Änderungen von Dokument BPE 16.5 Gegenseitige Lage von Ebenen und Geraden
Zuletzt geändert von Anna Kukin am 2026/05/25 18:19
Von Version 10.1
bearbeitet von Dirk Tebbe
am 2026/04/27 16:18
am 2026/04/27 16:18
Änderungskommentar:
Es gibt keinen Kommentar für diese Version
Auf Version 21.1
bearbeitet von Holger Engels
am 2026/04/28 13:07
am 2026/04/28 13:07
Änderungskommentar:
Neuen Anhang 3 Ebenen schneiden sich in einer Gerade.ggb hochladen
Zusammenfassung
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Seiteneigenschaften (2 geändert, 0 hinzugefügt, 0 gelöscht)
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... ... @@ -5,23 +5,50 @@ 5 5 [[Kompetenzen.K5]] Ich kann eine Gleichung der Schnittgerade zwischen zwei Ebenen bestimmmen. 6 6 [[Kompetenzen.K5]] [[Kompetenzen.K4]] Ich kann Geraden und Ebenen angeben, die gegebene Lagebeziehungen erfüllen. 7 7 8 -{{aufgabe id="Richtungsvektor und Spannvektoren" afb="I" kompetenzen="K1, K6" quelle="Holger Engels" zeit="3"}} 9 -Der Richtungsvektor einer Geraden //g// ist als Linearkombination der Spannvektoren der Ebene //E// darstellbar. Erläutere, was sich daraus über die Lage von //g// in Bezug auf //E// sagen lässt. 8 +{{aufgabe id="Aussagen" afb="II" kompetenzen="K1, K6" quelle="Holger Engels" zeit="6"}} 9 +(%class=abc%) 10 +1. Der Richtungsvektor einer Geraden //g// ist als Linearkombination der Spannvektoren der Ebene //E// darstellbar. Erläutere, was sich daraus über die Lage von //g// in Bezug auf //E// sagen lässt. 11 +1. Die Ebenen //E// und //F// teilen sich einen Spannvektor. Erläutere, was sich daraus über die Lage der beiden Ebenen zueinander sagen lässt. 12 +1. Die Ebenen //E// und //F// teilen sich den Punkt //P//. Erläutere, was sich daraus über die Lage der beiden Ebenen zueinander sagen lässt. 10 10 {{/aufgabe}} 11 11 12 -{{aufgabe id="Gemeinsamer Spannvektor" afb="I" kompetenzen="K1, K6" quelle="Holger Engels" zeit="3"}} 13 -Die Ebenen //E// und //F// teilen sich einen Spannvektor. Erläutere, was sich daraus über die Lage der beiden Ebenen zueinander sagen lässt. 14 -{{/aufgabe}} 15 - 16 -{{aufgabe id="Schnittgerade" afb="II" kompetenzen="K1, K6" quelle="Holger Engels" zeit="10"}} 17 -Es sind zwei Ebenen //E// und //F// gegeben: 18 -{{formula}}E: 2x_1-3x_2+x_3=0{{/formula}} 19 -{{formula}}F: 3x_1+2x_2=-1{{/formula}} 15 +{{aufgabe id="Schnittgerade" afb="II" kompetenzen="K1, K6" quelle="Martin Stern, Dirk Tebbe" zeit="15"}} 16 +Es sind zwei Ebenen //E// und //F// gegeben durch {{formula}}E: 2x_1-3x_2+x_3=0{{/formula}} und {{formula}}F: 3x_1+2x_2=-1{{/formula}}. 20 20 (%class=abc%) 21 -1. Zeige //E// und //F// schneiden sich. 22 22 1. Bestimme die Schnittgerade //g//. 23 -1. Nimm Stellung zu folgender Aussage: 24 -//Die Projektion von E auf F ergibt g. Die Projektion von F auf E ergibt ebenfalls g//. 19 +1. Welche besondere Lage haben die beiden Ebenen zueinander? 25 25 {{/aufgabe}} 26 26 27 - 22 +{{aufgabe id="Lösungsmenge geometrisch" afb="II" kompetenzen="K6" quelle="Frauke Beckstette" zeit="12"}} 23 +Ordne den folgenden linearen Gleichungssystemen jeweils die passende Abbildung zu. Begründe deine Entscheidung. 24 +Visualisiere das verbliebene LGS analog. 25 +(%class="abc horiz"%) 26 +1. (%style="vertical-align: top"%){{formula}} 27 +\begin{aligned} 28 + x_1 + x_2 &= 1 \\ 29 + - 3x_2 &= 8 \\ 30 + -x_1 + 2x_2 + x_3 &= 4 31 +\end{aligned} 32 +{{/formula}} 33 +1. (%style="vertical-align: top"%){{formula}} 34 +\begin{aligned} 35 + 3x_1 - 2x_2 + x_3 &= 7 \\ 36 + -6x_1 + 4x_2 - 2x_3 &= 3 \\ 37 + 15x_1 - 10x_2 + 5x_3 &= 5 38 +\end{aligned} 39 +{{/formula}} 40 +1. (%style="vertical-align: top"%){{formula}} 41 +\begin{aligned} 42 + 2x_1 - 2x_2 + 2x_3 &= 2 \\ 43 + -2x_1 - 6x_2 + 2x_3 &= 0 \\ 44 + 2x_1 + 2x_2 &= 1 45 +\end{aligned} 46 +{{/formula}} 47 +1. (%style="vertical-align: top"%){{formula}} 48 +\begin{aligned} 49 + x_1 + 3x_2 - 2x_3 &= 5 \\ 50 + -2x_1 - 6x_2 + 4x_3 &= 1 \\ 51 + 2x_1 + x_3 &= 3 52 +\end{aligned} 53 +{{/formula}} 54 +{{/aufgabe}}
- 3 Ebenen davon zwei parallel.ggb
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- 3 Ebenen davon zwei parallel.svg
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- 3 Ebenen schneiden sich in einer Gerade.ggb
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