Job type: 14

Condition

In a regular triangular pyramid DABC with base ABC, the side of the base is equal to 6\sqrt(3), and the height of the pyramid is 8 . Points M , N and K are respectively marked on edges AB , AC and AD such that AM=AN=\frac(3\sqrt(3))(2) and AK=\frac(5)(2).

a) Prove that the planes MNK and DBC are parallel.

b) Find the distance from point K to plane DBC.

Show Solution

Decision

a) Planes MNK and DBC are parallel if two intersecting lines in one plane are respectively parallel to two intersecting lines in the other plane. Let's prove it. Consider lines MN and KM of the plane MNK and lines BC and DB of the plane DBC.

In the triangle AOD : \angle AOD = 90^\circ and by the Pythagorean theorem AD=\sqrt(DO^2 +AO^2).

Find AO using \bigtriangleup ABC is correct.

AO=\frac(2)(3)AO_1, where AO_1 is the height of \bigtriangleup ABC, AO_1 = \frac(a\sqrt(3))(2), where a is the side of \bigtriangleup ABC.

AO_1 = \frac(6\sqrt(3) \cdot \sqrt(3))(2)=9, then AO=6, AD=\sqrt(8^2 + 6^2)=10.

1. Since \frac(AK)(AD)=\frac(5)(2) : 10=\frac(1)(4), \frac(AM)(AB)=\frac(3\sqrt(3))(2) : 6\sqrt(3)=\frac(1)(4) and \angle DAB is general, then \bigtriangleup AKM \sim ADB.

It follows from the similarity that \angle AKM = \angle ADB. These are the corresponding angles for the lines KM and BD and the secant AD . So KM \parallel BD.

2. Since \frac(AN)(AC)=\frac(3 \sqrt(3))(2 \cdot 6 \sqrt(3))=\frac(1)(4), \frac(AM)(AB)=\frac(1)(4) and \angle CAB is common, then \bigtriangleup ANM \sim \bigtriangleup ACB.

It follows from the similarity that \angle ANM = \angle ACB. These angles are corresponding to lines MN and BC and secant AC . So MN \parallel BC.

Conclusion: since two intersecting lines KM and MN of the plane MNK are respectively parallel to two intersecting lines BD and BC of the plane DBC , then these planes are parallel - MNK \parallel DBC.

b) Let's find the distance from the point K to the plane BDC.

Since the MNK plane is parallel to the DBC plane, the distance from the point K to the DBC plane is equal to the distance from the point O_2 to the DBC plane and it is equal to the length of the segment O_2 H. Let's prove it.

BC \perp AO_1 and BC \perp DO_1 (as the heights of triangles ABC and DBC ), so BC is perpendicular to the plane ADO_1, and then BC is perpendicular to any line of this plane, for example, O_2 H. By construction O_2H\perp DO_1, then O_2H is perpendicular two intersecting straight lines of the BCD plane, and then the segment O_2 H is perpendicular to the BCD plane and is equal to the distance from O_2 to the BCD plane.

In a triangle O_2HO_1:O_2H=O_(2)O_(1)\sin\angle HO_(1)O_(2).

O_(2)O_(1)=AO_(1)-AO_(2).\, \frac(AO_2)(AO_1)=\frac(1)(4), AO_(2)=\frac(AO_1)(4)=\frac(9)(4).

O_(2)O_(1)=9-\frac(9)(4)=\frac(27)(4).

\sin \angle DO_(1)A= \frac(DO)(DO_(1))= \frac(8)(\sqrt(64+3^2))= \frac(8)(\sqrt(73)).

O_2H=\frac(27)(4) \cdot \frac(8)(\sqrt(73))=\frac(54)(\sqrt(73)).

Answer

\frac(54)(\sqrt(73))

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 14
Topic: Distance from a point to a plane

Condition

ABCDA_1B_1C_1D_1 is a regular quadrangular prism.

a) Prove that the plane is BB_1D_1 \perp AD_1C .

b) Knowing AB = 5 and AA_1 = 6, find the distance from point B_1 to plane AD_1C.

Show Solution

Decision

a) Since this prism is regular, then BB_1 \perp ABCD , hence BB_1 \perp AC . Since ABCD is a square, then AC \perp BD . So AC \perp BD and AC \perp BB_1 . Since the lines BD and BB_1 intersect, then, according to the sign of perpendicularity of a line and a plane, AC \perp BB_1D_1D . Now on the basis of the perpendicularity of the planes AD_1C \perp BB_1D_1 .

b) Denote by O the intersection point of the diagonals AC and BD of the square ABCD. Planes AD_1C and BB_1D_1 intersect along the straight line OD_1 . Let B_1H be a perpendicular drawn in the plane BB_1D_1 to the line OD_1 . Then B_1H \perp AD_1C . Let E=OD_1 \cap BB_1 . For similar triangles D_1B_1E and OBE (the equality of the corresponding angles follows from the condition BO \parallel B_1D_1 ) we have \frac(B_1E)(BE)=\frac(B_1D_1)(BO)=\frac(2)1.

So B_1E=2BE=2 \cdot 6=12. Since B_1D_1=5\sqrt(2) , then the hypotenuse D_1E= \sqrt(B_1E^(2)+B_1D_1^(2))= \sqrt(12^(2)+(5\sqrt(2))^(2))= \sqrt(194). Next, we use the area method in triangle D_1B_1E to calculate the height of B_1H lowered to the hypotenuse D_1E :

S_(D_1B_1E)=\frac1(2)B_1E \cdot B_1D_1=\frac1(2)D_1E \cdot B_1H; 12 \cdot 5\sqrt(2)=\sqrt(194) \cdot B_1H;

B_1H=\frac(60\sqrt(2))(\sqrt(194))=\frac(60)(\sqrt(97))=\frac(60\sqrt(97))(97).

Answer

\frac(60\sqrt(97))(97)

Source: "Mathematics. Preparation for the exam-2016. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 14
Topic: Distance from a point to a plane

Condition

ABCDA_1B_1C_1D_1 is a rectangular box. Edges AB=24, BC=7, BB_(1)=4 .

a) Prove that the distances from points B and D to the plane ACD_(1) are the same.

b) Find this distance.

Show Solution

Decision

a) Consider a triangular pyramid D_1ACD .

In this pyramid, the distance from point D to the base plane ACD_1-DH is equal to the height of the pyramid drawn from point D to the base ACD_1 .

V_(D_1ABC)=\frac1(3)S_(ACD_1) \cdot DH, from this equality we get

DH=\frac(3V_(D_1ACD))(S_(ACD_1)).

Consider the pyramid D_1ABC . The distance from point B to the plane ACD_1 is equal to the height dropped from the top of B to the bottom of ACD_1 . Let's denote this distance BK . Then V_(D_1ABC)=\frac1(3)S_(ACD_1) \cdot BK, from this we get BK=\frac(3V_(D_1ABC))(S_(ACD_1)).\: But V_(D_1ACD) = V_(D_1ABC) , since if we consider the bases in the pyramids ADC and ABC , then the height D_1D is total and S_(ADC)=S_(ABC) ( \bigtriangleup ADC=\bigtriangleup ABC on two legs). So BK=DH .

b) Find the volume of the pyramid D_1ACD .

Height D_1D=4 .

S_(ACD)=\frac1(2)AD \cdot DC=\frac1(2) \cdot24 \cdot 7=84.

V=\frac1(3)S_(ACD) \cdot D_1D=\frac1(3) \cdot84 \cdot4=112.

The face area of ​​ACD_1 is equal to \frac1(2)AC \cdot D_1P.

AD_1= \sqrt(AD^(2)+DD_1^(2))= \sqrt(7^(2)+4^(2))= \sqrt(65), \:AC= \sqrt(AB^(2)+BC^(2))= \sqrt(24^(2)+7^(2))= 25

Knowing that the leg of a right triangle is the mean proportional for the hypotenuse and the segment of the hypotenuse enclosed between the leg and the height drawn from the vertex of the right angle, in triangle ADC we have AD^(2)=AC \cdot AP, \: AP=\frac(AD^(2))(AC)=\frac(7^(2))(25)=\frac(49)(25).

In a right triangle AD_1P by the Pythagorean theorem D_1P^(2)= AD_1^(2)-AP^(2)= 65-\left (\frac(49)(25) \right)^(2)= \frac(38\:224)(25^(2)), D_1P=\frac(4\sqrt(2\:389))(25).

S_(ACD_1)=\frac1(2) \cdot25 \cdot\frac(4\sqrt(2\:389))(25)=2\sqrt(2\:389).

DH=\frac(3V)(S_(ACD_1))=\frac(3 \cdot112)(2\sqrt(2\:389))=\frac(168)(\sqrt(2\:389)).

Any plane in the Cartesian coordinate system can be defined by the equation `Ax + By + Cz + D = 0`, where at least one of the numbers `A`, `B`, `C` is non-zero. Let the point `M (x_0;y_0;z_0)` be given, find the distance from it to the plane `Ax + By + Cz + D = 0`.

Let the line passing through the point `M` perpendicular to the `alpha` plane, intersects it at the point `K` with coordinates `(x; y; z)`. Vector `vec(MK)` perpendicular to the `alpha` plane, as is the vector `vecn` `(A;B;C)`, i.e. the vectors `vec(MK)` and `vecn` collinear, `vec(MK)=λvecn`.

Since `(x-x_0;y-y_0;z-z-0)` and `vecn(A,B,C)`, then `x-x_0=lambdaA`, `y-y_0=lambdaB`, `z-z_0=lambdaC`.

Dot `K` lies in the `alpha` plane (Fig. 6), its coordinates satisfy the equation of the plane. Substituting `x=x_0+lambdaA`, `y=y_0+lambdaB`, `z=z_0+lambdaC` into the equation `Ax+By+Cz+D=0`, we get

`A(x_0+lambdaA)+(B(y_0+lambdaB)+C(z_0+lambdaC)+D=0`,

whence `lambda=-(Ax_0+By_0+Cz_0+D)/(A^2+B^2+C^2)`.

Find the length of the vector `vec(MK)`, which is equal to the distance from the point `M(x_0;y_0;z_0)` to the plane `Ax + By + Cz + D` `|vec(MK)|=|lambdavecn|=|lambda|*sqrt(A^2+B^2+C^2)`.

So, the distance `h` from the point `M(x_0;y_0;z_0)` to the plane `Ax + By + Cz + D = 0` is

`h=(|Ax_0+By_0+Cz_0+D|)/(sqrt(A^2+B^2+C^2))`.

With the geometric method of finding the distance from the point `A` to the plane `alpha`, find the base of the perpendicular `A A^"`, lowered from the point `A` to the `alpha` plane. If the point `A^"` is outside the section of the `alpha` plane specified in the problem, then a line `c` is drawn through the point `A`, parallel to the plane `alpha`, and a more convenient point `C` is chosen on it, the orthogonal projection of which is `C^"` belongs to the given section of the `alpha` plane. Segment length `C C^"`will be equal to the desired distance from point `A`up to the `alpha` plane.

In a regular hexagonal prism `A...F_1`, all edges of which are equal to `1`, find the distance from the point `B` to the plane `AF F_1`.

Let `O` be the center of the lower base of the prism (Fig. 7). Line `BO` is parallel to line `AF` and, therefore, the distance from point `B` to plane `AF F_1` is equal to the distance `OH` from point `O` to plane `AF F_1`. In the triangle `AOF` we have `AO=OF=AF=1`. The height `OH` of this triangle is `(sqrt3)/2`. Therefore, the required distance is equal to `(sqrt3)/2`.

Let's show another way (auxiliary volume method) finding the distance from a point to a plane. It is known that the volume of the pyramid `V` , its base area `S`and height length `h`linked by the formula `h=(3V)/S`. But the length of the height of the pyramid is nothing but the distance from its top to the base plane. Therefore, to calculate the distance from a point to a plane, it is enough to find the volume and area of ​​the base of some pyramid with a vertex at this point and with a base lying in a given plane.

A regular prism `A...D_1` is given, in which `AB=a`, `A A_1=2a`. Find the distance from the point of intersection of the diagonals of the base `A_1B_1C_1D_1` to the plane `BDC_1`.

Consider the tetrahedron `O_1DBC_1` (Fig. 8). The desired distance `h` is the length of the height of this tetrahedron, lowered from the point `O_1` to the face plane `BDC_1` . To find it, it is enough to know the volume `V`tetrahedron `O_1DBC_1` and area triangle `DBC_1`. Let's calculate them. Note that the line `O_1C_1` perpendicular to plane `O_1DB`, since it is perpendicular to `BD` and `B B_1` . Hence, the volume of the tetrahedron `O_1DBC_1` equals

Determining the distance between: 1 - point and plane; 2 - straight and flat; 3 - planes; 4 - crossing lines are considered jointly, since the solution algorithm for all these problems is essentially the same and consists of geometric constructions that must be performed to determine the distance between the given point A and the plane α. If there is any difference, then it consists only in the fact that in cases 2 and 3, before starting to solve the problem, one should mark an arbitrary point A on the line m (case 2) or the plane β (case 3). distances between skew lines, we preliminarily enclose them in parallel planes α and β with subsequent determination of the distance between these planes.

Let's consider each of the noted cases of problem solving.

1. Determining the distance between a point and a plane.

The distance from a point to a plane is determined by the length of the perpendicular segment dropped from the point to the plane.

Therefore, the solution of this problem consists of sequential execution of the following graphical operations:

1) from point A we lower the perpendicular to the plane α (Fig. 269);

2) find the point M of intersection of this perpendicular with the plane M = a ∩ α;

3) determine the length of the segment.

If the plane α is in general position, then in order to drop a perpendicular onto this plane, it is necessary to first determine the direction of the projections of the horizontal and frontal of this plane. Finding the meeting point of this perpendicular with the plane also requires additional geometric constructions.

The solution of the problem is simplified if the plane α occupies a particular position relative to the projection planes. In this case, both the projection of the perpendicular and the finding of the point of its meeting with the plane are carried out without any additional auxiliary constructions.

EXAMPLE 1. Determine the distance from point A to the frontally projecting plane α (Fig. 270).

DECISION. Through A "we draw a horizontal projection of the perpendicular l" ⊥ h 0α, and through A "- its frontal projection l" ⊥ f 0α. We mark the point M" = l" ∩ f 0α . Since AM || π 2 , then [А" М"] == |AM| = d.

From the example considered, it can be seen how simply the problem is solved when the plane occupies a projecting position. Therefore, if a generic plane is specified in the initial data, then before proceeding with the solution, the plane should be transferred to a position perpendicular to any projection plane.

EXAMPLE 2. Determine the distance from point K to the plane given by ΔАВС (Fig. 271).

1. We transfer the plane ΔАВС to the projecting position *. To do this, we pass from the system xπ 2 / π 1 to x 1 π 3 / π 1: the direction of the new axis x 1 is chosen perpendicular to the horizontal projection of the horizontal plane of the triangle.

2. We project ΔАВС onto a new plane π 3 (the ΔАВС plane is projected onto π 3, in [С" 1 В" 1 ]).

3. We project the point K (K "→ K" 1) onto the same plane.

4. Through the point K "1 we draw (K" 1 M "1) ⊥ the segment [C" 1 B "1]. The desired distance d \u003d | K "1 M" 1 |.

The solution of the problem is simplified if the plane is given by traces, since there is no need to carry out projections of level lines.

EXAMPLE 3. Determine the distance from point K to the plane α, given by traces (Fig. 272).

* The most rational way of transferring the plane of the triangle to the projecting position is the method of replacing the projection planes, since in this case it is enough to build only one auxiliary projection.

DECISION. We replace the plane π 1 with the plane π 3, for this we draw a new axis x 1 ⊥ f 0α. On h 0α we mark an arbitrary point 1 "and determine its new horizontal projection on the plane π 3 (1" 1). Through the points X α 1 (X α 1 \u003d h 0α 1 ∩ x 1) and 1 "1 we draw h 0α 1. We define a new horizontal projection of the point K → K" 1. From the point K "1 we lower the perpendicular to h 0α 1 and mark the point of its intersection with h 0α 1 - M" 1. The length of the segment K "1 M" 1 will indicate the required distance.

2. Determining the distance between a straight line and a plane.

The distance between the straight line and the plane is determined by the length of the segment of the perpendicular dropped from an arbitrary point of the straight line to the plane (see Fig. 248).

Therefore, the solution of the problem of determining the distance between the line m and the plane α is no different from the examples considered in paragraph 1 for determining the distance between a point and a plane (see Fig. 270 ... 272). Any point belonging to the line m can be taken as a point.

3. Determination of the distance between the planes.

The distance between the planes is determined by the value of the segment of the perpendicular dropped from a point taken on one plane to another plane.

It follows from this definition that the algorithm for solving the problem of finding the distance between the planes α and β differs from the similar algorithm for solving the problem of determining the distance between the line m and the plane α only in that the line m must belong to the plane α, i.e., in order to determine the distance between the planes α and β follows:

1) take a line m in the plane α;

2) select an arbitrary point A on the line m;

3) from point A, lower the perpendicular l to the plane β;

4) determine the point M - the meeting point of the perpendicular l with the plane β;

5) determine the value of the segment.

In practice, it is advisable to use a different solution algorithm, which will differ from the one given only in that, before proceeding with the first step, the planes should be transferred to the projecting position.

The inclusion of this additional operation in the algorithm simplifies the implementation of all other points without exception, which ultimately leads to a simpler solution.

EXAMPLE 1. Determine the distance between the planes α and β (Fig. 273).

DECISION. We pass from the system xπ 2 /π 1 to x 1 π 1 /π 3 . With respect to the new plane π 3, the planes α and β occupy a projecting position, so the distance between the new frontal traces f 0α 1 and f 0β 1 is the required one.

In engineering practice, it is often necessary to solve the problem of constructing a plane parallel to a given one and at a given distance from it. Example 2 below illustrates the solution to such a problem.

EXAMPLE 2. It is required to construct projections of the plane β, parallel to the given plane α (m || n), if it is known that the distance between them is equal to d (Fig. 274).

1. In the plane α we draw an arbitrary horizontal h (1, 3) and a frontal f (1,2).

2. From point 1 we restore the perpendicular l to the plane α(l" ⊥ h", l" ⊥ f").

3. Mark an arbitrary point A on the perpendicular l.

4. Determine the length of the segment - (the position indicates the metrically undistorted direction of the straight line l on the diagram).

5. Set aside on a straight line (1 "A 0) from point 1" segment = d.

6. We mark on the projections l "and l" points B "and B", corresponding to the point B 0.

7. Draw a plane β through point B (h 1 ∩ f 1). So that β || α, it is necessary to observe the condition h 1 || h and f 1 || f.

4. Determining the distance between skew lines.

The distance between the skew lines is determined by the length of the perpendicular enclosed between the parallel planes to which the skew lines belong.

In order to draw mutually parallel planes α and β through intersecting lines m and f, it suffices to draw line p parallel to line f through point A (A ∈ m), and through point B (B ∈ f) - line k parallel to line m . Intersecting lines m and p, f and k define mutually parallel planes α and β (see Fig. 248, e). The distance between the planes α and β is equal to the desired distance between the skew lines m and f.

Another way can be proposed for determining the distance between skew lines, which consists in the fact that with the help of some method of transforming orthogonal projections, one of the skew lines is transferred to the projecting position. In this case, one projection of the line degenerates into a point. The distance between the new projections of skew lines (point A" 2 and segment C" 2 D" 2) is the required one.

On fig. 275 shows the solution to the problem of determining the distance between intersecting lines a and b, given segments [AB] and [CD]. The solution is carried out in the following sequence:

1. Transfer one of the intersecting lines (a) to a position parallel to the plane π 3; to do this, they move from the system of projection planes xπ 2 / π 1 to a new x 1 π 1 / π 3, the x 1 axis is parallel to the horizontal projection of the straight line a. Determine a" 1 [A" 1 B" 1 ] and b" 1 .

2. By replacing the plane π 1 with the plane π 4, a straight line is translated

and in position a "2, perpendicular to the plane π 4 (the new axis x 2 is drawn perpendicular to a" 1).

3. Build a new horizontal projection of the straight line b "2 - [ C" 2 D "2].

4. The distance from point A "2 to the straight line C" 2 D "2 (segment (A" 2 M "2] (is the desired one.

It should be borne in mind that the transfer of one of the intersecting lines to the projecting position is nothing more than the transfer of the planes of parallelism, in which lines a and b can be enclosed, also to the projecting position.

Indeed, by transferring the line a to a position perpendicular to the plane π 4 , we ensure that any plane containing the line a is perpendicular to the plane π 4 , including the plane α defined by the lines a and m (a ∩ m, m || b ). If we now draw a line n parallel to a and intersecting line b, then we get the plane β, which is the second plane of parallelism, which contains the intersecting lines a and b. Since β || α, then β ⊥ π 4 .

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Back forward

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

Goals:

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

Equipment:

• multimedia projector;
• a computer;
• task sheets

STUDY PROCESS

I. Organizational moment

II. The stage of updating knowledge(slide 2)

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

III. Lecture(slides 3-15)

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

First method: step-by-step computational

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

We will solve the following tasks:

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

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

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

Next method: volume method.

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

Let's solve the following problem:

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

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

Let's solve the following problem:

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

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

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

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

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

Let's solve the following problem:

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

Consider Application vector method.

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

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

IV. Group work

Try to solve the problem in different ways.

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

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

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

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

V. Lesson summary, homework, reflection