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Lambacher Schweizer Math Solutions for High School

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Lambacher Schweizer Math Solutions for High School

The Lambacher Schweizer Qualifikationsphase textbook covers advanced mathematics topics for upper secondary education in Germany. This summary focuses on solutions to complex geometry problems involving points, planes, and vectors in three-dimensional space.

Key concepts include:
• Calculating coordinates of points on planes
• Finding distances between points and planes
• Determining equations of planes
• Working with vector equations and parametric forms
• Applying the Pythagorean theorem in 3D space

Highlight: The solutions demonstrate step-by-step problem-solving techniques for challenging spatial geometry questions.

26.4.2021

2237

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Page 1 Summary:

This page covers solutions to problems involving points and planes in 3D space. It begins with finding the coordinates of point P(4,1,4,1,5) on a plane given by the equation x+y+2z=6. The solution involves setting up and solving a system of equations. The page also includes calculations of distances between points using the distance formula in three dimensions.

Example: For point P(4,1,4,1,5), the solution shows how to set up the equation 1(4+t) + 1(4+t) + 2(5+2t) = 6 and solve for t to find the coordinates.

Vocabulary: The distance formula in 3D space is given as |PD| = √(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)².

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

Page 3 Summary:

Page 3 focuses on solving problems related to the equation of a plane 3y+4z=0. It includes finding coordinates of multiple points on this plane and calculating distances between these points. The solutions demonstrate how to use vector equations and solve for the parameter t to find point coordinates.

Example: For point A(3,1,-1,1,7), the solution shows how to set up the equation 3(-1+3t) + 4(7+4t) = 0 and solve for t to find the coordinates.

Vocabulary: The vector equation of a line is often written in the form r = a + tb, where a is a point on the line, b is a direction vector, and t is a parameter.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

Page 4 Summary:

This page covers solutions to problems from page 245 of the Lambacher Schweizer Qualifikationsphase NRW PDF, specifically questions 6, 7, and 10. It includes finding points on planes given by equations like x+3y-5z=15 and 2x+10y+11z=252. The solutions demonstrate how to set up parametric equations and solve for coordinates and distances.

Highlight: The solutions show how to handle more complex plane equations with larger coefficients, requiring careful algebraic manipulation.

Example: For the plane 2x+10y+11z=252, the solution demonstrates how to set up the equation 2(3+2t) + 10(1+10t) + 11(1+11t) = 252 and solve for t.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

Page 5 Summary:

The final page continues with solutions to complex geometry problems. It includes finding coordinates of points on planes and calculating distances between points and planes. The solutions demonstrate advanced problem-solving techniques in three-dimensional geometry.

Vocabulary: The normal vector to a plane is a vector perpendicular to every vector contained in the plane.

Highlight: This page emphasizes the importance of systematic problem-solving approaches in spatial geometry, a key skill for students studying Mathe LK Buch Lambacher Schweizer.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

Page 2 Summary:

This page continues with more complex problems involving points and planes. It includes finding coordinates of point S(8,1,3,1,4) on a plane given by 2x+y+2z=9. The solution demonstrates how to set up and solve parametric equations. The page also covers finding another point with the same distance from a given plane.

Definition: A parametric equation represents a set of quantities as explicit functions of a number of independent variables, known as "parameters."

Highlight: The problem of finding another point with the same distance from a plane introduces the concept of symmetry in 3D geometry.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Öffnen

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Philipp, iOS User

Die App ist sehr einfach und gut gestaltet. Bis jetzt habe ich immer alles gefunden, was ich gesucht habe :D

Lena, iOS Userin

Ich liebe diese App ❤️, ich benutze sie eigentlich immer, wenn ich lerne.

Lambacher Schweizer Math Solutions for High School

The Lambacher Schweizer Qualifikationsphase textbook covers advanced mathematics topics for upper secondary education in Germany. This summary focuses on solutions to complex geometry problems involving points, planes, and vectors in three-dimensional space.

Key concepts include:
• Calculating coordinates of points on planes
• Finding distances between points and planes
• Determining equations of planes
• Working with vector equations and parametric forms
• Applying the Pythagorean theorem in 3D space

Highlight: The solutions demonstrate step-by-step problem-solving techniques for challenging spatial geometry questions.

26.4.2021

2237

 

11/12

 

Mathe

59

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Page 1 Summary:

This page covers solutions to problems involving points and planes in 3D space. It begins with finding the coordinates of point P(4,1,4,1,5) on a plane given by the equation x+y+2z=6. The solution involves setting up and solving a system of equations. The page also includes calculations of distances between points using the distance formula in three dimensions.

Example: For point P(4,1,4,1,5), the solution shows how to set up the equation 1(4+t) + 1(4+t) + 2(5+2t) = 6 and solve for t to find the coordinates.

Vocabulary: The distance formula in 3D space is given as |PD| = √(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)².

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Page 3 Summary:

Page 3 focuses on solving problems related to the equation of a plane 3y+4z=0. It includes finding coordinates of multiple points on this plane and calculating distances between these points. The solutions demonstrate how to use vector equations and solve for the parameter t to find point coordinates.

Example: For point A(3,1,-1,1,7), the solution shows how to set up the equation 3(-1+3t) + 4(7+4t) = 0 and solve for t to find the coordinates.

Vocabulary: The vector equation of a line is often written in the form r = a + tb, where a is a point on the line, b is a direction vector, and t is a parameter.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Page 4 Summary:

This page covers solutions to problems from page 245 of the Lambacher Schweizer Qualifikationsphase NRW PDF, specifically questions 6, 7, and 10. It includes finding points on planes given by equations like x+3y-5z=15 and 2x+10y+11z=252. The solutions demonstrate how to set up parametric equations and solve for coordinates and distances.

Highlight: The solutions show how to handle more complex plane equations with larger coefficients, requiring careful algebraic manipulation.

Example: For the plane 2x+10y+11z=252, the solution demonstrates how to set up the equation 2(3+2t) + 10(1+10t) + 11(1+11t) = 252 and solve for t.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Page 5 Summary:

The final page continues with solutions to complex geometry problems. It includes finding coordinates of points on planes and calculating distances between points and planes. The solutions demonstrate advanced problem-solving techniques in three-dimensional geometry.

Vocabulary: The normal vector to a plane is a vector perpendicular to every vector contained in the plane.

Highlight: This page emphasizes the importance of systematic problem-solving approaches in spatial geometry, a key skill for students studying Mathe LK Buch Lambacher Schweizer.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Page 2 Summary:

This page continues with more complex problems involving points and planes. It includes finding coordinates of point S(8,1,3,1,4) on a plane given by 2x+y+2z=9. The solution demonstrates how to set up and solve parametric equations. The page also covers finding another point with the same distance from a given plane.

Definition: A parametric equation represents a set of quantities as explicit functions of a number of independent variables, known as "parameters."

Highlight: The problem of finding another point with the same distance from a plane introduces the concept of symmetry in 3D geometry.

P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =
P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =
P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =
P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =
P(41415)
Ex+y+2z=6
9²-(5)-(2)
:x=
t
1⋅ ( 4 + t) + 1 ⋅ ( 4 + t) + 2・ (5 + 2t) = 6
4+t+4+t+10+ 4t=6
18+ 6t = 6
6t=-12
t = -2
0(21211)
| PDI =

Nichts passendes dabei? Erkunde andere Fachbereiche.

Knowunity ist die #1 unter den Bildungs-Apps in fünf europäischen Ländern

Knowunity wurde bei Apple als "Featured Story" ausgezeichnet und hat die App-Store-Charts in der Kategorie Bildung in Deutschland, Italien, Polen, der Schweiz und dem Vereinigten Königreich regelmäßig angeführt. Werde noch heute Mitglied bei Knowunity und hilf Millionen von Schüler:innen auf der ganzen Welt.

Ranked #1 Education App

Laden im

Google Play

Laden im

App Store

Knowunity ist die #1 unter den Bildungs-Apps in fünf europäischen Ländern

4.9+

Durchschnittliche App-Bewertung

13 M

Schüler:innen lieben Knowunity

#1

In Bildungs-App-Charts in 12 Ländern

950 K+

Schüler:innen haben Lernzettel hochgeladen

Immer noch nicht überzeugt? Schau dir an, was andere Schüler:innen sagen...

iOS User

Ich liebe diese App so sehr, ich benutze sie auch täglich. Ich empfehle Knowunity jedem!! Ich bin damit von einer 4 auf eine 1 gekommen :D

Philipp, iOS User

Die App ist sehr einfach und gut gestaltet. Bis jetzt habe ich immer alles gefunden, was ich gesucht habe :D

Lena, iOS Userin

Ich liebe diese App ❤️, ich benutze sie eigentlich immer, wenn ich lerne.