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Bestimme die Spiegelebene E , die den Punkt \( \mathrm{P}(4|-2|-2) \) auf \( \mathrm{P}^{\prime}(10|6| 8) \) abbildet! \( \mathrm{E}:+0 \square \mathrm{x}_{1}+0 \square \mathrm{x}_{2}+0 \square \mathrm{x}_{3}=+0 \square \) lösen

Ask by Griffiths Wood. in Germany
Feb 24,2025

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Answer

Die Spiegelebene E ist \( 3x + 4y + 5z - 44 = 0 \).

Solution

Um die Spiegelebene E zu bestimmen, über die P(4, –2, –2) zu P′(10, 6, 8) gespiegelt wird, gehen wir folgendermaßen vor: 1. Bestimme das Spiegelmittelpunkt M von P und P′:   M = ((4 + 10)/2, (–2 + 6)/2, (–2 + 8)/2) = (14/2, 4/2, 6/2) = (7, 2, 3). 2. Der Normalenvektor der Spiegelebene ist parallel zur Verbindungsstrecke von P nach P′. Berechne dazu:   v = P′ – P = (10 – 4, 6 – (–2), 8 – (–2)) = (6, 8, 10). 3. Die allgemeine Ebenengleichung lautet mit dem Normalenvektor (a, b, c)   a(x – x₀) + b(y – y₀) + c(z – z₀) = 0, wobei (x₀, y₀, z₀) ein Punkt der Ebene ist (hier M = (7, 2, 3)). Setze ein:   6(x – 7) + 8(y – 2) + 10(z – 3) = 0. 4. Jetzt lösen wir:   6x – 42 + 8y – 16 + 10z – 30 = 0,   6x + 8y + 10z – 88 = 0. 5. Zur Vereinfachung kann man noch durch 2 teilen:   3x + 4y + 5z – 44 = 0. Somit lautet die gesuchte Spiegelebene:   E: 3x + 4y + 5z – 44 = 0.

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Um die Spiegelebene \( E \) zu bestimmen, die den Punkt \( P(4, -2, -2) \) auf \( P'(10, 6, 8) \) abbildet, beginne damit, den Mittelpunkt \( M \) zwischen \( P \) und \( P' \) zu berechnen. Der Mittelpunkt ergibt sich durch den Durchschnitt der Koordinaten: \[ M = \left( \frac{4+10}{2}, \frac{-2+6}{2}, \frac{-2+8}{2} \right) = (7, 2, 3) \] Der Normalenvektor der Ebene lässt sich durch den Vektor \( \vec{n} = P' - P \) finden: \[ \vec{n} = (10 - 4, 6 - (-2), 8 - (-2)) = (6, 8, 10) \] Die allgemeine Form der Ebenengleichung lautet \( a \cdot x_1 + b \cdot x_2 + c \cdot x_3 = d \). Setze \( (a, b, c) = (6, 8, 10) \) und \( M(7, 2, 3) \) ein, um die Gleichung zu bestimmen: \[ 6 \cdot 7 + 8 \cdot 2 + 10 \cdot 3 = d \implies d = 42 + 16 + 30 = 88 \] Daher lautet die Spiegelebene \( E \): \[ 6 x_1 + 8 x_2 + 10 x_3 = 88 \]

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