Ryabushko A.P. IDZ 3.1 version 14

No. 1.14. There are four points: A1(3;5;4); A2(8;7;4); A3(5;10;4); A4(4;7;8). It is necessary to create equations:

a) Equation of the plane A1A2A3: Find the vector product of the vectors A1A2 and A1A3 to obtain the normal vector to the plane: A1A2 = (8-3; 7-5; 4-4) = (5; 2; 0) A1A3 = (5-3 ; 10-5; 4-4) = (2; 5; 0) n = A1A2 x A1A3 = (0; 0; 10) Thus, the equation of the plane A1A2A3 is: 0x + 0y + 10z + D = 0, where D = -104 = -40. Answer: 10z - 40 = 0.

b) Equation of line A1A2: Let's find the direction vector of the line: A1A2 = (8-3; 7-5; 4-4) = (5; 2; 0) Thus, the equation of line A1A2 has the form: x = 3 + 5t y = 5 + 2t z = 4 + 0t Answer: x = 3 + 5t, y = 5 + 2t, z = 4.

c) Equation of straight line A4M perpendicular to plane A1A2A3: Since straight line A4M is perpendicular to plane A1A2A3, its direction vector must be parallel to the normal vector of the plane: n = (0; 0; 10) Let point M have coordinates (x, y, z). Then the vector A4M will be equal to: A4M = (x-4; y-7; z-8) Multiply the vector A4M by the normal vector and equate the result to zero: 0*(x-4) + 0*(y-7) + 10 *(z-8) = 0 10z - 80 = 0 Thus, the equation of straight line A4M has the form: x = 4 + at y = 7 + bt z = 8 + 8t Answer: x = 4 + at, y = 7 + bt , z = 8 + 8t.

d) Equation of straight line A3N parallel to straight line A1A2: Find the directing vector of straight A1A2: A1A2 = (8-3; 7-5; 4-4) = (5; 2; 0) Since straight A3N is parallel to straight A1A2, then its directing the vector can be chosen the same, for example, v=(5,2,0). Let point N have coordinates (x, y, z). Then the vector A3N will be equal to: A3N = (x-5; y-10; z-4) Thus, the equation of the straight line A3N has the form: x = 5 + 5t y = 10 + 2t z = 4 + 0t Answer: x = 5 + 5t, y = 10 + 2t, z = 4.

e) Equation of a plane passing through point A4, perpendicular to line A1A2: Find the direction vector of line A1A2: A1A2 = (8-3; 7-5; 4-4) = (5; 2; 0) Since the plane must be perpendicular to the line A1A2, then its normal vector must be parallel to the vector product of the vectors A1A2 and (0,0,1): n = A1A2 x (0,0,1) = (-2, 5, 0) Then the equation of the plane has the form: -2x + 5y + D = 0, where D = -(-24 + 57) = -22. Answer: -2x + 5y - 22 = 0.

f) Sine of the angle between straight line A1A4 and plane A1A2A3: Find the vector connecting points A1 and A4: A1A4 = (4-3; 7-5; 8-4) = (1; 2; 4) Find the normal vector to the plane A1A2A3: n = (0; 0; 10) The angle between the vectors is determined by the formula: sin(angle) = |A1A4 x n| / (|A1A4| * |n|) where |...| denotes the length of the vector. Let's calculate the numerator: A1A4 x n = (20; -10; 0) |A1A4 x n| = sqrt(400 + 100) = 10sqrt(5) Calculate the denominator: |A1A4| = sqrt(1 + 4 + 16) = 3sqrt(2) |n| = 10 Then sin(angle) = (10sqrt(5)) / (3sqrt(2) * 10) = sqrt(5/18) Answer: sin(angle) = sqrt(5/18).

g) Cosine of the angle between the coordinate plane Oxy and the plane A1A2A3: Find the normal vector to the plane A1A2A3: n = (0; 0; 10) The coordinate plane Oxy is given by the equation z = 0. The angle between the planes is determined by the formula: cos (angle) = | n * (0, 0, 1)| / (|n| * |(0, 0, 1)|) where |...| denotes the length of the vector. Let's calculate the numerator: n * (0, 0, 1) = 10 Let's calculate the denominator: |n| = 10 |(0, 0, 1)| = 1 Then cos(angle) = 10 / (10 * 1) = 1 Answer: cos(angle) = 1.

No. 2.14. It is necessary to create an equation for a plane passing through points A(3;-1;2) and B(2;1;4) and parallel to the vector a = (5;-2;-1). Let's find the normal vector to the plane using the vector product of vectors AB and a: AB = (2-3; 1+1; 4-2) = (-1; 2; 2) n = AB x a = (3; 7; 11 ) Thus, the equation of the plane is: 3x + 7y + 11z + D = 0, where D = -33 - 7(-1) - 11*2 = -34. Answer: 3x + 7y + 11z - 34 =

Ryabushko A.P. IDZ 3.1 version 14

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Ryabushko A.P. IDZ 3.1 version 14 is a mathematics task containing several points in which it is necessary to create equations of planes and lines, as well as calculate the values ​​of the angles between them. The task presents the coordinates of various points for which it is necessary to solve the assigned problems. The assignment also contains the necessary formulas and instructions for their use. Solving the problem requires knowledge of mathematical concepts and vector algebra formulas.

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Ryabushko A.P. IDZ 3.1 option 14 is a geometry task that includes several points.

In the first task you need to create equations of a plane, a line and a perpendicular passing through given points. You also need to calculate the sine and cosine of the angles between some objects.

In the second task, you need to create an equation of a plane passing through two given points and parallel to a given vector.

In the third task, you need to create an equation of a line passing through a given point and parallel to two given lines.

If you have any questions about completing the assignment, you can contact the seller listed in the seller information.

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