IDZ Ryabushko 3.1 Option 4

№1.

Four points are given: A1(2;4;3), A2(1;1;5), A3(4;9;3), A4(3;6;7). It is necessary to create equations:

a) Equation of plane A1A2A3:

Let's find vectors A1A2 and A1A3:

A1A2 = (1-2; 1-4; 5-3) = (-1; -3; 2)

A1A3 = (4-2; 9-4; 3-3) = (2; 5; 0)

Let's find the vector product of vectors A1A2 and A1A3:

n = A1A2 × A1A3 = (-15; 4; 13)

Then the equation of the plane A1A2A3 will look like:

-15x + 4y + 13z + d = 0

To find d, we substitute the coordinates of point A1 into the equation:

-152 + 44 + 13*3 + d = 0

d = 152 - 44 - 13*3 = -23

So, the equation of the plane is A1A2A3:

-15x + 4y + 13z - 23 = 0

b) Equation of straight line A1A2:

Let's find the direction vector of straight line A1A2:

A1A2 = (-1; -3; 2)

Then the equation of line A1A2 will look like:

x = 2 - t

y = 4 - 3t

z = 3 + 2t

c) Equation of straight line A4M perpendicular to plane A1A2A3:

Let's find the direction vector for straight line A4M, which will be perpendicular to the normal vector of the plane A1A2A3:

n = (-15; 4; 13)

Let's find the coordinates of point M on line A4M. Let M(x, y, z). Then the vectors A4M and n will be collinear, and we can write the following system of equations:

(x - 3)/(-15) = (y - 6)/4 = (z - 7)/13

From here we can express x, y and z:

x = -5t + 3

y = (4/15)t + 6

z = (-13/15)t + 7

d) Equation of straight line A3N parallel to straight line A1A2:

Direction vector of straight line A1A2: (-1; -3; 2)

The direction vector of straight line A3N must be parallel to the direction vector of straight line A1A2. Then the equation of straight line A3N will look like:

x = 4 + a

y = 9 + b

z = 3 + 2a - 3b

e) Equation of a plane passing through point A4 and perpendicular to straight line A1A2:

Direction vector for straight line A1A2: (-1; -3; 2)

The normal vector for the desired plane must be perpendicular to this vector. Therefore, the equation of the desired plane will look like:

• x - 3y + 2z + d = 0

To find d, we substitute the coordinates of point A4:

-3 - 18 + 14 + d = 0

d = 7

So, the equation of a plane passing through point A4 and perpendicular to straight line A1A2:

-x - 3y + 2z + 7 = 0

f) Sine of the angle between straight line A1A4 and plane A1A2A3:

Let's find the direction vector for the straight line A1A4 and the normal vector for the plane A1A2A3:

A1A4 = (1; 2; 4)

n = (-15; 4; 13)

Then the sine of the angle between straight line A1A4 and plane A1A2A3 is calculated by the formula:

sin α = |(А1А4, n)| / |А1А4|*|n|

where |(A1A4, n)| - scalar product of vectors A1A4 and n, |A1A4| and |n| - lengths of vectors A1A4 and n.

Let's calculate the values:

|(A1A4, n)| = |-15 + 8 + 52| = 25

|A1A4| = √(1^2 + 2^2 + 4^2) = √21

|n| = √(15^2 + 4^2 + 13^2) = √370

Then:

sin α = 25 / (√21 * √370) ≈ 0.572

g) Cosine of the angle between the coordinate plane Oxy and the plane A1A2A3:

Normal vector for the Oxy plane: (0; 0; 1)

Normal vector for plane A1A2A3: (-15; 4; 13)

Then the cosine of the angle between the planes is calculated by the formula:

cos α = (Okhu, A1A2A3) / |Okhu|*|A1A2A3|

where (Ohu, A1A2A3) -

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Product description "IDZ Ryabushko 3.1 Option 4":

This is a digital product, which is a task from a series of individual homework (IH) in mathematics, compiled by the author Ryabushko. Option 4 of task 3.1 includes tasks on composing equations of planes and lines in three-dimensional space, calculating angles between lines and planes, as well as proving the perpendicularity of lines.

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Product description "IDZ Ryabushko 3.1 Option 4":

This product is a mathematics task from the "Individual Homework" (IH) series for school students. Option 4 is one of the options for tasks within the Ryabushko IDZ 3.1.

The task consists of three numbers. In the first issue you need to create equations of the plane, straight lines and calculate the sine and cosine of angles. In the second issue, you need to create an equation for a plane passing through a given point and parallel to the Oxy plane. In the third issue you need to prove the perpendicularity of two lines.

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IDZ Ryabushko 3.1 Option 4 is a set of geometry problems, which includes the following tasks:

1. Given four points A1(2;4;3); A2(1;1;5); A3(4;9;3) ; A4(3;6;7). Necessary:

a) draw up an equation of the plane passing through points A1, A2 and A3; b) compose an equation of a straight line passing through points A1 and A2; c) create an equation for a straight line passing through point A4 and perpendicular to the plane A1A2A3; d) draw up an equation of a straight line passing through point A3 and parallel to straight line A1A2; e) create an equation for a plane passing through point A4 and perpendicular to straight line A1A2; f) calculate the sine of the angle between straight line A1A4 and plane A1A2A3; g) calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.

1. Write an equation for a plane passing through point A(2;-3;5) and parallel to the plane Oxy.

2. Prove that the line .. is perpendicular to the line ... (specific lines and their equations are not indicated in the available description).

Please note that solving these problems requires knowledge of mathematical geometry and the ability to work with equations of lines and planes in three-dimensional space.

IDZ Ryabushko 3.1 Option 4 is a textbook for 3rd grade students, created on the basis of Ryabushko’s program. The manual contains tasks and exercises on mathematics, the Russian language, the world around us, as well as preparation for school Olympiads. It presents problems of varying difficulty, allowing students to choose tasks at their level and improve their knowledge and skills. Option 4 differs from other options in that it includes assignments and exercises that will help students consolidate the material they have learned and prepare for tests. IDZ Ryabushko 3.1 Option 4 is a useful guide for students who want to study successfully in elementary school.

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