Ryabushko A.P. IDZ 3.1 option 7

No. 1.7. In this problem, you need to create equations for various geometric objects based on given points. There are four points: A1(5;5;4); A2(1;-1;4); A3(3;5;1); A4(5;8;-1).

a) The equation of the plane passing through points A1, A2 and A3 can be calculated using the formula for the general equation of the plane:

(5-1)(y+1) - (5+1)(x-1) + (4-4)(x-1) = 0

Thus, the equation of the plane is A1A2A3: 4x - 6y + 2z - 14 = 0.

b) The equation of the straight line A1A2 can be found using the formula for the parametric equation of the straight line:

x = 5 - 4t y = 5 + 6t z = 4

c) The equation of straight line A4M can be found using the parametric equation of the straight line and the coordinates of point M:

x = 5 + t y = 8 + 3t z = -1 - 5t

d) To find the equation of line A3N parallel to line A1A2, you can use the parametric equation of line A1A2 and set a new point on the line, for example N(3;7;1):

x = 5 - 4t y = 5 + 6t z = 4 point N: x = 3, y = 7, z = 1

h) 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 a plane:

4(x-5) + 6(y-5) - 2(z-4) = 0

Thus, the equation of the plane passing through point A4 and perpendicular to line A1A2 is: 4x + 6y - 2z - 2 = 0.

f) To calculate the sine of the angle between straight line A1A4 and plane A1A2A3, you can use the formula for the sine of the angle between straight line and plane, which looks like this:

sin(angle) = |(n, d)| / (|n| * |d|),

where n is the normal to the plane, d is the direction vector of the line. Substituting the values, we get:

sin(angle) = |(A1A4, n)| / (|A1A4| * |n|),

where A1A4 is the vector connecting points A1 and A4.

Let's calculate the normal to the plane A1A2A3:

n = (A2-A1) x (A3-A1) = (-6,-12,12)

Let's calculate the direction vector of straight line A1A4:

d = A4-A1 = (0,3,-5)

Now you can calculate the sine of the angle between straight line A1A4 and plane A1A2A3:

sin(angle) = |(-18.6,-18)| / (|A1A4| * sqrt(360)) = 3sqrt(5)/10.

g) To calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3, you need to find the angle between the normal vector of the plane A1A2A3 and the vector connecting the point of intersection of the plane A1A2A3 with the coordinate plane and the origin.

The normal to the plane A1A2A3 was found in point a) and is equal to (-6,-12,12). The intersection point with the coordinate plane Oxy has coordinates (2,0,0), then the vector connecting the origin of coordinates with the intersection point is equal to (2,0,0).

Then the cosine of the angle between the plane A1A2A3 and the coordinate plane Oxy is equal to:

cos(angle) = (0,0,1) * (-6,-12,12) / (sqrt(6^2+12^2+12^2) * sqrt(2^2)) = -sqrt( 3)/3.

No. 2.7. To create an equation for a plane passing through point A(3;4;0) and a line, you must first find the direction vector of the line. The direction vector of the line can be found using the parametric equation of the line, which is given in paragraph b) of problem No. 1.7:

d = A2-A1 = (-4,-6,0)

Now, using the formula for the general equation of the plane, we can find the equation of the plane:

-4(x-3) - 6(y-4) + z = 0

No. 3.7. To find the point of intersection of a line and a plane, you need to solve a system of equations made up of the equation of the line and the equation of the plane.

The straight line is given by the parametric equation:

x = 5 - 4t y = 5 + 6t z = 4

The plane equation is given in standard form:

2x + 3y + z - 1 = 0

We substitute the parametric equations of the straight line into the equation of the plane and solve the resulting system:

2(5-4t) + 3(5+6t) + 4t - 1 = 0

Solving the equation for t, we get:

t = -5/26

Substituting the found value of t into the parametric equation of the line, we obtain the intersection point:

x = 5 - 4*(-5/26) = 135/26 y = 5 + 6*(-5/26) = 95/13 z = 4*(-5/26) = -10/13

Thus, the point of intersection of the line and the plane is (135/26, 95/13, -10/13).

Ryabushko A.P. IDZ 3.1 option 7

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Ryabushko A.P. IDZ 3.1 version 7 is a set of solutions to problems in mathematics related to equations of planes and lines in space. The product contains solutions to four problems, which include finding equations of planes, parametric equations of lines, finding angles between lines and planes, and finding the point of intersection of a line and a plane. The product is suitable for students who are preparing for exams or olympiads in mathematics. The product includes detailed solutions to problems with step-by-step explanations of the methods and formulas used to solve them.

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Ryabushko A.P. IDZ 3.1 option 7 is a mathematics task in which you need to solve several problems in analytical geometry. The task consists of three numbers:

1. It is necessary to compose equations of straight lines and planes, as well as calculate the sine and cosine of the angles between them.
2. You need to create an equation for a plane passing through a given point and a straight line.
3. You need to find the point of intersection of a given line and a plane.

This task is suitable for those who are studying analytical geometry and want to test their knowledge. If you have any questions or problems solving problems, you can contact the seller listed in the seller information.

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