IDZ Ryabushko 3.1 Option 6

No. 1. Four points are given: A1(0;7;1), A2(2;-1;5), A3(1;6;3), A4(3;-9;8). It is necessary to create equations:

a) Equation of a plane passing through points A1, A2 and A3:

To find the equation of a plane passing through three points, you can use the formula:

(x - x1)(y2 - y1)(z3 - z1) + (x2 - x1)(y - y1)(z3 - z1) + (x2 - x1)(y2 - y1)(z - z1) = 0,

where x, y, z are the coordinates of an arbitrary point on the plane, and x1, y1, z1, x2, y2, z2, x3, y3, z3 are the coordinates of given points.

Let's substitute the coordinates of the points and get the equation of the plane:

(2 - x)(-8) + (0 - 2)(13) + (0 - 7)(3 - 1) = 0

Let's simplify:

-16 + 26 - 12 = 0

Thus, the equation of the plane A1A2A3 has the form:

-8x + 13y - 12z + 6 = 0.

b) Equation of a straight line passing through points A1 and A2:

To find the equation of a line passing through two given points, you can use the formulas:

x = x1 + at y = y1 + bt z = z1 + ct,

where x1, y1, z1 and x2, y2, z2 are the coordinates of given points, a, b, c are guiding coefficients, t is a parameter.

Let's find the guiding coefficients:

a = x2 - x1 = 2 - 0 = 2 b = y2 - y1 = -1 - 7 = -8 c = z2 - z1 = 5 - 1 = 4

Let's substitute the values of the coordinates and direction coefficients and get the equation of the straight line:

x = 2t y = -8t + 7 z = 4t + 1

c) Equation of a straight line passing through point A4 and perpendicular to the plane A1A2A3:

To find the equation of a line perpendicular to a given plane and passing through a given point, you must use the following steps:

Find the equation of a plane passing through a given point and perpendicular to the given plane.
Find the point of intersection of the found plane and a line passing through a given point and parallel to a given plane.
Write an equation for a line passing through a given point and the found intersection point.

Let us find the equation of a plane perpendicular to the plane A1A2A3 and passing through point A4. To do this you can use the formula:

-8x + 13y - 12z + d = 0,

where d is the unknown coefficient that needs to be found. Let's substitute the coordinates of point A4 and get the equation of the plane:

-83 +13(-9) - 12*8 + d = 0

Let's solve the equation and find d:

d = 169

Thus, the equation of the plane passing through point A4 and perpendicular to the plane A1A2A3 has the form:

-8x + 13y - 12z + 169 = 0.

Let's find the point of intersection of the found plane and straight line A1A2. To do this, we substitute the equation of the straight line into the equation of the plane and solve the resulting system of equations:

-8(2t) + 13(-8t + 7) - 12(4t + 1) + 169 = 0

Solving the equation, we find the value of the parameter t:

t = -3/4

Now, let's find the coordinates of the intersection point:

x = 2(-3/4) = -3/2 y = -8(-3/4) + 7 = 13 z = 4(-3/4) + 1 = -2

Thus, the intersection point has coordinates (-3/2; 13; -2). It remains to create an equation for a straight line passing through points A4 and (-3/2; 13; -2). To do this, we use the formulas for the equation of a line passing through two points:

x = 3 - 5/4t y = -9 + 40/3t z = 8 - 11/2t

where t is a parameter.

d) Equation of a line parallel to straight line A1A2 and passing through point A3:

To find the equation of a line parallel to a given line and passing through a given point, you must use the formulas:

x = x1 + at y = y1 + bt z = z1 + ct,

where x1, y1, z1 are the coordinates of the given point, a, b, c are the guiding coefficients, which must be equal to the guiding coefficients of the given straight line.

Let's find the guiding coefficients of the given straight line A1A2:

a = 2 - 0 = 2 b = -1 - 7 = -8 c = 5 - 1 = 4

Thus, the equation of a line parallel to line A1A2 and passing through point A3 has the form:

x = 1 + 2t y = 6 - 8t z = 3 + 4t

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

To find the equation of a plane perpendicular to a given line and passing through a given point, you must use the formula:

a(x - x0) + b(y - y0) + c(z - z0) = 0,

where x0, y0, z0 are the coordinates of a given point, a, b, c are direction coefficients that must be perpendicular to the direction vector of a given straight line.

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

u = (2, -8, 4)

Thus, the guiding coefficients of the plane must be perpendicular

Ryabushko IDZ 3.1 Option 6 is a digital product, which is a collection of mathematics problems for schoolchildren. This product contains problems on various topics, such as geometry, algebra and trigonometry, with various levels of difficulty, which allows you to use it both for self-study and for preparing for olympiads and exams.

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IDZ Ryabushko 3.1 Option 6 is a math task that includes several tasks on finding equations of lines and planes passing through given points or parallel/perpendicular to given lines and planes.

The task gives four points: A1(0;7;1), A2(2;-1;5), A3(1;6;3), A4(3;-9;8), and it is necessary to create the equations:

a) equation of the plane passing through points A1, A2 and A3; b) the equation of a straight line passing through points A1 and A2; c) the equation of a straight line passing through point A4 and perpendicular to the plane A1A2A3; d) equation of a line parallel to straight line A1A2 and passing through point A3; e) equation of a plane passing through point A4 and perpendicular to straight line A1A2.

To solve each problem, the appropriate formulas and methods described in the task conditions are used.

IDZ Ryabushko 3.1 Option 6 is a set of tasks in mathematics that include solving equations of planes and lines, as well as finding directing coefficients and perpendicular planes.

The task gives four points: A1(0;7;1), A2(2;-1;5), A3(1;6;3), A4(3;-9;8). The following tasks need to be solved:

a) Find the equation of the plane passing through points A1, A2 and A3. To do this, you can use a formula that connects the coordinates of an arbitrary point on the plane with the coordinates of given points.

b) Find the equation of the line passing through points A1 and A2. To do this, you need to use formulas that connect the coordinates of an arbitrary point on a line with the coordinates of given points.

c) Find the equation of the line passing through point A4 and perpendicular to the plane A1A2A3. To do this, you first need to find the equation of a plane perpendicular to a given plane and passing through a given point. Then you need to find the point of intersection of this plane and a line passing through a given point and parallel to a given plane. And finally, you need to create an equation for a straight line passing through a given point and the found intersection point.

d) Find the equation of a line parallel to line A1A2 and passing through point A3. To do this, you need to use formulas that connect the coordinates of an arbitrary point on a line with the coordinates of a given point and the direction coefficients of a given line.

e) Find the equation of the plane passing through point A4 and perpendicular to line A1A2. To do this, you need to use a formula that connects the coordinates of an arbitrary point on the plane with the coordinates of a given point and guide coefficients, which must be perpendicular to the guide of a given straight line.

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IDZ Ryabushko 3.1 Option 6 is a geometry task containing several subtasks of varying complexity.

No. 1. In the first task, four points are given in three-dimensional space, and you also need to create equations of the plane and lines passing through these points, calculate angles and find perpendiculars and parallels.

a) To compile the equation of the plane A1A2A3, you can use the formula for the general equation of the plane Ax + By + Cz + D = 0, where the coefficients A, B and C are determined using the vector product of vectors lying in this plane. Thus, the equation of the plane A1A2A3 has the form:

-5x + 7y - 11z + 44 = 0

b) To compile the equation of the straight line A1A2, you can use the formula for the parametric equation of the straight line, which has the form:

x = 2t y = 7 - 4t z = 1 + 4t

c) To compile the equation of straight line A4M, you can use the formula for the parametric equation of a straight line that passes through points A4 and M. Point M is not specified, so it must be chosen arbitrarily. For example, you can take point M(0,0,0). Then the parametric equation of straight line A4M will have the form:

x = 3t y = -9 - 3t z = 8 - 2t

The perpendicular to the plane A1A2A3 can be found using the normal vector of this plane, which is determined by the coefficients of the equation of the plane. Thus, the normal vector of the plane A1A2A3 has coordinates (-5, 7, -11), and the straight line A4N, perpendicular to the plane A1A2A3, will have the equation:

x = 3 + 5t y = -9 - 7t z = 8 + 11t

d) Straight A3N is parallel to straight A1A2, which means that its direction vector will coincide with the direction vector of straight A1A2. The direction vector of straight line A1A2 has coordinates (2, -8, 4), so the equation of straight line A3N will have the form:

x = 1 + 2t y = 6 - 8t z = 3 + 4t

e) The equation of a plane passing through point A4 and perpendicular to line A1A2 can be found using the formula for the general equation of the plane and the normal vector, which will be directed perpendicular to line A1A2 and, therefore, coincide with the direction vector of line A1A4. The direction vector of straight line A1A4 has coordinates (3, -16, 7), so the normal vector of the plane will have coordinates (3, -16, 7), and the equation of the plane will be:

3x - 16y + 7z + 118 = 0

f) To calculate the sine of the angle between the straight line A1A4 and the plane A1A2A3, you need to find the cosine of the angle between the direction vector of the straight line A1A4 and the normal vector of the plane A1A2A3, and then take the sine of the angle additional to the calculated cosine. The direction vector of straight line A1A4 has coordinates (3, -16, 7), and the normal vector of plane A1A2A3 has coordinates (-5, 7, -11). Their dot product is -211, and the lengths of the vectors are √290.5 and √195, which gives the cosine of the angle between them equal to -0.924. Consequently, the sine of the angle between straight line A1A4 and plane A1A2A3 will be equal to √(1-0.924^2) ≈ 0.383.

g) To calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3, you need to find the angle between their normal vectors. The normal vector of the coordinate plane Oxy has coordinates (0, 0, 1), and the normal vector of the plane A1A2A3 was found earlier and has coordinates (-5, 7, -11). Their dot product is -11, and the lengths of the vectors are 1 and √195, which gives the cosine of the angle between them equal to -0.056. Consequently, the angle between the coordinate plane Oxy and the plane A1A2A3 will be approximately equal to acos(-0.056) ≈ 90.6 degrees.

No. 2. In the second task, you need to create an equation for a plane passing through two given points and parallel to the Oy axis.

In order for the plane to be parallel to the Oy axis, its normal vector must be directed along the Oy axis, that is, its coordinates must be (0, k, 0), where k is an arbitrary number. The normal vector of the plane must also be perpendicular to the vector connecting points A(2, 5, -1) and B(-3, 1, 3), that is, their dot product must be zero.

Thus, the equation of the plane is:

ky - 5k + 2*z - 4 = 0

where k is an arbitrary number.

No. 3. In the third task, you need to find the value of the parameter at which the line is parallel to a given line. To do this, it is necessary to express the coordinates of the direction vector of a given line through parameters and find the corresponding coordinates of the direction vector of the desired line. Then you need

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