Given four points A1(3;–1;2); A2(–1;0;1); A3(1;7;3); A4(8;5;8).
a) Let’s create an equation for the plane A1A2A3:
Vector lying on the plane A1A2A3:
A1A2 = (-1 - 3; 0 + 1; 1 - 2) = (-4; 1; -1)
A1A3 = (1 - 3; 7 + 1; 3 - 2) = (-2; 8; 1)
Normal vector of the plane A1A2A3:
n = [A1A2, A1A3] = (-1 - 3; 3 - 2; 0 - 8) = (-4; 1; -8)
Plane equation:
-4x + y - 8z + d = 0
to find d, substitute the coordinates of point A1:
-4 * 3 + (-1) * (-1) - 8 * 2 + d = 0
d = 26
Equation of plane A1A2A3:
-4x + y - 8z + 26 = 0
b) Let’s create an equation for straight line A1A2:
Direction vector of straight line A1A2:
A1A2 = (-4; 1; -1)
Equation of a straight line:
x = 3 - 4t, y = -1 + t, z = 2 - t, t ∈ R
c) Let’s create an equation for straight line A4M perpendicular to plane A1A2A3:
Normal vector of the plane A1A2A3:
n = (-4; 1; -8)
Directional vector A4M:
А4M = (x - 8; y - 5; z - 8)
Perpendicularity condition:
n * A4M = 0
-4(x - 8) + 1(y - 5) - 8(z - 8) = 0
Equation of line A4M:
x = 8 + 2t, y = 5 - t, z = 8 + 0.5t, t ∈ R
d) Let’s create an equation for straight line A3N parallel to straight line A1A2:
Direction vector of straight line A1A2:
A1A2 = (-4; 1; -1)
Equation of line A3N:
x = 1 + (-4)t, y = 7 + t, z = 3 - t, t ∈ R
e) Let’s create an equation for a plane passing through point A4 and perpendicular to straight line A1A2:
Direction vector of straight line A1A2:
A1A2 = (-4; 1; -1)
Normal plane vector:
n = (-4; 1; -1)
Plane equation:
-4x + y - z + d = 0
to find d, substitute the coordinates of point A4:
-4 * 8 + 5 - 8 + d = 0
d = 27
Plane equation:
-4x + y - z + 27 = 0
f) Find the sine of the angle between straight line A1A4 and plane A1A2A3:
Direction vector straight A1A4:
A1A4 = (5; 6; 6)
Normal vector of the plane A1A2A3:
n = (-4; 1; -8)
The sine of the angle between the vectors:
sin θ = |[А1А4, n]| / |А1А4| * |n|
sin θ = |(48; 38; 29)| / √(5^2 + 6^2 + 6^2) * √(16 + 1 + 64)
sin θ = 115 / √24545
g) Let’s find the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3:
Normal vector of the Oxy coordinate plane:
n = (0; 0; 1)
Normal vector of the plane A1A2A3:
n = (-4; 1; -8)
The cosine of the angle between the vectors:
cos θ = |n1 * n2| / |n1| * |n2|
cos θ = |-8| / √(16 + 1 + 64) * √1
cos θ = -8 / √81
Let's create an equation for a plane passing through the middle of the segment M1M2 perpendicular to this segment if M1(1;5;6); M2(–1;7;10).
Direction vector of segment M1M2:
M1M2 = (-2; 2; 4)
Coordinates of the middle of the segment M1M2:
Mm = ((1 + (-1)) / 2; (5 + 7) / 2; (6 + 10) / 2) = (0; 6; 8)
Normal vector of the desired plane:
n = M1M2 = (-2; 2; 4)
Plane equation:
-2x + 2y + 4z + d = 0
to find d, substitute the coordinates of point Mm:
-2 * 0 + 2 * 6 + 4 * 8 + d = 0
d = -44
Plane equation:
-2x + 2y + 4z - 44 = 0
Prove that the line ...... is parallel to the plane 2x + y – z = 0; and the straight line ..... lies in this plane.
It is necessary to supplement the problem statement with information about the lines so that the answer to the question posed can be given. Please specify the condition.
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The textbook includes more than 100 problems, each of which has detailed solutions and explanations, which will help students better understand the material and improve their performance.
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This digital product is a mathematics textbook containing more than 100 problems with detailed solutions and explanations. It covers various branches of mathematics, including algebra, geometry, probability theory and mathematical statistics. The textbook will be useful for students studying at universities, as well as for schoolchildren who are preparing to enter universities or take part in mathematics competitions.
The digital product is available in pdf format and can be downloaded immediately after payment. In case of problems with downloading or viewing, customers can contact the support team, which is always ready to help. Purchasing this textbook will help students and schoolchildren better understand the material and improve their performance, as well as save time on independently solving problems during exams and sessions.
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I present to you a description of the product:
IDZ 3.1 is a math task that consists of three numbers.
The first issue gives the coordinates of four points in space. It is necessary to write equations for a plane passing through three of these points, a line passing through two of these points, and a plane passing through one of these points and perpendicular to the line. You also need to find the sine and cosine of the angles between a given line and plane.
In the second issue, you need to create an equation for a plane passing through the middle of a segment, perpendicular to this segment, the given coordinates of which are also given.
In the third issue you need to prove that one line is parallel to a given plane, and the other lies in this plane. To do this, it is necessary to use knowledge about the parallelism of a line and a plane, as well as the fact that a point belongs to a plane if its coordinates satisfy the equation of this plane.
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Solution to problem 20.5.7 from the collection of Kepe O.E. - Задача 20.5.7Дано:Кинетический потенциал механической системы определяется выражением L = 16х2 + 20х ... ...