IDZ Ryabushko 3.1 Option 23

No. 1 Given four points A1(2;3;5); A2(5;3;–7); A3(1;2;7); A4(4;2;0). It is necessary to create equations:

a) Planes A1A2A3: In order to create an equation of a plane passing through three points A1, A2 and A3, it is necessary to find the vector product of vectors A1A2 and A1A3 and write it in the form of an equation of the plane: A1A2 = (5-2, 3-3, -7-5) = (3, 0, -12) A1A3 = (1-2, 2-3, 7-5) = (-1, -1, 2) n = A1A2 x A1A3 = (6, 30, 6) Plane equation: 6x + 30y + 6z - 60 = 0

b) Line A1A2: In order to create an equation of a line passing through two points A1 and A2, it is necessary to find the direction vector of the line and write it in the form of an equation of a line: A1A2 = (5-2, 3-3, -7-5) = (3, 0, -12) Line equation: x = 2 + 3t y = 3 z = -7 - 12t

c) Line A4M perpendicular to the plane A1A2A3: In order to create an equation for a line passing through two points A4 and M and perpendicular to the plane A1A2A3, it is necessary to find the vector product of vectors parallel to this plane and write it in the form of a directing vector of the line: A1A2 = (5-2, 3-3, -7-5) = (3, 0, -12) A1A3 = (1-2, 2-3, 7-5) = (-1, -1, 2) n = A1A2 x A1A3 = (6, 30, 6) A4M = (4-4, 2+3, 0-1) = (0, 5, -1) Line equation: x = 4 y = 2 + 5t z = 1 - t

d) Line A3N parallel to line A1A2: In order to create an equation for a line passing through two points A3 and N and parallel to line A1A2, it is necessary to use the direction vector of this line, which coincides with the direction vector of line A1A2: A1A2 = (5-2 , 3-3, -7-5) = (3, 0, -12) A3N = (x-1, y-2, z-7) Line equation: x = 2 + 3t y = 2 z = -7 - 12t

e) A plane passing through point A4 and perpendicular to line A1A2: In order to create an equation for a plane passing through point A4 and perpendicular to line A1A2, it is necessary to use the direction vector of this line, which coincides with the direction vector of line A1A2: A1A4 = ( 4-2, 2-3, 0-5) = (2, -1, -5) The normal vector of the plane will be perpendicular to the direction vector of straight line A1A4, so we find it through the vector product of the direction vector of straight line A1A2 and the direction vector of straight line A1A4: n = A1A2 x A1A4 = (33, -18, -3) Plane equation: 33x - 18y - 3z + 27 = 0

f) Sine of the angle between line A1A4 and plane A1A2A3: In order to find the sine of the angle between line A1A4 and plane A1A2A3, it is necessary to find the projection of the direction vector of line A1A4 onto the normal vector of the plane A1A2A3 and divide it by the length of the direction vector of line A1A4: A1A4 = ( 4-2, 2-3, 0-5) = (2, -1, -5) n = A1A2 x A1A3 = (6, 30, 6) sin(angle) = |proj(A1A4,n)| / |A1A4| |proj(A1A4,n)| = |A1A4| * sin(angle) = |A1A4| * |n| * sin(angle) / |A1A4| = |n| * sin(angle) |n| * sin(angle) = |A1A4 x n| = |(-174, 12, 42)| = 629 |n| = sqrt(6^2 + 30^2 + 6^2) = 6sqrt(11) sin(angle) = (629) / (6sqrt(11)*sqrt(30)) = 29 / (sqrt(11)5sqrt(2)) Answer: sin(angle) = 29 / (sqrt(11)5sqrt(2))

g) Cosine of the angle between the coordinate plane Oxy and the plane A1A2A3: In order to find the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3, it is necessary to find the scalar product of the normal vectors of these planes and divide it by the product of their lengths: Normal vector of the coordinate plane Oxy: n1 = (0, 0, 1) Normal vector of the plane A1A2A3: n2 = A1A2 x A1A3 = (6, 30, 6) cos(angle) = n1n2 / (|n1||n2|) |n1| = sqrt(0^2 + 0^2 + 1^2) = 1 |n2| = sqrt(6^2 + 30^2 + 6^2) = 6sqrt(11) n1n2 = 6 cos(angle) = 6 / (16sqrt(11)) = 1 / (sqrt(11)*sqrt(2)) Answer: cos(angle) = 1 / (sqrt(11)*sqrt(2))

No. 2 It is necessary to find the projection of the point M(4;-3;1) onto the plane x - 2y - z - 15 = 0.

The projection of a point onto a plane is equal to its orthogonal projection onto this plane. Since the plane is given by the equation, we can find the normal vector of the plane and the direction

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The product "IDZ Ryabushko 3.1 Option 23" is a textbook in mathematics containing problems and solutions on the following topics: drawing up equations of planes and lines, finding the projection of a point onto a plane, calculating the sine and cosine of angles between lines and planes. The manual presents 2 problems with a detailed description of the solution.

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IDZ Ryabushko 3.1 Option 23 is a set of problems in mathematics or geometry, which includes various tasks on composing equations of planes and lines, as well as calculating angles between them and finding projections of points on a plane. In particular, this set of tasks contains the following tasks:

1. Given four points A1(2;3;5); A2(5;3;–7); A3(1;2;7); A4(4;2;0). It is necessary to create equations: a) plane A1A2A3; b) straight A1A2; c) straight line A4M, perpendicular to the plane A1A2A3; d) straight line A3N parallel to straight line A1A2; e) a plane passing through point A4 and perpendicular to straight line A1A2. You also need to calculate: f) sine of the angle between straight line A1A4 and plane A1A2A3; g) cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.

2. It is necessary to find the projection of the point M(4;–3;1) onto the plane x – 2y – z – 15 = 0.

3. It is necessary to create an equation for a plane passing through the point K(2;–5;3) and parallel to the Oxz plane.

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