IDZ Ryabushko 3.1 Option 7

No. 1. Given four points A1(5;5;4); A2(1;–1;4); A3(3;5;1); A4(5;8;–1). It is necessary to create equations:

a) Find the vectors AB1 = A1 - A2 and AC1 = A1 - A3:

AB1 = (5-1; 5-(-1); 4-4) = (4; 6; 0) AC1 = (5-3; 5-5; 4-1) = (2; 0; 3)

Then the vector product of AB1 and AC1 will give the normal vector of the plane:

n1 = AB1 x AC1 = (63 - 00; 04 - 34; 42 - 60) = (18; -12; 8)

Now let’s find the coefficient D of the plane by substituting the coordinates of point A1:

18(x-5) - 12(y-5) + 8(z-4) = 0

To simplify:

6x - 4y + 2z - 2 = 0

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

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

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

c) Find the vector of the line A4M, where M is an arbitrary point on the desired perpendicular line. Take, for example, M(0;0;0):

AM = M - A4 = (-5; -8; 1) Then the desired vector will be equal to the projection of AM onto the normal vector of the plane A1A2A3:

n1 = (18; -12; 8) proj_AMn1 = (AM * n1 / |n1|^2) * n1 = ((-518) + (-8(-12)) + (18)) / (18^2 + (-12)^2 + 8^2) * (18; -12; 8) = (-78/332)(18; -12; 8) = (-39/166; 13/83; -39/83)

Then the equation of the desired line has the form:

x = 5 + (-39/166)t y = 8 + (13/83)t z = -1 + (-39/83)t

d) Since straight line A3N is parallel to straight line A1A2, its direction vector will be equal to AB1:

AB1 = (4; 6; 0)

Let's find the equation of straight line A3N using the parametric equation:

x = 3 + 4t y = 5 + 6t z = 1

e) The equation of the plane passing through point A4 and perpendicular to the straight line A1A2 can also be found as the equation of the plane A1A2A3, using a normal vector equal to the projection of the vector A4A1 onto the plane described by the straight line A1A2:

AB2 = A2 - A1 = (-4; -6; 0) A4A1 = A1 - A4 = (0; -3; 5)

proj_A4A1n1 = (A4A1 * AB1 / |AB1|^2) * n1 = ((04) + (-36) + (5*0)) / (4^2 + 6^2 + 0^2) * (18; -12; 8) = (-54/52; 36/52; 24/13)

Then the normal vector of the desired plane will be equal to:

n2 = proj_A4A1n1 = (-54/52; 36/52; 24/13)

Now let’s find the coefficient D of the plane by substituting the coordinates of point A4:

(-54/52)(x-5) + (36/52)(y-8) + (24/13)(z+1) = 0

To simplify:

-27x + 18y + 24z - 64 = 0

Thus, the equation of the desired plane has the form: -27x + 18y + 24z - 64 = 0.

f) Find the direction vector of straight line A1A4:

AA4 = A4 - A1 = (0; 3; -5)

Then the sine of the angle between straight line A1A4 and plane A1A2A3 is equal to the projection of vector AA4 onto the normal vector of the plane, divided by the modulus of vector AA4:

sin(angle) = |proj_AA4n1| / |AA4| = ((018) + (3(-12)) + ((-5)*8)) / sqrt(0^2 + 3^2 + (-5)^2) / sqrt(18^2 + (-12)^2 + 8^2 ) = -11/29

Answer: sin(angle) = -11/29.

g) Find the normal vector of the plane A1A2A3:

n1 = (18; -12; 8)

Then the cosine of the angle between the plane A1A2A3 and the coordinate plane Oxy is equal to the projection of the normal vector of the plane onto the Ox axis, divided by the modulus of the normal vector:

cos(angle) = |proj_n1_Ox| / |n1| = |18| / sqrt(18^2 + (-12)^2 + 8^2) = 3/7

Answer: cos(angle) = 3/7.

No. 2. It is necessary to create an equation for a plane passing through point A(3;4;0) and a straight line defined by parametric equations:

x = 2 + t y = 3 - 2t z = 1 + 3t

Let's find the directing vector of the line:

v = (1; -2; 3)

Then the normal vector of the plane will be perpendicular to the vector v and you can find it by taking the cross product with an arbitrary vector, for example, with the vector (1; 0; 0):

n = v x (1; 0; 0) = (-2; -3; -2)

Now let’s find the coefficient D of the plane by substituting the coordinates of point A:

-2(x-3) - 3(y-4) - 2z = 0

To simplify:

-2x - 3y - 2z + 18 = 0

Thus, the equation of the desired plane has the form: -2x - 3y - 2z + 18 = 0.

No. 3. It is necessary to find the intersection point of the line specified by the parametric equations:

x = 2 + t y = 1 - 2t z = -1 + 3t

and planes 2x + 3y + z - 1 = 0.

Note that the coordinates of the intersection point must satisfy the plane equation

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IDZ Ryabushko 3.1 Option 7 is a task for solving various geometric problems related to straight lines and planes in three-dimensional space. The task is given four points in three-dimensional space, and it is required to create equations for planes and lines passing through these points or parallel/perpendicular to them, as well as calculate the values ​​of the sine and cosine of angles between some lines and planes. The assignment also gives an equation of a plane and a line, and requires finding their point of intersection.

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