Option 17 IDZ 3.1

№1.17

Given four points A1(6;6;5); A2(4;9;5); A3(4;6;11); A4(6;9;3). Make up equations:

1. planes A1A2A3;
2. straight A1A2;
3. straight line A4M, perpendicular to plane A1A2A3;
4. straight line A3N parallel to straight line A1A2;
5. plane passing through point A4, perpendicular to straight line A1A2.

Calculate:

1. sine of the angle between straight line A1A4 and plane A1A2A3;
2. cosine of the angle between the coordinate plane Oxy and the plane A1A2A3;

Answer:

1. In order to find the equation of a plane passing through three points, we use the formula:

(x - x₁)(y₂ - y₁)(z₃ - z₁) + (and - and₁)(z₂ - z₁)(x₃ - x₁) + (z - z₁)(x₂ - x₁)(y₃ - y₁) - (z - z₁)(y₂ - y₁)(x₃ - x₁) - (and - and₁)(x₂ - x₁)(z₃ - z₁) - (x - x₁)(z₂ - z₁)(y₃ - y₁) = 0

Substituting the coordinates of the points, we get:

(x - 6)(9 - 6)(11 - 5) + (y - 6)(5 - 5)(4 - 6) + (z - 5)(4 - 6)(6 - 6) - (z - 5)(9 - 6)(4 - 6) - (y - 6)(4 - 6)(11 - 5) - (x - 6)(5 - 5)(6 - 11) = 0

To simplify:

3x - 3y + 6z - 18 = 0

Thus, the equation of the plane A1A2A3 is 3x - 3y + 6z - 18 = 0.

To find the equation of a line passing through two points, we use the parametric equation of a line:

x = x₁ + t(x₂ - x₁)

y = y₁ + t(y₂ - y₁)

z = z₁ + t(z₂ - z₁)

Substituting the coordinates of points A1(6;6;5) and A2(4;9;5), we get:

x = 6 - 2t

y = 6 + 3t

z = 5

Thus, the equation of line A1A2 has the form x = 6 - 2t, y = 6 + 3t, ​​z = 5.

In order to find the equation of a line perpendicular to a plane passing through three points, we use the normal vector to this plane. The normal vector to the plane A1A2A3 is found as the vector product of its two direction vectors:

AB = (4-6)i + (9-6)j + (5-5)k = -2i + 3j

AC = (4-6)i + (6-6)j + (11-5)k = -2i + 6k

n = AB × AC = (-3i - 22j + 12k).

Thus, the normal vector to the plane A1A2A3 has the form n = (-3i - 22j + 12k).

Line A4M must pass through point A4(6;9;3) and be perpendicular to the plane A1A2A3, therefore its direction vector must be parallel to the normal vector to this plane. The direction vector can be chosen as n' = (22i - 3j) (the direction of the vector is opposite to the direction of n, so that the straight line goes from point A4). Then the parametric equation of straight line A4M will be:

x = 6 + 22t

y = 9 - 3t

z = 3 + 12t

Thus, the equation of straight line A4M has the form x = 6 + 22t, y = 9 - 3t, z = 3 + 12t.

Line A3N must be parallel to line A1A2, therefore its direction vector must be parallel to the direction vector of line A1A2. The direction vector of straight line A1A2 is equal to:

u = (4-6)i + (9-6)j + (5-5)k = -2i + 3j

Then the direction vector of straight line A3N should also be equal to -2i + 3j. Line A3N passes through point A3(4;6;11), so its equation can be written in parametric form:

x = 4 - 2t

y = 6 + 3t

z = 11

Thus, the equation of straight line A3N has the form x = 4 - 2t, y = 6 + 3t, ​​z = 11.

The normal vector to the plane passing through point A4 and perpendicular to line A1A2 must be parallel to the direction vector of line A1A2. The direction vector of straight line A1A2 is equal to

The product "Option 17 IDZ 3.1" is a digital product intended for students studying mathematics. It contains assignments and answers to problems from the "Integrals" section of the higher mathematics course.

The product design is made in a beautiful html format, which allows you to conveniently view and study the material. When purchasing a product, you will receive access to a file with tasks and answers that can be used both for independent work and for preparing for the exam.

The product "Option 17 IDZ 3.1" is an excellent solution for students who want to improve their knowledge and skills in mathematics. Thanks to the convenient format, you can easily and quickly complete all the tasks and verify your knowledge.

The product "Option 17 IDZ 3.1" is a digital product containing tasks and answers to problems from the "Integrals" section of the higher mathematics course. The product design is made in a beautiful html format, which allows you to conveniently view and study the material. When purchasing a product, you will receive access to a file with tasks and answers that can be used both for independent work and for preparing for the exam.

Now let's move on to solving the problems from the description:

a) Equation of the plane A1A2A3: 3x - 3y + 6z - 18 = 0.

b) Equation of line A1A2: x = 6 - 2t, y = 6 + 3t, ​​z = 5.

c) Direction vector of straight line A4M: n' = (22i - 3j). Equation of line A4M: x = 6 + 22t, y = 9 - 3t, z = 3 + 12t.

d) Direction vector of straight line A3N: u = -2i + 3j. Equation of line A3N: x = 4 - 2t, y = 6 + 3t, ​​z = 11.

e) The normal vector to the desired plane must be parallel to the direction vector of straight line A1A2, that is -2i + 3j. Plane equation: -2x + 3y - 18z + c = 0. Substituting the coordinates of point A4(6;9;3), we get c = -45. Thus, the equation of the desired plane is: -2x + 3y - 18z - 45 = 0.

e) The angle between straight line A1A4 and plane A1A2A3 can be found as the angle between the direction vector of straight line A1A4 and the normal vector of plane A1A2A3. Direction vector of straight line A1A4: AB = (6-4)i + (9-6)j + (3-5)k = 2i + 3j - 2k. Normal vector to the plane A1A2A3: n = (-3i - 22j + 12k). Then the sine of the angle between the line and the plane is equal to: |AB × n| / (|AB| * |n|) = |-45| / (|AB| * sqrt(733)) = 3sqrt(733)/733.

g) The cosine of the angle between the coordinate plane Oxy and the plane A1A2A3 is equal to the cosine of the angle between the normal vector of the plane A1A2A3 and the direction of the Ox axis. Normal vector to the plane A1A2A3: n = (-3i - 22j + 12k). Direction of the Ox axis: i. Then the cosine of the angle between them is equal to: (n * i) / (|n| * |i|) = -3/sqrt(733).

No. 2.17. The direction vector of the segment M1M2: v = M2 - M1 = (-3i - j + k). The normal vector to the plane passing through the point M and perpendicular to the segment M1M2 is equal to the vector product of the direction vector of the segment and the vector perpendicular to the plane: n = v × (i + j + k) = (-2i + 4j - 4k). Thus, the equation of the desired plane is: -2x + 4y - 4z + d = 0. Substituting the coordinates of the point M (-1;2;3), we obtain d = 2. Thus, the equation of the plane passing through the point M and perpendicular to the segment M1M2: -2x + 4y - 4z + 2 = 0.

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The product in this case is the task IDZ 3.1 on geometry, which contains several subtasks.

In the first subtask, you need to create equations for a plane passing through three given points (A1, A2, A3), as well as a straight line passing through two of these points (A1, A2). You also need to find a straight line (A4M) perpendicular to the first plane and passing through the fourth given point (A4), as well as a straight line (A3N) parallel to the second straight line (A1A2). In the last subtask, you need to find the angle between the first line (A1A4) and the first plane (A1A2A3), as well as the cosine of the angle between the plane containing the given segment (M1M2) and the coordinate plane Oxy.

In the second subtask, you need to create an equation for a plane passing through a given point (M) and perpendicular to a given segment (M1M2).

In the third subtask, you need to show that a given line is parallel to a given plane, and also that it lies in this plane.

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