This article talks about the topic « distance from a point to a line », Discusses the definition of the distance from a point to a line with illustrated examples using the coordinate method. Each theory block at the end has shown examples of solving similar problems.

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The distance from a point to a line is found by determining the distance from point to point. Let's take a closer look.

Let there be a line a and a point M 1 that does not belong to the given line. Through it we draw a straight line b, located perpendicular to the straight line a. Let's take the point of intersection of the lines as H 1. We obtain that M 1 H 1 is a perpendicular that was lowered from point M 1 to straight line a.

Definition 1

Distance from point M 1 to straight line a is called the distance between points M 1 and H 1.

There are definitions that include the length of the perpendicular.

Definition 2

Distance from point to line is the length of the perpendicular drawn from a given point to a given line.

The definitions are equivalent. Consider the figure below.

It is known that the distance from a point to a line is the smallest of all possible. Let's look at this with an example.

If we take a point Q lying on a straight line a, which does not coincide with the point M 1, then we obtain that the segment M 1 Q is called an inclined segment, lowered from M 1 to a straight line a. It is necessary to indicate that the perpendicular from point M 1 is less than any other inclined line drawn from the point to the straight line.

To prove this, consider the triangle M 1 Q 1 H 1, where M 1 Q 1 is the hypotenuse. It is known that its length is always greater than the length of any of the legs. This means we have that M 1 H 1< M 1 Q . Рассмотрим рисунок, приведенный ниже.

The initial data for finding from a point to a line allows you to use several solution methods: through the Pythagorean theorem, determination of sine, cosine, tangent of an angle and others. Most tasks of this type are solved at school during geometry lessons.

When, when finding the distance from a point to a line, it is possible to introduce a rectangular coordinate system, then the coordinate method is used. In this paragraph, we will consider the main two methods of finding the required distance from a given point.

The first method involves searching for the distance as a perpendicular drawn from M 1 to straight line a. The second method uses the normal equation of straight line a to find the required distance.

If there is a point on the plane with coordinates M 1 (x 1 , y 1), located in a rectangular coordinate system, straight line a, and you need to find the distance M 1 H 1, you can make the calculation in two ways. Let's look at them.

First way

If there are coordinates of point H 1 equal to x 2, y 2, then the distance from the point to the line is calculated using the coordinates from the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2.

Now let's move on to finding the coordinates of point H 1.

It is known that a straight line in O x y corresponds to the equation of a straight line on the plane. Let's take the method of defining a straight line a by writing a general equation of a straight line or an equation with an angular coefficient. We compose the equation of a straight line that passes through point M 1 perpendicular to a given straight line a. Let's denote the straight line by the letter b. H 1 is the point of intersection of lines a and b, which means to determine the coordinates you need to use the article, which deals with the coordinates of the points of intersection of two lines.

It can be seen that the algorithm for finding the distance from a given point M 1 (x 1, y 1) to straight line a is carried out according to the points:

Definition 3

• finding the general equation of a straight line a, having the form A 1 x + B 1 y + C 1 = 0, or an equation with an angular coefficient, having the form y = k 1 x + b 1;
• obtaining a general equation of line b, having the form A 2 x + B 2 y + C 2 = 0 or an equation with an angular coefficient y = k 2 x + b 2, if line b intersects point M 1 and is perpendicular to a given line a;
• determination of the coordinates x 2, y 2 of the point H 1, which is the intersection point of a and b, for this purpose the system of linear equations is solved A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 y + C 2 = 0 or y = k 1 x + b 1 y = k 2 x + b 2 ;
• calculating the required distance from a point to a line using the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2.

Second way

The theorem can help answer the question of finding the distance from a given point to a given straight line on a plane.

Theorem

The rectangular coordinate system has O x y has a point M 1 (x 1, y 1), from which a straight line is drawn to the plane, given by the normal equation of the plane, having the form cos α x + cos β y - p = 0, equal to The absolute value obtained on the left side of the normal equation of the line, calculated at x = x 1, y = y 1, means that M 1 H 1 = cos α · x 1 + cos β · y 1 - p.

Proof

Line a corresponds to the normal equation of the plane, having the form cos α x + cos β y - p = 0, then n → = (cos α, cos β) is considered the normal vector of line a at a distance from the origin to line a with p units . It is necessary to display all the data in the figure, add a point with coordinates M 1 (x 1, y 1), where the radius vector of the point M 1 - O M 1 → = (x 1, y 1). It is necessary to draw a straight line from a point to a straight line, which we denote as M 1 H 1 . It is necessary to show the projections M 2 and H 2 of the points M 1 and H 2 onto a straight line passing through the point O with a direction vector of the form n → = (cos α, cos β), and denote the numerical projection of the vector as O M 1 → = (x 1, y 1) to the direction n → = (cos α , cos β) as n p n → O M 1 → .

The variations depend on the location of the M1 point itself. Let's look at the figure below.

We fix the results using the formula M 1 H 1 = n p n → O M → 1 - p. Then we bring the equality to this form M 1 H 1 = cos α · x 1 + cos β · y 1 - p in order to obtain n p n → O M → 1 = cos α · x 1 + cos β · y 1 .

The scalar product of vectors results in a transformed formula of the form n → , O M → 1 = n → · n p n → O M 1 → = 1 · n p n → O M 1 → = n p n → O M 1 → , which is a product in coordinate form of the form n → , O M 1 → = cos α · x 1 + cos β · y 1 . This means that we get that n p n → O M 1 → = cos α · x 1 + cos β · y 1 . It follows that M 1 H 1 = n p n → O M 1 → - p = cos α · x 1 + cos β · y 1 - p. The theorem has been proven.

We find that to find the distance from point M 1 (x 1 , y 1) to straight line a on the plane, you need to perform several actions:

Definition 4

• obtaining the normal equation of the straight line a cos α · x + cos β · y - p = 0, provided that it is not in the task;
• calculation of the expression cos α · x 1 + cos β · y 1 - p, where the resulting value takes M 1 H 1.

Let's apply these methods to solve problems with finding the distance from a point to a plane.

Example 1

Find the distance from the point with coordinates M 1 (- 1, 2) to the straight line 4 x - 3 y + 35 = 0.

Solution

Let's use the first method to solve.

To do this, it is necessary to find the general equation of the line b, which passes through a given point M 1 (- 1, 2), perpendicular to the line 4 x - 3 y + 35 = 0. From the condition it is clear that line b is perpendicular to line a, then its direction vector has coordinates equal to (4, - 3). Thus, we have the opportunity to write down the canonical equation of line b on the plane, since there are coordinates of the point M 1, which belongs to line b. Let's determine the coordinates of the directing vector of the straight line b. We get that x - (- 1) 4 = y - 2 - 3 ⇔ x + 1 4 = y - 2 - 3. The resulting canonical equation must be converted to a general one. Then we get that

x + 1 4 = y - 2 - 3 ⇔ - 3 · (x + 1) = 4 · (y - 2) ⇔ 3 x + 4 y - 5 = 0

Let us find the coordinates of the points of intersection of the lines, which we will take as the designation H 1. The transformations look like this:

4 x - 3 y + 35 = 0 3 x + 4 y - 5 = 0 ⇔ x = 3 4 y - 35 4 3 x + 4 y - 5 = 0 ⇔ x = 3 4 y - 35 4 3 3 4 y - 35 4 + 4 y - 5 = 0 ⇔ ⇔ x = 3 4 y - 35 4 y = 5 ⇔ x = 3 4 5 - 35 4 y = 5 ⇔ x = - 5 y = 5

From what was written above, we have that the coordinates of point H 1 are equal to (- 5; 5).

It is necessary to calculate the distance from point M 1 to straight line a. We have that the coordinates of the points M 1 (- 1, 2) and H 1 (- 5, 5), then we substitute them into the formula to find the distance and get that

M 1 H 1 = (- 5 - (- 1) 2 + (5 - 2) 2 = 25 = 5

Second solution.

In order to solve in another way, it is necessary to obtain the normal equation of the line. We calculate the value of the normalizing factor and multiply both sides of the equation 4 x - 3 y + 35 = 0. From here we get that the normalizing factor is equal to - 1 4 2 + (- 3) 2 = - 1 5, and the normal equation will be of the form - 1 5 4 x - 3 y + 35 = - 1 5 0 ⇔ - 4 5 x + 3 5 y - 7 = 0 .

According to the calculation algorithm, it is necessary to obtain the normal equation of the line and calculate it with the values ​​x = - 1, y = 2. Then we get that

4 5 · - 1 + 3 5 · 2 - 7 = - 5

From this we obtain that the distance from point M 1 (- 1, 2) to the given straight line 4 x - 3 y + 35 = 0 has the value - 5 = 5.

Answer: 5 .

It can be seen that in this method it is important to use the normal equation of the line, since this method is the shortest. But the first method is convenient because it is consistent and logical, although it has more calculation points.

Example 2

On the plane there is a rectangular coordinate system O x y with point M 1 (8, 0) and straight line y = 1 2 x + 1. Find the distance from a given point to a straight line.

Solution

The first method involves reducing a given equation with an angular coefficient to a general equation. To simplify, you can do it differently.

If the product of the angular coefficients of perpendicular lines has a value of - 1, then the angular coefficient of a line perpendicular to a given one y = 1 2 x + 1 has a value of 2. Now we get the equation of a line passing through a point with coordinates M 1 (8, 0). We have that y - 0 = - 2 · (x - 8) ⇔ y = - 2 x + 16 .

We proceed to finding the coordinates of point H 1, that is, the intersection points y = - 2 x + 16 and y = 1 2 x + 1. We compose a system of equations and get:

y = 1 2 x + 1 y = - 2 x + 16 ⇔ y = 1 2 x + 1 1 2 x + 1 = - 2 x + 16 ⇔ y = 1 2 x + 1 x = 6 ⇔ ⇔ y = 1 2 · 6 + 1 x = 6 = y = 4 x = 6 ⇒ H 1 (6, 4)

It follows that the distance from the point with coordinates M 1 (8, 0) to the straight line y = 1 2 x + 1 is equal to the distance from the start point and end point with coordinates M 1 (8, 0) and H 1 (6, 4) . Let's calculate and find that M 1 H 1 = 6 - 8 2 + (4 - 0) 2 20 = 2 5.

The solution in the second way is to move from an equation with a coefficient to its normal form. That is, we get y = 1 2 x + 1 ⇔ 1 2 x - y + 1 = 0, then the value of the normalizing factor will be - 1 1 2 2 + (- 1) 2 = - 2 5. It follows that the normal equation of the line takes the form - 2 5 1 2 x - y + 1 = - 2 5 0 ⇔ - 1 5 x + 2 5 y - 2 5 = 0. Let's carry out the calculation from the point M 1 8, 0 to a line of the form - 1 5 x + 2 5 y - 2 5 = 0. We get:

M 1 H 1 = - 1 5 8 + 2 5 0 - 2 5 = - 10 5 = 2 5

Answer: 2 5 .

Example 3

It is necessary to calculate the distance from the point with coordinates M 1 (- 2, 4) to the lines 2 x - 3 = 0 and y + 1 = 0.

Solution

We obtain the equation of the normal form of the straight line 2 x - 3 = 0:

2 x - 3 = 0 ⇔ 1 2 2 x - 3 = 1 2 0 ⇔ x - 3 2 = 0

Then we proceed to calculating the distance from the point M 1 - 2, 4 to the straight line x - 3 2 = 0. We get:

M 1 H 1 = - 2 - 3 2 = 3 1 2

The equation of the straight line y + 1 = 0 has a normalizing factor with a value equal to -1. This means that the equation will take the form - y - 1 = 0. We proceed to calculating the distance from the point M 1 (- 2, 4) to the straight line - y - 1 = 0. We find that it is equal to - 4 - 1 = 5.

Answer: 3 1 2 and 5.

Let's take a closer look at finding the distance from a given point on the plane to the coordinate axes O x and O y.

In a rectangular coordinate system, the O axis y has an equation of a straight line, which is incomplete and has the form x = 0, and O x - y = 0. The equations are normal for the coordinate axes, then it is necessary to find the distance from the point with coordinates M 1 x 1, y 1 to the lines. This is done based on the formulas M 1 H 1 = x 1 and M 1 H 1 = y 1. Let's look at the figure below.

Example 4

Find the distance from the point M 1 (6, - 7) to the coordinate lines located in the O x y plane.

Solution

Since the equation y = 0 refers to the straight line O x, you can find the distance from M 1 with given coordinates to this straight line using the formula. We get that 6 = 6.

Since the equation x = 0 refers to the straight line O y, you can find the distance from M 1 to this straight line using the formula. Then we get that - 7 = 7.

Answer: the distance from M 1 to O x has a value of 6, and from M 1 to O y has a value of 7.

When in three-dimensional space we have a point with coordinates M 1 (x 1, y 1, z 1), it is necessary to find the distance from point A to straight line a.

Let's consider two methods that allow you to calculate the distance from a point to a straight line a located in space. The first case considers the distance from point M 1 to a line, where a point on the line is called H 1 and is the base of a perpendicular drawn from point M 1 to line a. The second case suggests that the points of this plane must be sought as the height of the parallelogram.

First way

From the definition we have that the distance from point M 1 located on straight line a is the length of the perpendicular M 1 H 1, then we obtain that with the found coordinates of point H 1, then we find the distance between M 1 (x 1, y 1, z 1 ) and H 1 (x 1 , y 1 , z 1) , based on the formula M 1 H 1 = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.

We find that the whole solution goes towards finding the coordinates of the base of the perpendicular drawn from M 1 to the straight line a. This is done as follows: H 1 is the point where straight line a intersects with the plane that passes through the given point.

This means that the algorithm for determining the distance from point M 1 (x 1, y 1, z 1) to line a in space implies several points:

Definition 5

• drawing up the equation of the plane χ as an equation of the plane passing through a given point located perpendicular to the line;
• determination of the coordinates (x 2, y 2, z 2) belonging to the point H 1, which is the intersection point of straight line a and plane χ;
• calculating the distance from a point to a line using the formula M 1 H 1 = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.

Second way

From the condition we have a straight line a, then we can determine the direction vector a → = a x, a y, a z with coordinates x 3, y 3, z 3 and a certain point M 3 belonging to straight a. If you have the coordinates of the points M 1 (x 1, y 1) and M 3 x 3, y 3, z 3, you can calculate M 3 M 1 →:

M 3 M 1 → = (x 1 - x 3, y 1 - y 3, z 1 - z 3)

We should set aside the vectors a → = a x , a y , a z and M 3 M 1 → = x 1 - x 3 , y 1 - y 3 , z 1 - z 3 from point M 3 , connect them and get a parallelogram figure. M 1 H 1 is the height of the parallelogram.

Let's look at the figure below.

We have that the height M 1 H 1 is the required distance, then it is necessary to find it using the formula. That is, we are looking for M 1 H 1.

Let us denote the area of ​​the parallelogram by the letter S, found by the formula using the vector a → = (a x, a y, a z) and M 3 M 1 → = x 1 - x 3. y 1 - y 3, z 1 - z 3. The area formula is S = a → × M 3 M 1 → . Also, the area of ​​the figure is equal to the product of the lengths of its sides and the height, we get that S = a → · M 1 H 1 with a → = a x 2 + a y 2 + a z 2, which is the length of the vector a → = (a x, a y, a z), which is equal to the side of the parallelogram. This means that M 1 H 1 is the distance from the point to the line. It is found using the formula M 1 H 1 = a → × M 3 M 1 → a → .

To find the distance from a point with coordinates M 1 (x 1, y 1, z 1) to a straight line a in space, you need to perform several steps of the algorithm:

Definition 6

• determination of the direction vector of the straight line a - a → = (a x, a y, a z);
• calculating the length of the direction vector a → = a x 2 + a y 2 + a z 2 ;
• obtaining coordinates x 3 , y 3 , z 3 belonging to point M 3 located on straight line a;
• calculating the coordinates of the vector M 3 M 1 → ;
• finding the vector product of vectors a → (a x , a y , a z) and M 3 M 1 → = x 1 - x 3 , y 1 - y 3 , z 1 - z 3 as a → × M 3 M 1 → = i → j → k → a x a y a z x 1 - x 3 y 1 - y 3 z 1 - z 3 to obtain the length using the formula a → × M 3 M 1 → ;
• calculating the distance from a point to a line M 1 H 1 = a → × M 3 M 1 → a → .

Solving problems of finding the distance from a given point to a given line in space

Example 5

Find the distance from the point with coordinates M 1 2, - 4, - 1 to the line x + 1 2 = y - 1 = z + 5 5.

Solution

The first method begins with writing the equation of the plane χ passing through M 1 and perpendicular to a given point. We get an expression like:

2 (x - 2) - 1 (y - (- 4)) + 5 (z - (- 1)) = 0 ⇔ 2 x - y + 5 z - 3 = 0

It is necessary to find the coordinates of the point H 1, which is the point of intersection with the χ plane to the line specified by the condition. You should move from the canonical view to the intersecting one. Then we obtain a system of equations of the form:

x + 1 2 = y - 1 = z + 5 5 ⇔ - 1 · (x + 1) = 2 · y 5 · (x + 1) = 2 · (z + 5) 5 · y = - 1 · (z + 5) ⇔ 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

It is necessary to calculate 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 by Cramer’s method, then we get that:

∆ = 1 2 0 5 0 - 2 2 - 1 5 = - 60 ∆ x = - 1 2 0 5 0 - 2 3 - 1 5 = - 60 ⇔ x = ∆ x ∆ = - 60 - 60 = 1 ∆ y = 1 - 1 0 5 5 2 2 3 5 = 60 ⇒ y = ∆ y ∆ = 60 - 60 = - 1 ∆ z = 1 2 - 1 5 0 5 2 - 1 3 = 0 ⇒ z = ∆ z ∆ = 0 - 60 = 0

From here we have that H 1 (1, - 1, 0).

M 1 H 1 = 1 - 2 2 + - 1 - - 4 2 + 0 - - 1 2 = 11

The second method must begin by searching for coordinates in the canonical equation. To do this, you need to pay attention to the denominators of the fraction. Then a → = 2, - 1, 5 is the direction vector of the line x + 1 2 = y - 1 = z + 5 5. It is necessary to calculate the length using the formula a → = 2 2 + (- 1) 2 + 5 2 = 30.

It is clear that the straight line x + 1 2 = y - 1 = z + 5 5 intersects the point M 3 (- 1 , 0 , - 5), hence we have that the vector with the origin M 3 (- 1 , 0 , - 5) and its end at the point M 1 2, - 4, - 1 is M 3 M 1 → = 3, - 4, 4. Find the vector product a → = (2, - 1, 5) and M 3 M 1 → = (3, - 4, 4).

We get an expression of the form a → × M 3 M 1 → = i → j → k → 2 - 1 5 3 - 4 4 = - 4 i → + 15 j → - 8 k → + 20 i → - 8 · j → = 16 · i → + 7 · j → - 5 · k →

we find that the length of the vector product is equal to a → × M 3 M 1 → = 16 2 + 7 2 + - 5 2 = 330.

We have all the data to use the formula for calculating the distance from a point for a straight line, so let’s apply it and get:

M 1 H 1 = a → × M 3 M 1 → a → = 330 30 = 11

Answer: 11 .

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The distance from a point to a line is the length of the perpendicular drawn from the point to the line. In descriptive geometry, it is determined graphically using the algorithm given below.

Algorithm

1. The straight line is moved to a position in which it will be parallel to any projection plane. For this purpose, methods of transforming orthogonal projections are used.
2. From a point a perpendicular is drawn to a line. This construction is based on the theorem about the projection of a right angle.
3. The length of a perpendicular is determined by transforming its projections or using the right triangle method.

The following figure shows a complex drawing of point M and line b, defined by segment CD. You need to find the distance between them.

According to our algorithm, the first thing to do is to move the line to a position parallel to the projection plane. It is important to understand that after the transformations have been carried out, the actual distance between the point and the line should not change. That is why it is convenient here to use the plane replacement method, which does not involve moving figures in space.

The results of the first stage of construction are shown below. The figure shows how an additional frontal plane P 4 is introduced parallel to b. In the new system (P 1, P 4), points C"" 1, D"" 1, M"" 1 are at the same distance from the X 1 axis as C"", D"", M"" from the axis X.

Carrying out the second part of the algorithm, from M"" 1 we lower the perpendicular M"" 1 N"" 1 to the straight line b"" 1, since the right angle MND between b and MN is projected onto the plane P 4 in full size. Using the communication line, we determine the position of point N" and carry out the projection M"N" of the segment MN.

At the final stage, you need to determine the size of the segment MN from its projections M"N" and M"" 1 N"" 1. To do this, we build a right triangle M"" 1 N"" 1 N 0, whose leg N"" 1 N 0 is equal to the difference (Y M 1 – Y N 1) of the distance of points M" and N" from the X 1 axis. The length of the hypotenuse M"" 1 N 0 of the triangle M"" 1 N"" 1 N 0 corresponds to the desired distance from M to b.

Second solution

• Parallel to CD, we introduce a new frontal plane P 4. It intersects P 1 along the X 1 axis, and X 1 ∥C"D". In accordance with the method of replacing planes, we determine the projections of points C"" 1, D"" 1 and M"" 1, as shown in the figure.
• Perpendicular to C"" 1 D"" 1 we build an additional horizontal plane P 5, onto which straight line b is projected to point C" 2 = b" 2.
• The distance between point M and line b is determined by the length of the segment M" 2 C" 2, indicated in red.

Similar tasks:

Formula for calculating the distance from a point to a line on a plane

If the equation of the line Ax + By + C = 0 is given, then the distance from the point M(M x , M y) to the line can be found using the following formula

Examples of problems for calculating the distance from a point to a line on a plane

Example 1.

Find the distance between the line 3x + 4y - 6 = 0 and the point M(-1, 3).

Solution. Let's substitute the coefficients of the line and the coordinates of the point into the formula

Answer: the distance from the point to the line is 0.6.

equation of a plane passing through points perpendicular to a vectorGeneral equation of a plane

A nonzero vector perpendicular to a given plane is called normal vector (or, in short, normal ) for this plane.

Let the following be given in coordinate space (in a rectangular coordinate system):

a) point ;

b) non-zero vector (Fig. 4.8, a).

You need to create an equation for a plane passing through a point perpendicular to the vector End of proof.

Let us now consider various types of equations of a straight line on a plane.

1) General equation of the planeP .

From the derivation of the equation it follows that at the same time A, B And C are not equal to 0 (explain why).

The point belongs to the plane P only if its coordinates satisfy the equation of the plane. Depending on the odds A, B, C And D plane P occupies one position or another:

- the plane passes through the origin of the coordinate system, - the plane does not pass through the origin of the coordinate system,

- plane parallel to the axis X,

X,

- plane parallel to the axis Y,

- the plane is not parallel to the axis Y,

- plane parallel to the axis Z,

- the plane is not parallel to the axis Z.

Prove these statements yourself.

Equation (6) is easily derived from equation (5). Indeed, let the point lie on the plane P. Then its coordinates satisfy the equation. Subtracting equation (7) from equation (5) and grouping the terms, we obtain equation (6). Let us now consider two vectors with coordinates respectively. From formula (6) it follows that their scalar product is equal to zero. Therefore, the vector is perpendicular to the vector. The beginning and end of the last vector are located, respectively, at points that belong to the plane P. Therefore, the vector is perpendicular to the plane P. Distance from point to plane P, whose general equation determined by the formula The proof of this formula is completely similar to the proof of the formula for the distance between a point and a line (see Fig. 2).
Rice. 2. To derive the formula for the distance between a plane and a straight line.

Indeed, the distance d between a straight line and a plane is equal

where is a point lying on the plane. From here, as in lecture No. 11, the above formula is obtained. Two planes are parallel if their normal vectors are parallel. From here we obtain the condition for parallelism of two planes - coefficients of general equations of planes. Two planes are perpendicular if their normal vectors are perpendicular, hence we obtain the condition for the perpendicularity of two planes if their general equations are known

Corner f between two planes is equal to the angle between their normal vectors (see Fig. 3) and can therefore be calculated using the formula
Determining the angle between planes.

(11)

Distance from a point to a plane and methods for finding it

Distance from point to plane– the length of the perpendicular dropped from a point onto this plane. There are at least two ways to find the distance from a point to a plane: geometric And algebraic.

With the geometric method You must first understand how the perpendicular from a point to a plane is located: maybe it lies in some convenient plane, is a height in some convenient (or not so convenient) triangle, or maybe this perpendicular is generally a height in some pyramid.

After this first and most complex stage, the problem breaks down into several specific planimetric problems (perhaps in different planes).

With the algebraic method in order to find the distance from a point to a plane, you need to enter a coordinate system, find the coordinates of the point and the equation of the plane, and then apply the formula for the distance from a point to the plane.

Oh-oh-oh-oh-oh... well, it’s tough, as if he was reading out a sentence to himself =) However, relaxation will help later, especially since today I bought the appropriate accessories. Therefore, let's proceed to the first section, I hope that by the end of the article I will maintain a cheerful mood.

The relative position of two straight lines

This is the case when the audience sings along in chorus. Two straight lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : Please remember the mathematical intersection sign, it will appear very often. The notation means that the line intersects with the line at point .

How to determine the relative position of two lines?

Let's start with the first case:

Two lines coincide if and only if their corresponding coefficients are proportional, that is, there is a number “lambda” such that the equalities are satisfied

Let's consider the straight lines and create three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by –1 (change signs), and all coefficients of the equation cut by 2, you get the same equation: .

The second case, when the lines are parallel:

Two lines are parallel if and only if their coefficients of the variables are proportional: , But.

As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables:

However, it is quite obvious that.

And the third case, when the lines intersect:

Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NO such value of “lambda” that the equalities are satisfied

So, for straight lines we will create a system:

From the first equation it follows that , and from the second equation: , which means the system is inconsistent(no solutions). Thus, the coefficients of the variables are not proportional.

Conclusion: lines intersect

In practical problems, you can use the solution scheme just discussed. By the way, it is very reminiscent of the algorithm for checking vectors for collinearity, which we looked at in class The concept of linear (in)dependence of vectors. Basis of vectors. But there is a more civilized packaging:

Example 1

Find out the relative position of the lines:

Solution based on the study of directing vectors of straight lines:

a) From the equations we find the direction vectors of the lines: .

, which means that the vectors are not collinear and the lines intersect.

Just in case, I’ll put a stone with signs at the crossroads:

The rest jump over the stone and follow further, straight to Kashchei the Immortal =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or coincident. There is no need to count the determinant here.

It is obvious that the coefficients of the unknowns are proportional, and .

Let's find out whether the equality is true:

Thus,

c) Find the direction vectors of the lines:

Let's calculate the determinant made up of the coordinates of these vectors:
, therefore, the direction vectors are collinear. The lines are either parallel or coincident.

The proportionality coefficient “lambda” is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

Now let's find out whether the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number in general satisfies it).

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the problem discussed verbally literally in a matter of seconds. In this regard, I don’t see any point in offering anything for an independent solution; it’s better to lay another important brick in the geometric foundation:

How to construct a line parallel to a given one?

For ignorance of this simplest task, the Nightingale the Robber severely punishes.

Example 2

The straight line is given by the equation. Write an equation for a parallel line that passes through the point.

Solution: Let's denote the unknown line by the letter . What does the condition say about her? The straight line passes through the point. And if the lines are parallel, then it is obvious that the direction vector of the straight line “tse” is also suitable for constructing the straight line “de”.

We take the direction vector out of the equation:

Answer:

The example geometry looks simple:

Analytical testing consists of the following steps:

1) We check that the lines have the same direction vector (if the equation of the line is not simplified properly, then the vectors will be collinear).

2) Check whether the point satisfies the resulting equation.

In most cases, analytical testing can be easily performed orally. Look at the two equations, and many of you will quickly determine the parallelism of the lines without any drawing.

Examples for independent solutions today will be creative. Because you will still have to compete with Baba Yaga, and she, you know, is a lover of all sorts of riddles.

Example 3

Write an equation for a line passing through a point parallel to the line if

There is a rational and not so rational way to solve it. The shortest way is at the end of the lesson.

We worked a little with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let’s consider a problem that is very familiar to you from the school curriculum:

How to find the point of intersection of two lines?

If straight intersect at point , then its coordinates are the solution systems of linear equations

How to find the point of intersection of lines? Solve the system.

Here you go geometric meaning of a system of two linear equations with two unknowns- these are two intersecting (most often) lines on a plane.

Example 4

Find the point of intersection of lines

Solution: There are two ways to solve - graphical and analytical.

The graphical method is to simply draw the given lines and find out the intersection point directly from the drawing:

Here is our point: . To check, you should substitute its coordinates into each equation of the line, they should fit both there and there. In other words, the coordinates of a point are a solution to the system. Essentially, we looked at a graphical solution systems of linear equations with two equations, two unknowns.

The graphical method is, of course, not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to create a correct and ACCURATE drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point using the analytical method. Let's solve the system:

To solve the system, the method of term-by-term addition of equations was used. To develop relevant skills, take a lesson How to solve a system of equations?

Answer:

The check is trivial - the coordinates of the intersection point must satisfy each equation of the system.

Example 5

Find the point of intersection of the lines if they intersect.

This is an example for you to solve on your own. It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Write down the equation of the straight line.
2) Write down the equation of the straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.

Full solution and answer at the end of the lesson:

Not even a pair of shoes were worn out before we got to the second section of the lesson:

Perpendicular lines. Distance from a point to a line.
Angle between straight lines

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to this one, and now the hut on chicken legs will turn 90 degrees:

How to construct a line perpendicular to a given one?

Example 6

The straight line is given by the equation. Write an equation perpendicular to the line passing through the point.

Solution: By condition it is known that . It would be nice to find the directing vector of the line. Since the lines are perpendicular, the trick is simple:

From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.

Let's compose the equation of a straight line using a point and a direction vector:

Answer:

Let's expand the geometric sketch:

Hmmm... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) We take out the direction vectors from the equations and with the help scalar product of vectors we come to the conclusion that the lines are indeed perpendicular: .

By the way, you can use normal vectors, it's even easier.

2) Check whether the point satisfies the resulting equation .

The test, again, is easy to perform orally.

Example 7

Find the point of intersection of perpendicular lines if the equation is known and period.

This is an example for you to solve on your own. There are several actions in the problem, so it is convenient to formulate the solution point by point.

Our exciting journey continues:

Distance from point to line

In front of us is a straight strip of the river and our task is to get to it by the shortest route. There are no obstacles, and the most optimal route will be to move along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

Distance in geometry is traditionally denoted by the Greek letter “rho”, for example: – the distance from the point “em” to the straight line “de”.

Distance from point to line expressed by the formula

Example 8

Find the distance from a point to a line

Solution: all you need to do is carefully substitute the numbers into the formula and carry out the calculations:

Answer:

Let's make the drawing:

The found distance from the point to the line is exactly the length of the red segment. If you draw up a drawing on checkered paper on a scale of 1 unit. = 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Let's consider another task based on the same drawing:

The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line . I suggest performing the steps yourself, but I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to the line.

2) Find the point of intersection of the lines: .

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the midpoint of a segment we find .

It would be a good idea to check that the distance is also 2.2 units.

Difficulties may arise in calculations here, but a microcalculator is a great help in the tower, allowing you to calculate ordinary fractions. I have advised you many times and will recommend you again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for you to decide on your own. I’ll give you a little hint: there are infinitely many ways to solve this. Debriefing at the end of the lesson, but it’s better to try to guess for yourself, I think your ingenuity was well developed.

Angle between two straight lines

Every corner is a jamb:


In geometry, the angle between two straight lines is taken to be the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered the angle between intersecting lines. And his “green” neighbor or oppositely oriented"raspberry" corner.

If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How are the angles different? Orientation. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .

Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing, for a negative angle, be sure to indicate its orientation with an arrow (clockwise).

How to find the angle between two straight lines? There are two working formulas:

Example 10

Find the angle between lines

Solution And Method one

Let's consider two straight lines defined by equations in general form:

If straight not perpendicular, That oriented The angle between them can be calculated using the formula:

Let us pay close attention to the denominator - this is exactly scalar product directing vectors of straight lines:

If , then the denominator of the formula becomes zero, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of straight lines in the formulation.

Based on the above, it is convenient to formalize the solution in two steps:

1) Let's calculate the scalar product of the direction vectors of the lines:
, which means the lines are not perpendicular.

2) Find the angle between straight lines using the formula:

Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):

Answer:

In your answer, we indicate the exact value, as well as an approximate value (preferably in both degrees and radians), calculated using a calculator.

Well, minus, minus, no big deal. Here is a geometric illustration:

It is not surprising that the angle turned out to be of a negative orientation, because in the problem statement the first number is a straight line and the “unscrewing” of the angle began precisely with it.

If you really want to get a positive angle, you need to swap the lines, that is, take the coefficients from the second equation , and take the coefficients from the first equation. In short, you need to start with a direct .

First level

Coordinates and vectors. The Comprehensive Guide (2019)

In this article, we will begin to discuss one “magic wand” that will allow you to reduce many geometry problems to simple arithmetic. This “stick” can make your life much easier, especially when you feel unsure of constructing spatial figures, sections, etc. All this requires a certain imagination and practical skills. The method that we will begin to consider here will allow you to almost completely abstract 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. Constructing a vector from two points
4. Vector length (distance between two points)​
5. Coordinates of the middle of the segment
6. Dot product of vectors
7. Angle between two vectors​

I think you've already guessed why the coordinate method is called that? That's right, it got this name because it operates not with geometric objects, but with their numerical characteristics (coordinates). And the transformation itself, which allows us 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 goal 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 on planimetry in Part B of the Unified State Exam). The next two sections on this topic are devoted to a discussion of methods for solving problems C2 (the problem of stereometry).

Where would it be logical to start discussing the coordinate method? Probably from the concept of a coordinate system. Remember when you first encountered her. It seems to me that in 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 it that way. For example, if, then, if, then, etc. What did you get in the end? And you received points with coordinates: and. Next, you drew a “cross” (coordinate system), chose a scale on it (how many cells you will have as a unit segment) and marked the points you obtained on it, which you then connected with a straight line; the resulting line is the graph of the function.

There are a few points here that should be explained to you in a little more detail:

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

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

3. They intersect at right angles, and the point of their intersection is called the origin. It is indicated by a letter.

4. In writing the coordinates of a point, for example, on the left in parentheses there is the coordinate of the point along the axis, and on the right, along the axis. In particular, it simply means that at the point

5. In order to specify any point on the coordinate axis, you need to indicate 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: mark two points. Let's connect these two points with a segment. And we’ll put the arrow as if we were drawing a segment from point to point: that is, we’ll make our segment directed!

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

So if we connect dot to 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 8th grade, remember?

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

Thus, since in a vector the point is the beginning and the point is 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 the point, and the end will be at the point. Then:

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

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

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

Examination:

Now solve a slightly more difficult problem:

A vector with a beginning at a point has a co-or-di-na-you. Find the abs-cis-su points.

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

I compiled the system based on the definition of what vector coordinates 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 can’t divide, but you can multiply in two ways, one of which we will discuss here a little later)

1. Vectors can be added to 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 by each other

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

A vector stretches or contracts 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 the amount of co-or-di-nat century-to-ra.

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

Answer:

Now solve the following problem yourself:

· Find the sum of vector coordinates

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 us denote the distance between them by. Let's make the following drawing for clarity:

What I've done? Firstly, I connected the points and, also, from the point I drew a line parallel to the axis, and from the point I drew a line parallel to the axis. Did they intersect at a point, forming a remarkable figure? What's so special about her? Yes, you and I know almost everything about the right triangle. Well, the Pythagorean theorem for sure. The required 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 by, respectively, 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 of the 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 points does not depend on the direction. Then:

From here we draw three conclusions:

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

For example, if, then the distance between and is equal to

Or let's go another way: find the coordinates of the vector

And find the length of the vector:

As you can see, it's the same thing!

Now practice a little yourself:

Task: find the distance between the indicated points:

We check:

Here are a couple more problems using the same formula, although they sound a little different:

1. Find the square of the length of the eyelid.

2. Find the square of the length of the eyelid

I think you dealt with them without difficulty? We check:

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

2. Find the coordinates of the vector

Then the square of its length is

Nothing complicated, right? Simple arithmetic, nothing more.

The following problems cannot be classified unambiguously; they are more about general erudition and the ability to draw simple pictures.

1. Find the sine of the angle from the cut, connecting the point, with the abscissa axis.

And

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

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

What's left for us to do? Find the hypotenuse. You can do this in two ways: using the Pythagorean theorem (the legs are known!) or using the formula for the distance between two points (in fact, the same thing as the first method!). I'll go the second way:

Answer:

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

Task 2. From the point the per-pen-di-ku-lyar is lowered onto the ab-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 a 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. In 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 will remind you:

So, in my drawing just above, have I already drawn one such perpendicular? Which axis is it on? 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 task 2, find the ordinate of a point symmetrical to the point relative to the abscissa axis.

I think it is intuitively clear to you what symmetry is? Many objects have it: many buildings, tables, airplanes, many geometric figures: ball, cylinder, square, rhombus, etc. Roughly speaking, symmetry can be understood as follows: a figure consists of two (or more) identical halves. This symmetry is called axial symmetry. What then is an axis? This is exactly the line along which the figure can, relatively speaking, be “cut” into equal 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 symmetrical about the axis. Then this axis is the axis of symmetry. This means that we need to mark a point such that the axis cuts the segment into two equal parts. Try to mark such a point yourself. Now compare with my solution:

Did it work out the same way for you? Fine! We are interested in the ordinate of the found point. It is equal

Answer:

Now tell me, after thinking for a few seconds, what will be the abscissa of a point symmetrical to point A relative to the ordinate? What is your answer? Correct answer: .

In general, the rule can be written like this:

A point symmetrical to a point relative to the abscissa axis has the coordinates:

A point symmetrical to a point relative to the ordinate axis has coordinates:

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

Answer:

Now parallelogram problem:

Task 5: The points appear ver-shi-na-mi pa-ral-le-lo-gram-ma. Find or-di-on-that point.

You can solve this problem in two ways: logic and the coordinate method. I'll use the coordinate method first, and then I'll tell you how you can solve it differently.

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

We lower the perpendicular connecting the point to the axis. I will denote the intersection point with a letter.

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

The length of a segment coincides exactly with its ordinate.

Answer: .

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

Solution progress:

1. Conduct

2. Find the coordinates of the point and length

3. Prove that.

Another one segment length problem:

The points appear on top of the triangle. Find the length of its midline, parallel.

Do you remember what the middle line of a triangle is? Then this task is elementary for you. If you don’t remember, I’ll remind you: the middle line of a triangle is the 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 middle line is half as large 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 problems for you, practice on them, they are very simple, but they help you get better at using the coordinate method!

1. The points are the top of the tra-pe-tions. Find the length of its midline.

2. Points and appearances ver-shi-na-mi pa-ral-le-lo-gram-ma. Find or-di-on-that point.

3. Find the length from the cut, connecting the point and

4. Find the area behind the colored figure on the co-ordi-nat plane.

5. A circle with a center in na-cha-le ko-or-di-nat passes through the point. Find her ra-di-us.

6. Find-di-te ra-di-us of the circle, describe-san-noy about the right-angle-no-ka, the tops of something have a co-or -di-na-you are so-responsible

Solutions:

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

Answer:

2. The easiest way to solve this problem is to note that (parallelogram rule). Calculating the coordinates of vectors is not difficult: . When adding vectors, the coordinates are added. Then it has coordinates. The point also has these coordinates, since the origin of the vector is the point with the coordinates. We are interested in the ordinate. She is equal.

Answer:

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

Answer:

4. Look at the picture and tell me which two figures the shaded area is “sandwiched” between? 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 a 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

We find the area of ​​the desired figure using the formula:

Answer:

5. If a 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). Let's 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 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 cope with everything? It wasn't very difficult to figure it out, was it? There is only one rule here - 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 midpoint 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 = the 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-ny from-cut, connect-the-point and

2. The points appear to be the top of the world. Find-di-te or-di-na-tu points per-re-se-che-niya of his dia-go-na-ley.

3. Find-di-te abs-cis-su center of the circle, describe-san-noy about the rectangular-no-ka, the tops of something have co-or-di-na-you so-responsibly-but.

Solutions:

1. The first problem is simply a classic. We proceed immediately to determine the middle of the segment. It has coordinates. The ordinate is equal.

Answer:

2. It is easy to see that this quadrilateral is a parallelogram (even a rhombus!). You can prove this yourself by calculating the lengths of the sides and comparing them with each other. What do I know about parallelograms? Its diagonals are divided in half by the point of intersection! Yeah! So what is the point of intersection of the diagonals? 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 does the center of the circle circumscribed about the rectangle coincide with? It coincides with the intersection point of its diagonals. What do you know about the diagonals of a rectangle? They are equal and the point of intersection divides them in half. The task was reduced to the previous one. Let's take, for example, the diagonal. Then if is the center of the circumcircle, then is the midpoint. I'm looking for coordinates: The abscissa is equal.

Answer:

Now practice a little on your own, I’ll just give the answers to each problem so you can test yourself.

1. Find-di-te ra-di-us of the circle, describe-san-noy about the tri-angle-no-ka, the tops of something have a co-or-di -no misters

2. Find-di-te or-di-on-that center of the circle, describe-san-noy about the triangle-no-ka, the tops of which have coordinates

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

4. Find-di-those or-di-on-that point of re-se-ce-tion of the axis and from-cut, connect-the-point and

Answers:

Was everything successful? I really hope for it! Now - the last push. Now be especially careful. The material that I will now explain is directly related not only to simple problems on the coordinate method from Part B, but is also found everywhere 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 ultimately introduced? Are you sure I haven't forgotten anything? Forgot! I forgot to explain what vector multiplication means.

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

The cross product is done quite cleverly. We will discuss how to do it and why it is needed in the next article. And in this one we will focus on the scalar product.

There are two ways that allow us to calculate it:

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

Dot product via coordinates

Find: - generally accepted notation for scalar product

The formula for calculation is as follows:

That is, the scalar product = the sum of the products of vector coordinates!

Example:

Find-di-te

Solution:

Let's find the coordinates of each of the vectors:

We calculate the scalar product using the formula:

Answer:

See, absolutely nothing complicated!

Well, now try it yourself:

· Find a scalar pro-iz-ve-de-nie of centuries and

Did you manage? Maybe you noticed a small catch? Let's check:

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

In addition to the coordinate one, 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 it is needed so that from the first and second formulas you and I can deduce how to find the angle between vectors!

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

Then if I substitute this data into the scalar product formula, I get:

But in other way:

So what did you and I get? We now have a formula that allows us to calculate the angle between two vectors! Sometimes it is also written like this for brevity:

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

1. Calculate the scalar product through coordinates
2. Find the lengths of the 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 and. Give the answer in grad-du-sah.

2. In 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 yourself! Agree? Then let's begin!

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

Then we look 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 will give just a very short solution:

2. has coordinates, has coordinates.

Let be the angle between the vectors and, then

Answer:

It should be noted that problems directly on vectors and the coordinate method in Part B of the exam 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 the foundation on the basis of which we will make quite clever constructions that we will need to solve complex problems.

COORDINATES AND VECTORS. AVERAGE LEVEL

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

1. Find vector coordinates
2. Find the length of a vector (alternatively: the distance between two points)
3. Add and 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 become familiar with at university. I just want to build a foundation that will allow you to solve problems in a single state. exam. We have dealt with the tasks of Part B. Now it’s time to move to a whole 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 is required 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 straight line and a plane
3. Find the angle between two straight 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 problem statement is a body of rotation (ball, cylinder, cone...)

Suitable figures for the coordinate method are:

1. Rectangular parallelepiped
2. Pyramid (triangular, quadrangular, hexagonal)

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

1. Finding cross-sectional areas
2. Calculation of volumes of bodies

However, it should immediately be noted that the 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 good at three-dimensional constructions (which can sometimes be quite intricate).

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

All of them are mutually perpendicular and intersect at one point, which we will call the origin of coordinates. As before, we will denote the abscissa axis, the ordinate axis - , and the introduced applicate axis - .

If previously 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, and the applicate. For example:

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

Sometimes the abscissa of a point is also called the projection of a point onto the abscissa axis, the ordinate - the projection of a point onto the ordinate axis, and the applicate - the projection of a 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 fair and have the same appearance. For a small detail. I think you've already guessed which one it is. 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 midpoint of the segment has coordinates

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

• Their scalar product is equal to:
• The cosine of the angle between the vectors is equal to:

However, space is not so simple. As you understand, adding one more coordinate introduces significant diversity into the spectrum of figures “living” in this space. And for further narration I will 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 “sheet” stuck into space. “Infinity” should be understood that the plane extends in all directions, that is, its area is equal to infinity. However, this “hands-on” explanation does not give the slightest idea about the structure of the plane. And it is she who will be interested in us.

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

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

Or its analogue in space:

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

You took this in 7th grade. In space, the equation of a line looks like this: let us be given two points with coordinates: , then the equation of the 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 line, but we need to pay attention to the very important concept of the direction vector of a 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 line and let be its direction vector. Then the equation of the line can be written in the following form:

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

Withdraw equation of a plane based on three given points is no longer so trivial, and the issue is not usually addressed in high school courses. But in vain! This technique is vital when we resort to the coordinate method to solve complex problems. However, I assume that you are eager 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 a technique that is usually studied in an analytical geometry course. 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 you and I argued? We said that if we have three points that do not lie on the same line, then the equation of the plane can be uniquely reconstructed from them. But how? I'll try to explain it to you.

Since the equation of the plane is:

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

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

However, we will not solve such a system, but will write out the mysterious 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 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 very often encounter these same 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 we mean the column number. For example, it means that this number is at the intersection of the second row and third column. Let's pose the following question: how exactly will we calculate such a determinant? That is, what specific number will we compare to it? For the third-order determinant there is a heuristic (visual) triangle rule, it looks like this:

1. The product of the elements of the main diagonal (from the upper left corner to the lower right) the product of the elements forming the first triangle “perpendicular” to the main diagonal the product of the elements forming the second triangle “perpendicular” to the main diagonal
2. The product of the elements of the secondary diagonal (from the upper right corner to the lower left) the product of the elements forming the first triangle “perpendicular” to the secondary diagonal the product of the elements forming 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 down in numbers, we get the following expression:

However, you don’t need to remember the method of calculation in this form; it’s enough to just keep in your head the triangles and the very idea of ​​what adds up 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 equal to

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

Second triangle, "perpendicular to the main diagonal: the product of the elements is equal to

Add up three numbers:

Terms that come with a minus

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

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

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

Add up three numbers:

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

Thus,

As you can see, there is nothing complicated or supernatural in calculating third-order determinants. It’s just important to remember about triangles and not make arithmetic errors. Now try to calculate it yourself:

We check:

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

Here are a couple more determinants, calculate their values ​​yourself and compare them with the answers:

Answers:

Well, did everything coincide? Great, then you can move on! If there are difficulties, then my advice is this: on the Internet there are a lot 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 begin to coincide. I am sure this moment will not take long to arrive!

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

All you need is to calculate its value directly (using the triangle method) and set the result to zero. Naturally, since these 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 the same straight line!

Let's illustrate this with a simple example:

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

We compile a determinant for these three points:

Let's simplify:

Now we calculate it directly using the triangle rule:

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

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 now discuss the solution:

Let's create 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 coincide? Again, if there are certain difficulties, then my advice is this: take three points from your head (with a high degree of probability they will not lie on the same straight line), build a plane based on them. And then you 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 not only dot product is defined for vectors. There is also a vector product, as well as a mixed product. And if the scalar product of two vectors is 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 module will be equal to the area of ​​a 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 vector product of vectors and, if their coordinates are given? The third-order determinant comes to our aid again. However, before I move on to the algorithm for calculating the vector product, I have to make a small digression.

This digression concerns basis vectors.

They are shown schematically 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 artwork

Now I can start introducing the cross product:

The vector product of two vectors is a vector, which 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 up a determinant:

And I calculate it:

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

Thus:

Now try it.

Ready? We check:

And traditionally two tasks for control:

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

Answers:

Mixed product of three vectors

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

Namely, let us be given 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 independent solutions:

Answers:

Selecting a coordinate system

Well, now we have all the necessary foundation of knowledge to solve complex stereometric geometry problems. However, before proceeding directly to 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.

Let me remind you that in this section we consider the following figures:

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

For a rectangular parallelepiped or cube, I recommend you the following construction:

That is, I will place the figure “in the corner”. The cube and parallelepiped 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 coordinates of the vertices are as follows:

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

Straight prism

The prism is a more harmful figure. It can be positioned in space in different ways. However, the following option seems to me the most acceptable:

Triangular prism:

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

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:

The situation is similar to a cube: we align two sides of the base with the coordinate axes, and align one of the vertices with the origin of coordinates. The only slight 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 to find the coordinates of the vertex.

Tetrahedron (triangular pyramid)

The situation is very similar to the one I gave for a 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 are divided into 2 categories: angle problems and distance problems. First, we will look at the problems of finding an angle. They are in turn divided into the following categories (as complexity increases):

Problems for finding angles

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

Let's look at these problems sequentially: let's start by finding the angle between two straight lines. Well, remember, haven’t you and I solved similar examples before? Do you remember, we already had something similar... We were looking for the angle between two vectors. Let me remind you, if two vectors are given: and, then the angle between them is found from the relation:

Now our goal is to find the angle between two straight lines. Let's look at the “flat picture”:

How many angles did we get when two straight lines intersected? Just a few things. True, only two of them are not equal, while the others are vertical to them (and therefore coincide with them). So which 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 two straight lines is equal. In order not to bother each time with finding the smallest of two angles, cunning mathematicians suggested using a modulus. Thus, the angle between two straight lines is determined by the formula:

You, as an attentive reader, should have had a question: where, exactly, 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 straight 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 straight line
3. We 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 accurately calculate the angle, we look for it
9. Otherwise we write through arc cosine

Well, now it’s time to move on to the problems: I will demonstrate the solution to the first two in detail, I will present the solution to another one in a brief form, and to the last two problems I will only give the answers; you must carry out all the calculations for them yourself.

Tasks:

1. In the right tet-ra-ed-re, find the angle between the height of the tet-ra-ed-ra and the middle side.

2. In the right-hand six-corner pi-ra-mi-de, the hundred os-no-va-niyas are equal, and the side edges are equal, find the angle between the lines and.

3. The lengths of all the edges of the right four-coal pi-ra-mi-dy are equal to each other. Find the angle between the straight lines and if from the cut - you are with the given pi-ra-mi-dy, the point is se-re-di-on its bo-co- second ribs

4. On the edge of the cube there is a point so that Find the angle between the straight lines and

5. Point - on the edges of the cube Find the angle between the straight lines and.

It is no coincidence that I arranged the tasks in this order. While you have not yet begun to navigate the coordinate method, I will analyze the most “problematic” figures myself, and I will leave you to deal with the simplest cube! Gradually you will 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, all its faces (including the base) are regular triangles. Since we are not given the length of the side, I can take it to be equal. I think you understand that the angle will not actually depend on how much our tetrahedron is “stretched”?. I will also draw the height and median in the tetrahedron. Along the way, I will draw its base (it will also be useful to us).

I need to find the angle between and. What do we know? We only know the coordinate of the point. This means that we need to find the coordinates of the points. Now we think: a point is the point of intersection of the altitudes (or bisectors or medians) of the triangle. And a point is a raised point. The point is the middle of the segment. Then we finally need to find: the coordinates of the points: .

Let's start with the simplest thing: the coordinates of a point. Look at the figure: It is clear that the applicate of a point is equal to zero (the point lies on the plane). Its ordinate is equal (since it is the median). It is more difficult to find its abscissa. However, this is easily done based on the Pythagorean theorem: Consider a triangle. Its hypotenuse is equal, and one of its 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 quite trivially if you remember that the heights of an equilateral triangle by the point of intersection are divided in proportion, counting from the top. Since: , then the required 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 applicate 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 sought for reasons that I have highlighted in bold:

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

Thus,

Answer:

You shouldn’t be scared by such “scary” answers: for C2 problems this is common practice. I would rather be surprised by the “beautiful” answer in this part. Also, as you noticed, I practically did not resort to anything other than the Pythagorean theorem and the property of altitudes 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. Let us depict 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 comes down to finding the coordinates of the points: . We will find the coordinates of the last three using a small drawing, and we will find the coordinate of the vertex through the coordinate of the point. There's a lot of work to do, but we need to get started!

a) Coordinate: it is clear that its applicate and ordinate are equal to zero. Let's find the abscissa. To do this, consider a right triangle. Alas, in it we only know the hypotenuse, which is equal. We will try to find the leg (for it is clear that double the length of the leg will give us the abscissa of the point). How can we look for it? 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 angle. 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 equal to 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 equal to degrees. Then:

Then where from.

Thus, 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 dots and designate the point of intersection of the line as, say, . (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 let's find the coordinates of the point. Consider the 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 the applica. Since, then. Consider a right triangle. According to the conditions of the problem, a side edge. This is the hypotenuse of my triangle. Then the height of the pyramid is a leg.

Then the point has coordinates:

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

We are looking for the angle between these vectors:

Answer:

Again, in solving this problem I did not use any sophisticated techniques other than 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 there is a square, and the side faces are regular triangles. Let us draw such a pyramid, as well as its base on a plane, noting 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 search for the coordinates of the points. You will need to “decipher” them:

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

c) I will find the length of the segment using the Pythagorean theorem in a triangle. I can find it using 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 the angle:

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

Finding the angle between a straight line and a plane

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

1. Using three points we construct an equation of the plane
   ,
   using a third order determinant.
2. Using two points, we look 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 angles between two straight lines. The structure on the right side is simply the same, and on the left we are now looking for the sine, not the cosine as before. Well, one nasty action was added - searching for the equation of the plane.

Let's not procrastinate solution examples:

1. The main-but-va-ni-em direct prism-we are an equal-to-poor triangle. Find the angle between the straight line and the plane

2. In a rectangular par-ral-le-le-pi-pe-de from the West Find the angle between the straight line and the plane

3. In a right six-corner 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-no-va-ni-em of the known ribs Find a corner, ob-ra-zo-van -flat in base and straight, passing through the gray ribs and

5. The lengths of all the edges of a right quadrangular pi-ra-mi-dy with a vertex are equal to each other. Find the angle between the straight line and the plane if the point is on the side of the pi-ra-mi-dy’s edge.

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

Solutions:

1. Let us depict a prism, as well as its base. Let's combine it with the coordinate system and note all the data that is given in the problem statement:

I apologize for some non-compliance with the proportions, but for solving the problem this is, in fact, not so important. The plane is simply 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 be shown directly:

Let's choose arbitrary three points on this plane: for example, .

Let's create the equation of the plane:

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

Or simply

Thus,

To solve the example, I need to find the coordinates of the direction vector of the straight line. Since the point coincides with the origin of coordinates, 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 the height (also known as the median and bisector) from the vertex. Since, the ordinate of the point is equal to. In order to find the abscissa of this point, we need to calculate the length of the segment. According to the Pythagorean theorem we have:

Then the point has coordinates:

A dot is a "raised" dot:

Then the vector coordinates are:

Answer:

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

2. 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 guiding 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 look for the desired angle:

Answer:

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

Here it’s even problematic to draw a plane, not to mention solving this problem, but the coordinate method doesn’t care! Its versatility is its main advantage!

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

1) . Find out the coordinates for the last two points yourself. You'll need to solve the hexagonal pyramid problem for this!

2) We construct the equation of the plane:

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

3) 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. I will only give answers to the last two problems:

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 certain formulas. We still have to consider one more class of problems for calculating angles, namely:

Calculating angles between two planes

The solution algorithm will be as follows:

1. Using three points we look for the equation of the first plane:
2. Using the other three points we look for the equation of the second plane:
3. We apply the formula:

As you can see, the formula is very similar to the two previous ones, with the help of which we looked for angles between straight lines and between a straight line and a plane. So it won’t be difficult for you to remember this one. Let's move on to the analysis of the tasks:

1. The side of the base 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 axis of the prism.

2. In the right four-corner pi-ra-mi-de, all the edges of which are equal, find the sine of the angle between the plane and the plane bone, passing through the point per-pen-di-ku-lyar-but straight.

3. In a regular four-corner prism, the sides of the base are equal, and the side edges are equal. There is a point on the edge from-me-che-on so that. Find the angle between the planes and

4. In a right quadrangular prism, the sides of the base are equal, and the side edges are equal. There is a point on the edge from the point so that Find the angle between the planes and.

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

Problem solutions:

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

We need to find the equations of two planes: The equation of the base is trivial: you can compose the corresponding determinant using three points, but I will compose the equation right away:

Now let’s find the equation Point has coordinates Point - Since is the median and altitude of the triangle, it is easily found using the Pythagorean theorem in the triangle. Then the point has coordinates: Let's 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 this is, passing perpendicularly through the point. Well, the main thing is, what is it? The main thing is attentiveness! In fact, the line is perpendicular. The straight line is also perpendicular. Then the plane passing through these two lines will be perpendicular to the line, and, by the way, pass through the point. This plane also passes through the top of the pyramid. Then the desired plane - And the plane has already been given to us. We are looking for the coordinates of the points.

We find the coordinate of the point through the point. From the small picture it is easy to deduce that the coordinates of the point will be as follows: What now remains to be found to find the coordinates of the top of the pyramid? You also 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. Without difficulty you will receive:

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

Now let's find the equation of the plane:

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

We calculate the determinant:

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

Now let's calculate the angle:

We need to find the sine:

Answer:

3. Tricky question: what do you think a rectangular prism is? This is just a parallelepiped that you know well! Let's make a drawing right away! You don’t even have to depict the base separately; it’s of little use here:

The plane, as we noted earlier, is written in the form of an equation:

Now let's create a plane

We immediately create 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 little 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 calculation problems. Namely, we will consider the following cases:

1. Calculation of the distance between intersecting lines.

I have ordered these assignments in order of increasing difficulty. It turns out to be easiest to find distance from point to plane, and the most difficult thing is to find distance between crossing lines. Although, of course, nothing is impossible! Let's not procrastinate and immediately proceed to consider 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 receive all the necessary data, we apply the formula:

You should already know how we construct the equation of a plane from the previous problems that I discussed in the last part. Let's get straight to the tasks. The scheme is as follows: 1, 2 - I help you decide, and in some detail, 3, 4 - only the answer, you carry out the solution yourself and compare. Let's start!

Tasks:

1. Given a cube. The length of the edge of the cube is equal. Find the distance from the se-re-di-na from the cut to the plane

2. Given the right four-coal pi-ra-mi-yes, the side of the side is equal to the base. Find the distance from the point to the plane where - se-re-di-on the edges.

3. In the right triangular pi-ra-mi-de with the os-no-va-ni-em, the side edge is equal, and the hundred-ro-on the os-no-va- nia is equal. Find the distance from the top to the plane.

4. In a right hexagonal prism, all edges are equal. Find the distance from a point to a plane.

Solutions:

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

.

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

Now we compose the equation of the plane using 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 with its paw will not prevent us from solving this problem with ease!

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

Since the coordinates of the point, then

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

Without any problems, we can find the coordinates of two more points on the plane. We create an equation for 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: , we calculate the distance:

Answer (very rare!):

Well, did you figure it out? It seems to me that everything here is just as technical as in the examples that we looked at 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 straight line to a plane

In fact, there is nothing new here. How can a straight line and a plane be positioned relative to each other? They have only one possibility: to intersect, or a straight line is parallel to the plane. What do you think is the distance from a straight line to the plane with which this straight line intersects? It seems to me that it is clear here that such a distance is equal to zero. Not an interesting 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:

Thus:

This means that my task has been reduced to the previous one: we are looking for the coordinates of any point on a straight line, looking for the equation of the plane, and calculating the distance from the point to the plane. In fact, such tasks are extremely rare in the Unified State Examination. I managed to find only one problem, and the data in it were 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 do we need?

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

2. Coordinates of any point lying on a line

3. Coordinates of the directing vector of the straight line

What formula do we use?

What the denominator of this fraction means should be clear to you: this is the length of the directing vector of the straight line. This is a very tricky numerator! The expression means the modulus (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, we will need it very much 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. Construct a vector

4. Construct a directing vector of a straight line

5. Calculate the vector product

6. We look for the length of the resulting vector:

7. Calculate the distance:

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

1. Given a right triangular pi-ra-mi-da with a top. The hundred-ro-on the basis of the pi-ra-mi-dy is equal, you are equal. Find the distance from the gray edge to the straight line, where the points and are the gray edges and from veterinary.

2. The lengths of the ribs and the straight-angle-no-go par-ral-le-le-pi-pe-da are equal accordingly and Find the distance from the top to the straight line

3. In a right hexagonal prism, all edges are equal, find the 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 to do! First, I would 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. Length of the vector product

8. Distance from to

Well, we have a lot of work ahead of us! Let's get to it with our sleeves rolled up!

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 its ordinate is equal to its abscissa is equal to the length of the segment. Since is the height of an equilateral triangle, it is divided in the ratio, counting from the vertex, from here. Finally, we got the coordinates:

Point coordinates

2. - middle of the segment

3. - middle of the segment

Midpoint of the segment

4.Coordinates

Vector coordinates

5. Calculate the vector product:

6. Vector length: the easiest way to replace is that the segment is the midline of the triangle, which means it is equal to half the base. So.

7. Calculate the length of the vector product:

8. Finally, we find the distance:

Ugh, that's it! I’ll tell you honestly: solving this problem using traditional methods (through construction) would be much faster. But here I reduced everything to a ready-made algorithm! I think the solution algorithm is clear to you? Therefore, I will ask you to solve the remaining two problems yourself. Let's compare the answers?

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

Finally, consider the last class of problems:

Calculating the distance between intersecting 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 line:

How do we find the distance between lines?

The formula is as follows:

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

I'll remind you that

Then the formula for the distance can be rewritten as:

This is a determinant divided by a determinant! Although, to be honest, I have no time for jokes here! This formula is, in fact, very cumbersome and leads to quite complex calculations. If I were you, I would resort to it only as a last resort!

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

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

2. Given a right triangular prism, all the edges of the base are equal to the section passing through the body rib and se-re-di-well ribs are a square. Find the distance between the straight lines and

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

1. I draw a prism and mark straight lines and

Coordinates of point C: 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 calculate the vector product between 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 calculate its length:

Answer:

Now try to complete the second task carefully. 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.
A vector is denoted by or.

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

Vector coordinates:

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

Sum of vectors: .

Product of vectors:

Dot product of vectors: