=for timestamp
Di Mär 28 18:16:11 CEST 2006

=head2 67. Hausaufgabe

=head3 Geometrie-Buch Seite 191, Aufgabe 6

M<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)\!;>

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

M<k{:}\, \vec X = \vec A + \beta \vec w;>

=over

=item *

Gleichsetzen von M<\vec X_k> mit M<\vec X_g> und M<\vec X_h> bringt:

M<\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}>

=item *

Elimination von M<\lambda> (I.):

M<\lambda = \dfrac{A_1 - G_1 + \beta_g w_1}{g_1};>

=item *

Elimination von M<\beta_g> (II.):

M<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);>

⇔ M<\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};>

=item *

Elimination von M<w_3> (III.):

M<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};>

=item *

Elimination von M<\mu> (IV.):

M<\mu = \dfrac{A_1 - H_1 + \beta_h w_1}{h_1};>

=item *

Elimination von M<w_2> (V.):

M<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};>

=item *

Elimination von M<\beta_h> (VI.):

M<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;>

M<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;>

[...]

M<p := \frac{h_2}{h_1} - \frac{g_2}{g_1};>

M<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;>

=item *

Auflösen nach M<\beta_h>:

M<\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};>

=item *

Speziell für M<w_1 = 0>:

M<\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);>

M<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\}\!;>

=back

=begin comment

Maxima:

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=end comment

=head3 Geometrie-Buch Seite 191, Aufgabe 7

M<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)\!;>

Welche Schargerade ist parallel zu M<E>? Ist sie echt parallel?

M<\begin{vmatrix}1&1&-1-a\\1&-1&-1+a\\0&3&-1\end{vmatrix} = 2 - 6a = 0;> ⇔ M<a
= \frac{1}{3};>

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

M<\begin{vmatrix}1&1&-3\\1&-1&-2\\0&3&1\end{vmatrix} = -5;> ⇔ M<g_{\frac{1}{3}}
\cap E = \varnothing;>