«K12/K13» 67. Hausaufgabe

- - 0.0.1 67. Hausaufgabe
    - 0.0.1.1 Geometrie-Buch Seite 191, Aufgabe 6
    - 0.0.1.2 Geometrie-Buch Seite 191, Aufgabe 7

0.0.1 ↑ 67. Hausaufgabe

0.0.1.1 ↑ Geometrie-Buch Seite 191, Aufgabe 6

A(2, -1, 0); \quad g{:}\, \vec X = \left(\!\begin{smallmatrix}-2\\6\\1\end{smallmatrix}\!\right) + \lambda \left(\!\begin{smallmatrix}2\\-1\\3\end{smallmatrix}\!\right)\!; \quad h{:}\, \vec X = \left(\!\begin{smallmatrix}3\\0\\-4\end{smallmatrix}\!\right) + \mu \left(\!\begin{smallmatrix}-2\\4\\5\end{smallmatrix}\!\right)\!; $A (2, - 1, 0); g : \vec{X} = (\begin{matrix} - 2 \\ 6 \\ 1 \end{matrix}) + λ (\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix}); h : \vec{X} = (\begin{matrix} 3 \\ 0 \\ - 4 \end{matrix}) + μ (\begin{matrix} - 2 \\ 4 \\ 5 \end{matrix});$

Stelle eine Gleichung der Gerade k $k$ auf, die durch A $A$ geht und g $g$ und h $h$ schneidet. Berechne die Schnittpunkte.

k{:}\, \vec X = \vec A + \beta \vec w; $k : \vec{X} = \vec{A} + β \vec{w};$

Gleichsetzen von \vec X_k ${\vec{X}}_{k}$ mit \vec X_g ${\vec{X}}_{g}$ und \vec X_h ${\vec{X}}_{h}$ bringt:

\begin{array}{lrcl} {} \text{I.} & A_1 + \beta_g w_1 &=& G_1 + \lambda g_1; \\ {} \text{II.} & A_2 + \beta_g w_2 &=& G_2 + \lambda g_2; \\ {} \text{III.} & A_3 + \beta_g w_3 &=& G_3 + \lambda g_3; \\ {} \text{IV.} & A_1 + \beta_h w_1 &=& H_1 + \mu h_1; \\ {} \text{V.} & A_2 + \beta_h w_2 &=& H_2 + \mu h_2; \\ {} \text{VI.} & A_3 + \beta_h w_3 &=& H_3 + \mu h_3; \end{array} $\begin{matrix} I. & A_{1} + β_{g} w_{1} & = & G_{1} + λ g_{1}; \\ II. & A_{2} + β_{g} w_{2} & = & G_{2} + λ g_{2}; \\ III. & A_{3} + β_{g} w_{3} & = & G_{3} + λ g_{3}; \\ IV. & A_{1} + β_{h} w_{1} & = & H_{1} + μ h_{1}; \\ V. & A_{2} + β_{h} w_{2} & = & H_{2} + μ h_{2}; \\ VI. & A_{3} + β_{h} w_{3} & = & H_{3} + μ h_{3}; \end{matrix}$
Elimination von \lambda $λ$ (I.):

\lambda = \dfrac{A_1 - G_1 + \beta_g w_1}{g_1}; $λ = \frac{A_{1} - G_{1} + β_{g} w_{1}}{g_{1}};$
Elimination von \beta_g $β_{g}$ (II.):

A_2 + \beta_g w_2 = G_2 + \lambda g_2 = G_2 + \frac{g_2}{g_1} \left(A_1 - G_1 + \beta_g w_1\right); $A_{2} + β_{g} w_{2} = G_{2} + λ g_{2} = G_{2} + \frac{g_{2}}{g_{1}} (A_{1} - G_{1} + β_{g} w_{1});$

⇔ \beta_g = \dfrac{\overbrace{G_2 - A_2 + \frac{g_2}{g_1} A_1 - \frac{g_2}{g_1} G_1}^o}{w_2 - \frac{g_2}{g_1} w_1}; $β_{g} = \frac{\overset{o}{\overset{︷}{G_{2} - A_{2} + \frac{g_{2}}{g_{1}} A_{1} - \frac{g_{2}}{g_{1}} G_{1}}}}{w_{2} - \frac{g_{2}}{g_{1}} w_{1}};$
Elimination von w_3 $w_{3}$ (III.):

w_3 = \dfrac{\left(w_2 - \frac{g_2}{g_1} w_1\right)\left(G_3 - A_3 + \lambda g_3\right)}{\underbrace{G_2 - A_2 + \frac{g_2}{g_1} A_1 - \frac{g_2}{g_1} G_1}_k}; $w_{3} = \frac{(w_{2} - \frac{g_{2}}{g_{1}} w_{1}) (G_{3} - A_{3} + λ g_{3})}{\underset{k}{\underset{︸}{G_{2} - A_{2} + \frac{g_{2}}{g_{1}} A_{1} - \frac{g_{2}}{g_{1}} G_{1}}}};$
Elimination von \mu $μ$ (IV.):

\mu = \dfrac{A_1 - H_1 + \beta_h w_1}{h_1}; $μ = \frac{A_{1} - H_{1} + β_{h} w_{1}}{h_{1}};$
Elimination von w_2 $w_{2}$ (V.):

w_2 = \dfrac{H_2 - A_2 + \mu h_2}{\beta_h} = \dfrac{\overbrace{H_2 - A_2 + \frac{h_2}{h_1} A_1 - \frac{h_2}{h_1} H1}^l + \frac{h_2}{h_1} \beta_h w_1}{\beta_h}; $w_{2} = \frac{H_{2} - A_{2} + μ h_{2}}{β_{h}} = \frac{\overset{l}{\overset{︷}{H_{2} - A_{2} + \frac{h_{2}}{h_{1}} A_{1} - \frac{h_{2}}{h_{1}} H 1}} + \frac{h_{2}}{h_{1}} β_{h} w_{1}}{β_{h}};$
Elimination von \beta_h $β_{h}$ (VI.):

A_3 + \beta_h w_3 = \underbrace{H_3 + \frac{h_3}{h_1} A_1 - \frac{h_3}{h_1} H_1}_m + \frac{h_3}{h_1} \beta_h w_1; $A_{3} + β_{h} w_{3} = \underset{m}{\underset{︸}{H_{3} + \frac{h_{3}}{h_{1}} A_{1} - \frac{h_{3}}{h_{1}} H_{1}}} + \frac{h_{3}}{h_{1}} β_{h} w_{1};$

A_3 + \dfrac{\beta_h \left(w_2 - \frac{g_2}{g_1} w_1\right)\left(\overbrace{G_3 + \frac{g_3}{g_1} A_1 - \frac{g_3}{g_1} G_1 - A_3}^n + \frac{g_3}{g_1} \beta_g w_1\right)}{k} = m + \frac{h_3}{h_1} \beta_h w_1; $A_{3} + \frac{β_{h} (w_{2} - \frac{g_{2}}{g_{1}} w_{1}) (\overset{n}{\overset{︷}{G_{3} + \frac{g_{3}}{g_{1}} A_{1} - \frac{g_{3}}{g_{1}} G_{1} - A_{3}}} + \frac{g_{3}}{g_{1}} β_{g} w_{1})}{k} = m + \frac{h_{3}}{h_{1}} β_{h} w_{1};$

[...]

p := \frac{h_2}{h_1} - \frac{g_2}{g_1}; $p : = \frac{h_{2}}{h_{1}} - \frac{g_{2}}{g_{1}};$

A_3 l + A_3 \beta_h p + \frac{l^2}{k} n + \frac{l}{k} n \beta_h p + \frac{g_2}{g_1} \frac{l}{k} w_1 o + \beta_h \frac{w_1}{k} p n + \beta_h \frac{w_1}{k} p \frac{g_2}{g_1} w_1 o = l m + \beta_h p m + \frac{h_3}{h_1} \beta_h + w_1 l + \frac{h_3}{h_1} \beta_h^2 w_1 p; $A_{3} l + A_{3} β_{h} p + \frac{l^{2}}{k} n + \frac{l}{k} n β_{h} p + \frac{g_{2}}{g_{1}} \frac{l}{k} w_{1} o + β_{h} \frac{w_{1}}{k} p n + β_{h} \frac{w_{1}}{k} p \frac{g_{2}}{g_{1}} w_{1} o = l m + β_{h} p m + \frac{h_{3}}{h_{1}} β_{h} + w_{1} l + \frac{h_{3}}{h_{1}} β_{h}^{2} w_{1} p;$
Auflösen nach \beta_h $β_{h}$ :

\beta_h = \dfrac{5 w_1^2 + 36 w_1 - 57 \pm \sqrt{25 w_1^4 + 360 w_1^3 - 274 w_1^2 + 7296 w_1 + 3249}}{50 w_1}; $β_{h} = \frac{5 w_{1}^{2} + 36 w_{1} - 57 \pm \sqrt{25 w_{1}^{4} + 360 w_{1}^{3} - 274 w_{1}^{2} + 7296 w_{1} + 3249}}{50 w_{1}};$
Speziell für w_1 = 0 $w_{1} = 0$ :

\left(\beta_g, \beta_h, \lambda, \mu, w_1, w_2, w_3\right) = \left(\frac{10}{3}, 2, 2, \frac{1}{2}, 0, \frac{3}{2}, \frac{21}{10}\right); $(β_{g}, β_{h}, λ, μ, w_{1}, w_{2}, w_{3}) = (\frac{10}{3}, 2, 2, \frac{1}{2}, 0, \frac{3}{2}, \frac{21}{10});$

k \cap g = \left\{ \left(\!\begin{smallmatrix}2\\4\\7\end{smallmatrix}\!\right) \right\}\!; \quad k \cap h = \left\{ \left(\!\begin{smallmatrix}2\\2\\-\frac{3}{2}\end{smallmatrix}\!\right) \right\}\!; $k \cap g = \{(\begin{matrix} 2 \\ 4 \\ 7 \end{matrix})\}; k \cap h = \{(\begin{matrix} 2 \\ 2 \\ - \frac{3}{2} \end{matrix})\};$

0.0.1.2 ↑ Geometrie-Buch Seite 191, Aufgabe 7

E{:}\, \vec X = \left(\!\begin{smallmatrix}3\\2\\-1\end{smallmatrix}\!\right) + \lambda \left(\!\begin{smallmatrix}1\\1\\0\end{smallmatrix}\!\right) + \mu \left(\!\begin{smallmatrix}1\\-1\\3\end{smallmatrix}\!\right)\!; \quad g_a{:}\, \vec X= \sigma \left(\!\begin{smallmatrix}1+a\\1-a\\1\end{smallmatrix}\!\right)\!; $E : \vec{X} = (\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}) + λ (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 3 \end{matrix}); g_{a} : \vec{X} = σ (\begin{matrix} 1 + a \\ 1 - a \\ 1 \end{matrix});$

Welche Schargerade ist parallel zu E $E$ ? Ist sie echt parallel?

\begin{vmatrix}1&1&-1-a\\1&-1&-1+a\\0&3&-1\end{vmatrix} = 2 - 6a = 0; $|\begin{matrix} 1 & 1 & - 1 - a \\ 1 & - 1 & - 1 + a \\ 0 & 3 & - 1 \end{matrix}| = 2 - 6 a = 0;$ ⇔ a = \frac{1}{3}; $a = \frac{1}{3};$

Verbindungsvektor der Aufpunkte: \vec d = \left(\!\begin{smallmatrix}-3\\-2\\1\end{smallmatrix}\!\right)\!; $\vec{d} = (\begin{matrix} - 3 \\ - 2 \\ 1 \end{matrix});$

\begin{vmatrix}1&1&-3\\1&-1&-2\\0&3&1\end{vmatrix} = -5; $|\begin{matrix} 1 & 1 & - 3 \\ 1 & - 1 & - 2 \\ 0 & 3 & 1 \end{matrix}| = - 5;$ ⇔ g_{\frac{1}{3}} \cap E = \varnothing; $g_{\frac{1}{3}} \cap E = \emptyset;$

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0.0.1 ↑ 67. Hausaufgabe

0.0.1.1 ↑ Geometrie-Buch Seite 191, Aufgabe 6

0.0.1.2 ↑ Geometrie-Buch Seite 191, Aufgabe 7