates of the points: .

Let's start with the simplest: point coordinates. Look at the figure: It is clear that the applicate of a point is equal to zero (the point lies on a plane). Its ordinate is equal (because it is the median). It is more difficult to find its abscissa. However, this is easily done on the basis of the Pythagorean theorem: Consider a triangle. Its hypotenuse is equal, and one of the legs is equal Then:

Finally we have:

Now let's find the coordinates of the point. It is clear that its applicate is again equal to zero, and its ordinate is the same as that of a point, that is. Let's find its abscissa. This is done rather trivially if one remembers that the heights of an equilateral triangle are divided by the intersection point in the proportion counting from the top. Since:, then the desired abscissa of the point, equal to the length of the segment, is equal to:. Thus, the coordinates of the point are:

Let's find the coordinates of the point. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. And the applique is equal to the length of the segment. - this is one of the legs of the triangle. The hypotenuse of a triangle is a segment - a leg. It is searched for the reasons that I highlighted in bold:

The point is the midpoint of the segment. Then we need to remember the formula for the coordinates of the middle of the segment:

That's it, now we can look for the coordinates of the direction vectors:

Well, everything is ready: we substitute all the data into the formula:

In this way,

Answer:

You should not be afraid of such "terrible" answers: for problems C2 this is a common practice. I would rather be surprised by the "beautiful" answer in this part. Also, as you noted, I practically did not resort to anything other than the Pythagorean theorem and the property of the heights of an equilateral triangle. That is, to solve the stereometric problem, I used the very minimum of stereometry. The gain in this is partially "extinguished" by rather cumbersome calculations. But they are quite algorithmic!

2. Draw a regular hexagonal pyramid along with the coordinate system, as well as its base:

We need to find the angle between the lines and. Thus, our task is reduced to finding the coordinates of points: . We will find the coordinates of the last three from the small drawing, and we will find the coordinate of the vertex through the coordinate of the point. Lots of work, but gotta get started!

a) Coordinate: it is clear that its applicate and ordinate are zero. Let's find the abscissa. To do this, consider a right triangle. Alas, in it we only know the hypotenuse, which is equal to. We will try to find the leg (because it is clear that twice the length of the leg will give us the abscissa of the point). How can we look for her? Let's remember what kind of figure we have at the base of the pyramid? This is a regular hexagon. What does it mean? This means that all sides and all angles are equal. We need to find one such corner. Any ideas? There are a lot of ideas, but there is a formula:

The sum of the angles of a regular n-gon is .

Thus, the sum of the angles of a regular hexagon is degrees. Then each of the angles is equal to:

Let's look at the picture again. It is clear that the segment is the bisector of the angle. Then the angle is degrees. Then:

Then where.

So it has coordinates

b) Now we can easily find the coordinate of the point: .

c) Find the coordinates of the point. Since its abscissa coincides with the length of the segment, it is equal. Finding the ordinate is also not very difficult: if we connect the points and and denote the point of intersection of the line, say for. (do it yourself simple construction). Then Thus, the ordinate of point B is equal to the sum of the lengths of the segments. Let's look at the triangle again. Then

Then since Then the point has coordinates

d) Now find the coordinates of the point. Consider a rectangle and prove that Thus, the coordinates of the point are:

e) It remains to find the coordinates of the vertex. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. Let's find an app. Since then. Consider a right triangle. By the condition of the problem, the lateral edge. This is the hypotenuse of my triangle. Then the height of the pyramid is the leg.

Then the point has coordinates:

That's it, I have the coordinates of all points of interest to me. I am looking for the coordinates of the directing vectors of the straight lines:

We are looking for the angle between these vectors:

Answer:

Again, when solving this problem, I did not use any sophisticated tricks, except for the formula for the sum of the angles of a regular n-gon, as well as the definition of the cosine and sine of a right triangle.

3. Since we are again not given the lengths of the edges in the pyramid, I will consider them equal to one. Thus, since ALL edges, and not just the side ones, are equal to each other, then at the base of the pyramid and me lies a square, and the side faces are regular triangles. Let's depict such a pyramid, as well as its base on a plane, marking all the data given in the text of the problem:

We are looking for the angle between and. I will make very brief calculations when I am looking for the coordinates of points. You will need to "decrypt" them:

b) - the middle of the segment. Her coordinates:

c) I will find the length of the segment using the Pythagorean theorem in a triangle. I will find by the Pythagorean theorem in a triangle.

Coordinates:

d) - the middle of the segment. Its coordinates are

e) Vector coordinates

f) Vector coordinates

g) Looking for an angle:

The cube is the simplest figure. I'm sure you can figure it out on your own. The answers to problems 4 and 5 are as follows:

Finding the angle between a line and a plane

Well, the time for simple puzzles is over! Now the examples will be even more difficult. To find the angle between a line and a plane, we will proceed as follows:

1. Using three points, we build the equation of the plane
   ,
   using a third order determinant.
2. By two points we are looking for the coordinates of the directing vector of the straight line:
3. We apply the formula to calculate the angle between a straight line and a plane:

As you can see, this formula is very similar to the one we used to find the angles between two lines. The structure of the right side is just the same, and on the left we are now looking for a sine, and not a cosine, as before. Well, one nasty action was added - the search for the equation of the plane.

Let's not shelve solving examples:

1. Os-no-va-ni-em straight-my prize-we are-la-et-xia equal-but-poor-ren-ny triangle-nick you-with-that prize-we are equal. Find the angle between the straight line and the plane

2. In a rectangular pa-ral-le-le-pi-pe-de from the West Nai-di-te the angle between the straight line and the plane

3. In the right-handed six-coal prism, all edges are equal. Find the angle between the straight line and the plane.

4. In the right triangular pi-ra-mi-de with the os-but-va-ni-em from the west of the rib Nai-di-te angle, ob-ra-zo-van -ny plane of the os-no-va-niya and straight-my, passing through the se-re-di-na of the ribs and

5. The lengths of all edges of the right quadrangular pi-ra-mi-dy with the top are equal to each other. Find the angle between the straight line and the plane, if the point is se-re-di-on the bo-ko-in-th edge of the pi-ra-mi-dy.

Again, I will solve the first two problems in detail, the third - briefly, and I leave the last two for you to solve on your own. In addition, you already had to deal with triangular and quadrangular pyramids, but not yet with prisms.

Solutions:

1. Draw a prism, as well as its base. Let's combine it with the coordinate system and mark all the data that are given in the problem statement:

I apologize for some non-observance of proportions, but for solving the problem this, in fact, is not so important. The plane is just the "back wall" of my prism. It is enough to simply guess that the equation of such a plane has the form:

However, this can also be shown directly:

We choose arbitrary three points on this plane: for example, .

Let's make the equation of the plane:

Exercise for you: calculate this determinant yourself. Did you succeed? Then the equation of the plane has the form:

Or simply

In this way,

To solve the example, I need to find the coordinates of the directing vector of the straight line. Since the point coincided with the origin, the coordinates of the vector will simply coincide with the coordinates of the point. To do this, we first find the coordinates of the point.

To do this, consider a triangle. Let's draw a height (it is also a median and a bisector) from the top. Since, then the ordinate of the point is equal. In order to find the abscissa of this point, we need to calculate the length of the segment. By the Pythagorean theorem we have:

Then the point has coordinates:

A dot is a "raised" on a dot:

Then the coordinates of the vector:

Answer:

As you can see, there is nothing fundamentally difficult in solving such problems. In fact, the “straightness” of a figure such as a prism simplifies the process a little more. Now let's move on to the next example:

2. We draw a parallelepiped, draw a plane and a straight line in it, and also separately draw its lower base:

First, we find the equation of the plane: The coordinates of the three points lying in it:

(the first two coordinates are obtained in an obvious way, and you can easily find the last coordinate from the picture from the point). Then we compose the equation of the plane:

We calculate:

We are looking for the coordinates of the direction vector: It is clear that its coordinates coincide with the coordinates of the point, isn't it? How to find coordinates? These are the coordinates of the point, raised along the applicate axis by one! . Then we are looking for the desired angle:

Answer:

3. Draw a regular hexagonal pyramid, and then draw a plane and a straight line in it.

Here it is even problematic to draw a plane, not to mention the solution of this problem, but the coordinate method does not care! It is in its versatility that its main advantage lies!

The plane passes through three points: . We are looking for their coordinates:

one) . Display the coordinates for the last two points yourself. You will need to solve the problem with a hexagonal pyramid for this!

2) We build the equation of the plane:

We are looking for the coordinates of the vector: . (See triangular pyramid problem again!)

3) We are looking for an angle:

Answer:

As you can see, there is nothing supernaturally difficult in these tasks. You just need to be very careful with the roots. To the last two problems, I will give only answers:

As you can see, the technique for solving problems is the same everywhere: the main task is to find the coordinates of the vertices and substitute them into some formulas. It remains for us to consider one more class of problems for calculating angles, namely:

Calculating angles between two planes

The solution algorithm will be as follows:

1. For three points we are looking for the equation of the first plane:
2. For the other three points, we are looking for the equation of the second plane:
3. We apply the formula:

As you can see, the formula is very similar to the previous two, with the help of which we were looking for angles between straight lines and between a straight line and a plane. So remembering this one will not be difficult for you. Let's jump right into the problem:

1. A hundred-ro-on the basis of the right triangular prism is equal, and the dia-go-nal of the side face is equal. Find the angle between the plane and the plane of the base of the prize.

2. In the right-forward four-you-re-coal-noy pi-ra-mi-de, all the edges of someone are equal, find the sine of the angle between the plane and the plane Ko-Stu, passing through the point of per-pen-di-ku-lyar-but straight-my.

3. In a regular four-coal prism, the sides of the os-no-va-nia are equal, and the side edges are equal. On the edge from-me-che-to the point so that. Find the angle between the planes and

4. In the right quadrangular prism, the sides of the bases are equal, and the side edges are equal. On the edge from-me-che-to a point so that Find the angle between the planes and.

5. In the cube, find the co-si-nus of the angle between the planes and

Problem solutions:

1. I draw a regular (at the base - an equilateral triangle) triangular prism and mark on it the planes that appear in the condition of the problem:

We need to find the equations of two planes: The base equation is obtained trivially: you can make the corresponding determinant for three points, but I will make the equation right away:

Now let's find the equation The point has coordinates The point - Since - the median and the height of the triangle, it is easy to find by the Pythagorean theorem in a triangle. Then the point has coordinates: Find the applicate of the point To do this, consider a right triangle

Then we get the following coordinates: We compose the equation of the plane.

We calculate the angle between the planes:

Answer:

2. Making a drawing:

The most difficult thing is to understand what kind of mysterious plane it is, passing through a point perpendicularly. Well, the main thing is what is it? The main thing is attentiveness! Indeed, the line is perpendicular. The line is also perpendicular. Then the plane passing through these two lines will be perpendicular to the line, and, by the way, will pass through the point. This plane also passes through the top of the pyramid. Then the desired plane - And the plane is already given to us. We are looking for coordinates of points.

We find the coordinate of the point through the point. It is easy to deduce from a small drawing that the coordinates of the point will be as follows: What is now left to find in order to find the coordinates of the top of the pyramid? Still need to calculate its height. This is done using the same Pythagorean theorem: first, prove that (trivially from small triangles forming a square at the base). Since by condition, we have:

Now everything is ready: vertex coordinates:

We compose the equation of the plane:

You are already an expert in calculating determinants. Easily you will receive:

Or otherwise (if we multiply both parts by the root of two)

Now let's find the equation of the plane:

(You didn’t forget how we get the equation of the plane, right? If you don’t understand where this minus one came from, then go back to the definition of the equation of the plane! It just always turned out before that that my plane belonged to the origin!)

We calculate the determinant:

(You may notice that the equation of the plane coincided with the equation of the straight line passing through the points and! Think why!)

Now we calculate the angle:

We need to find the sine:

Answer:

3. A tricky question: what is a rectangular prism, what do you think? It's just a well-known parallelepiped to you! Drawing right away! You can even not separately depict the base, there is little use from it here:

The plane, as we noted earlier, is written as an equation:

Now we make a plane

We immediately compose the equation of the plane:

Looking for an angle

Now the answers to the last two problems:

Well, now is the time to take a break, because you and I are great and have done a great job!

Coordinates and vectors. Advanced level

In this article, we will discuss with you another class of problems that can be solved using the coordinate method: distance problems. Namely, we will consider the following cases:

1. Calculating the distance between skew lines.

I have ordered the given tasks as their complexity increases. The easiest is to find point to plane distance and the hardest part is finding distance between intersecting lines. Although, of course, nothing is impossible! Let's not procrastinate and immediately proceed to the consideration of the first class of problems:

Calculating the distance from a point to a plane

What do we need to solve this problem?

1. Point coordinates

So, as soon as we get all the necessary data, we apply the formula:

You should already know how we build the equation of the plane from the previous problems that I analyzed in the last part. Let's get down to business right away. The scheme is as follows: 1, 2 - I help you decide, and in some detail, 3, 4 - only the answer, you make the decision yourself and compare. Started!

Tasks:

1. Given a cube. The edge length of the cube is Find-di-te distance from se-re-di-ny from cut to flat

2. Given the right-vil-naya four-you-rekh-coal-naya pi-ra-mi-da Bo-ko-voe edge hundred-ro-on the os-no-va-nia is equal. Find-di-those distances from a point to a plane where - se-re-di-on the edges.

3. In the right triangular pi-ra-mi-de with os-but-va-ni-em, the other edge is equal, and one hundred-ro-on os-no-va- niya is equal. Find-di-those distances from the top to the plane.

4. In the right-handed six-coal prism, all edges are equal. Find-di-those distances from a point to a plane.

Solutions:

1. Draw a cube with single edges, build a segment and a plane, denote the middle of the segment by the letter

.

First, let's start with an easy one: find the coordinates of a point. Since then (remember the coordinates of the middle of the segment!)

Now we compose the equation of the plane on three points

\[\left| (\begin(array)(*(20)(c))x&0&1\\y&1&0\\z&1&1\end(array)) \right| = 0\]

Now I can start finding the distance:

2. We start again with a drawing, on which we mark all the data!

For a pyramid, it would be useful to draw its base separately.

Even the fact that I draw like a chicken paw will not prevent us from easily solving this problem!

Now it's easy to find the coordinates of a point

Since the coordinates of the point

2. Since the coordinates of the point a are the middle of the segment, then

We can easily find the coordinates of two more points on the plane. We compose the equation of the plane and simplify it:

\[\left| (\left| (\begin(array)(*(20)(c))x&1&(\frac(3)(2))\\y&0&(\frac(3)(2))\\z&0&(\frac( (\sqrt 3 ))(2))\end(array)) \right|) \right| = 0\]

Since the point has coordinates: , then we calculate the distance:

Answer (very rare!):

Well, did you understand? It seems to me that everything here is just as technical as in the examples that we considered with you in the previous part. So I am sure that if you have mastered that material, then it will not be difficult for you to solve the remaining two problems. I'll just give you the answers:

Calculating the Distance from a Line to a Plane

In fact, there is nothing new here. How can a line and a plane be located relative to each other? They have all the possibilities: to intersect, or a straight line is parallel to the plane. What do you think is the distance from the line to the plane with which the given line intersects? It seems to me that it is clear that such a distance is equal to zero. Uninteresting case.

The second case is trickier: here the distance is already non-zero. However, since the line is parallel to the plane, then each point of the line is equidistant from this plane:

In this way:

And this means that my task has been reduced to the previous one: we are looking for the coordinates of any point on the line, we are looking for the equation of the plane, we calculate the distance from the point to the plane. In fact, such tasks in the exam are extremely rare. I managed to find only one problem, and the data in it was such that the coordinate method was not very applicable to it!

Now let's move on to another, much more important class of problems:

Calculating the Distance of a Point to a Line

What will we need?

1. The coordinates of the point from which we are looking for the distance:

2. Coordinates of any point lying on a straight line

3. Direction vector coordinates of the straight line

What formula do we use?

What does the denominator of this fraction mean to you and so it should be clear: this is the length of the directing vector of the straight line. Here is a very tricky numerator! The expression means the module (length) of the vector product of vectors and How to calculate the vector product, we studied in the previous part of the work. Refresh your knowledge, it will be very useful to us now!

Thus, the algorithm for solving problems will be as follows:

1. We are looking for the coordinates of the point from which we are looking for the distance:

2. We are looking for the coordinates of any point on the line to which we are looking for the distance:

3. Building a vector

4. We build the direction vector of the straight line

5. Calculate the cross product

6. We are looking for the length of the resulting vector:

7. Calculate the distance:

We have a lot of work, and the examples will be quite complex! So now focus all your attention!

1. Dana is a right-handed triangular pi-ra-mi-da with a vertex. One hundred-ro-on the os-no-va-niya pi-ra-mi-dy is equal, you-so-ta is equal. Find-di-those distances from the se-re-di-ny of the bo-ko-th edge to the straight line, where the points and are the se-re-di-ny of the ribs and co-from- vet-stven-but.

2. The lengths of the ribs and the right-angle-no-para-ral-le-le-pi-pe-da are equal, respectively, and Find-di-te distance from top-shi-ny to straight-my

3. In the right six-coal prism, all the edges of a swarm are equal find-di-those distance from a point to a straight line

Solutions:

1. We make a neat drawing, on which we mark all the data:

We have a lot of work for you! I would first like to describe in words what we will look for and in what order:

1. Coordinates of points and

2. Point coordinates

3. Coordinates of points and

4. Coordinates of vectors and

5. Their cross product

6. Vector length

7. The length of the vector product

8. Distance from to

Well, we have a lot of work to do! Let's roll up our sleeves!

1. To find the coordinates of the height of the pyramid, we need to know the coordinates of the point. Its applicate is zero, and the ordinate is equal to its abscissa. Finally, we got the coordinates:

Point coordinates

2. - middle of the segment

3. - the middle of the segment

midpoint

4.Coordinates

Vector coordinates

5. Calculate the vector product:

6. The length of the vector: the easiest way is to replace that the segment is the middle line of the triangle, which means it is equal to half the base. So that.

7. We consider the length of the vector product:

8. Finally, find the distance:

Phew, that's all! Honestly, I'll tell you: solving this problem by traditional methods (through constructions) would be much faster. But here I reduced everything to a ready-made algorithm! I think that the solution algorithm is clear to you? Therefore, I will ask you to solve the remaining two problems on your own. Compare answers?

Again, I repeat: it is easier (faster) to solve these problems through constructions, rather than resorting to the coordinate method. I demonstrated this way of solving only to show you a universal method that allows you to “do not finish anything”.

Finally, consider the last class of problems:

Calculating the distance between skew lines

Here the algorithm for solving problems will be similar to the previous one. What we have:

3. Any vector connecting the points of the first and second lines:

How do we find the distance between lines?

The formula is:

The numerator is the module of the mixed product (we introduced it in the previous part), and the denominator - as in the previous formula (the module of the vector product of the directing vectors of the lines, the distance between which we are looking for).

I will remind you that

then the distance formula can be rewritten as:

Divide this determinant by the determinant! Although, to be honest, I'm not in the mood for jokes here! This formula, in fact, is very cumbersome and leads to rather complicated calculations. If I were you, I would only use it as a last resort!

Let's try to solve a few problems using the above method:

1. In the right triangular prism, all the edges are somehow equal, find the distance between the straight lines and.

2. Given a right-fore-shaped triangular prism, all the edges of the os-no-va-niya of someone are equal to Se-che-tion, passing through the other rib and se-re-di-nu ribs are yav-la-et-sya square-ra-tom. Find-di-te dis-sto-I-nie between straight-we-mi and

I decide the first, and based on it, you decide the second!

1. I draw a prism and mark the lines and

Point C coordinates: then

Point coordinates

Vector coordinates

Point coordinates

Vector coordinates

Vector coordinates

\[\left((B,\overrightarrow (A(A_1)) \overrightarrow (B(C_1)) ) \right) = \left| (\begin(array)(*(20)(l))(\begin(array)(*(20)(c))0&1&0\end(array))\\(\begin(array)(*(20) (c))0&0&1\end(array))\\(\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2))&( - \frac(1) (2))&1\end(array))\end(array)) \right| = \frac((\sqrt 3 ))(2)\]

We consider the cross product between the vectors and

\[\overrightarrow (A(A_1)) \cdot \overrightarrow (B(C_1)) = \left| \begin(array)(l)\begin(array)(*(20)(c))(\overrightarrow i )&(\overrightarrow j )&(\overrightarrow k )\end(array)\\\begin(array )(*(20)(c))0&0&1\end(array)\\\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2))&( - \ frac(1)(2))&1\end(array)\end(array) \right| - \frac((\sqrt 3 ))(2)\overrightarrow k + \frac(1)(2)\overrightarrow i \]

Now we consider its length:

Answer:

Now try to carefully complete the second task. The answer to it will be:.

Coordinates and vectors. Brief description and basic formulas

A vector is a directed segment. - the beginning of the vector, - the end of the vector.
The vector is denoted by or.

Absolute value vector - the length of the segment representing the vector. Designated as.

Vector coordinates:

,
where are the ends of the vector \displaystyle a .

Sum of vectors: .

The product of vectors:

Dot product of vectors:

• Through point A we build a planeβ II α .
• We build the third planeperpendicular to parallel planes α and β
• On the line of intersection of the planes, select point B and drop the perpendicular from point B.
• Segment BN - the distance between the planes is equal to the distance from point A to the planeα . AH=BN.

2. Given a cube ABCDA 1 B 1 C 1 D 1 . The length of the edge of the cube is 1. Find the distance from point A to plane CB 1 D 1 .
Solution [ , 250Kb]. The following algorithm will help us in this task:

• Through point A we construct a plane perpendicular to the plane α
• We lower the perpendicular to the line of intersection of the planes AH. AR - the desired distance from point A to the plane α .

3. It is often very difficult to show the distance from a point to a plane in a drawing, and it is extremely difficult to apply geometric methods. There is another way to find the desired distance by calculating the volume of a polyhedron or some part of a given polyhedron.

For example, in the above problem, I found the distance from point A to plane A 1 BT, expressing twice the volume of the pyramid ABTA 1 with the base of ABT.

Given a cube ABCDA 1 B 1 C 1 D 1 with edge 1. Find the distance from point A to plane A 1 BT, where T is the midpoint of segment AD.
Solution [ , 193Kb].

4. In a regular quadrangular prism ABCDA 1 B 1 C 1 D 1 with a base side of 12 and a height of 21, a point M is taken on the edge AA 1 so that AM \u003d 8. A point K is taken on the edge BB 1 so that B 1 K=8. Find the distance from point A 1 to plane D 1 MK.
Solution [ , 347Kb].

5. In a regular triangular prism ABCA 1 B 1 C 1, the sides of the base are equal to 2, and the side edges are equal to 3. Point D is the midpoint of the edge CC 1. Find the distance from vertex C to plane ADB 1.
Solution [ , 285Kb].

6. The base of the straight prism ABCA 1 B 1 C 1 is an isosceles triangle ABC, AB = AC = 5, BC = 6. The height of the prism is 3. Find the distance from the middle of the edge B 1 C 1 to the plane BCA 1.
Solution [ , 103Kb].

7. The base of a straight prism ABCA 1 B 1 C 1 is a right triangle ABC with a right angle C. BC \u003d 3. The height of the prism is 4. Find the distance from point B to the plane ACB 1.
Solution [ , 127Kb].

8. The base of the prism ABCDA 1 B 1 C 1 D 1 is the rhombus ABCD, AB = 10, BD = 12. The height of the prism is 6. Find the distance from the center of the face A 1 B 1 C 1 D 1 to the plane BDC 1.
Solution [ , 148Kb].

9. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1, all edges are equal to 1. Find the distance from point B to the plane DEA 1.
Solution [ , 194Kb].

10. Given a regular tetrahedron ABCD with edge . Find the distance from vertex A to plane BDC.
Solution [ , 119Kb].

11. In the pyramid DABC all edges are equal to a. Through O we denote the center of the base of ABC, and through K - the middle of the height DO of the pyramid. Find the distance from point K to face ABD.
Solution [