Ryabushko A.P. IDZ 3.1 option 22

IDZ - 3.1 №1.22

Given four points A1(4;2;10); A2(1;2;0); A3(3;5;7); A4(2;–3;5).

Make up equations:

1. planes A1A2A3;
2. straight A1A2;
3. straight line A4M, perpendicular to plane A1A2A3;
4. straight line A3N parallel to straight line A1A2;
5. plane passing through point A4 and perpendicular to line A1A2;

Calculate:

1. sine of the angle between straight line A1A4 and plane A1A2A3;
2. cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.

a) Find the vectors A1A2 and A1A3:

A1A2 = (1-4; 2-2; 0-10) = (-3;0;-10)

A1A3 = (3-4; 5-2; 7-10) = (-1;3;-3)

Let's find the vector product of these vectors:

n = A1A2 x A1A3 = (-30;-24;6)

Plane equation:

-30x - 24y + 6z + d = 0

Let's substitute the coordinates of point A1:

-30·4 - 24·2 + 6·10 + d = 0

d = 72

Answer: -30x - 24y + 6z + 72 = 0.

b) Line vector A1A2:

A1A2 = (-3;0;-10)

Equation of a straight line:

x = 4 - 3t

y = 2

z = 10 - 10t

Answer: x = 4 - 3t, y = 2, z = 10 - 10t.

c) Line vector A4M:

A4M = (2-2; -3-3; 5+1) = (0;-6;6)

Perpendicular to plane A1A2A3:

n = (-30;-24;6)

Equation of a straight line:

x = 2

y = -3-6t

z = 5+6t

Answer: x = 2, y = -3-6t, z = 5+6t.

d) Line vector A1A2:

A1A2 = (-3;0;-10)

Equation of a straight line:

x = 4 - 3t

y = 2

z = 10 - 10t

Direct vector A3N:

А3N = (3-4; 5-2; 7-7) = (-1;3;0)

Equation of a straight line:

x = 4 - t

y = 2 + 3t

z = 10 - 10t

Answer: x = 4 - t, y = 2 + 3t, ​​z =10 - 10t.

e) Find the vector of the line A1A2:

A1A2 = (-3;0;-10)

Normal vector to the plane:

n = (-30;-24;6)

Plane equation:

-30x - 24y + 6z + d = 0

Let's substitute the coordinates of point A4:

-30·2 - 24·(-3) + 6·5 + d = 0

d = -12

Answer: -30x - 24y + 6z - 12 = 0.

f) Line vector A1A4:

A1A4 = (-2; -5; -5)

Normal vector to plane A1A2A3:

n = (-30;-24;6)

Angle between vectors:

sin α = |n·А1А4| / (|n|·|А1А4|) = |-30·(-2) - 24·(-5) + 6·(-5)| / (√(302+242+62) · √((-2)2+(-5)2+(-5)2)) = 24/3

Answer: sin α = 24/35.

g) Normal vector to plane A1A2A3:

n = (-30;-24;6)

Coordinates of the normal vector to the Oxy plane:

n₀ = (0;0;1)

Angle between vectors:

cos α = |n·n₀| / (|n|·|n₀|) = |6| / (√(30²+24²+6²) · 1) = 1/35

Answer: cos α = 1/35.

№2.22

Write an equation for a plane passing through the point M(2;3;–1) and the straight line x=t–3; y=2t+5; z=–3t + 1.

Straight direction vector:

a = (1;2;-3)

The second vector lying in the plane:

b = (1;2;0)

Normal vector to the plane:

n = a x b = (6;-3;-4)

Plane equation:

6(x-2) - 3(y-3) - 4(z+1) = 0

Answer: 6x - 3y - 4z - 8 = 0.

№3.22

Find the intersection point of the straight line (x-7)/5 = (y-1)/1 = (z-5)/4 and the plane 3x–y+2z–8=0.

Equation of a straight line:

x-7 = 5t

y-1 = t

z-5 = 4t

Let's substitute the plane into the equation:

3(5t+7) - (t+1) + 2(4t+5) - 8 = 0

23t = -20

t = -20/23

Intersection point:

x = -3/23

y = -20/23

z = 12/23

Answer: (-3/23; -20/23; 12/23).

Thanks for

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The digital product "Ryabushko A.P. IDZ 3.1 version 22" is a collection of solutions to geometry problems created by the author A.P. Ryabushko. The collection includes solutions to problems that will help students deepen their knowledge in the field of geometry and successfully cope with exams.

A) Equation of the plane A1A2A3: -30x - 24y + 6z + 72 = 0.

B) Equation of straight line A1A2: x = 4 - 3t, y = 2, z = 10 - 10t.

B) Equation of straight line A4M: x = 2, y = -3-6t, z = 5+6t.

Perpendicular to the plane A1A2A3 has a direction vector (-30, -24, 6).

D) Equation of straight line A3N: x = 4 - t, y = 2 + 3t, ​​z = 10 - 10t.

Parallel line A1A2 has a direction vector (-3, 0, -10).

E) Equation of a plane passing through point A4 and perpendicular to line A1A2: -30x - 24y + 6z - 12 = 0.

E) Sine of the angle between straight line A1A4 and plane A1A2A3: sin α = 24/35.

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Ryabushko A.P. IDZ 3.1 option 22 is a training task that consists of three numbers.

In the first issue, four points are given in three-dimensional space, and you need to create equations for the plane and straight lines, as well as calculate the sine and cosine of the angles between some of them.

In the second issue, you need to create an equation for a plane passing through a given point and a line defined parametrically.

In the third number you need to find the point of intersection of a given line and a plane.

After completing the task, if you have any questions, you can contact the seller at the specified email address.

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