=for timestamp
Mo Okt  2 17:52:14 CEST 2006

=head2 98. Hausaufgabe

=head3 Geometrie-Buch 208, Aufgabe 1

Berechne die Beträge von

=over

=item a)

M<\left|\left(\!\begin{smallmatrix}3\\4\\12\end{smallmatrix}\!\right)\right| =
\sqrt{3^2 + 4^2 + 12^2} = 13;>

=item b)

M<\left|\left(\!\begin{smallmatrix}4\\-12\\-3\end{smallmatrix}\!\right)\right| =
\left|\left(\!\begin{smallmatrix}3\\4\\12\end{smallmatrix}\!\right)\right| =
13;>

=item c)

M<\left|\left(\!\begin{smallmatrix}12\\-15\\16\end{smallmatrix}\!\right)\right|
= \sqrt{12^2 + 15^2 + 16^2} = 25;>

=back

=head3 Geometrie-Buch 208, Aufgabe 3

Berechne die Einheitsvektoren in Richtung

=for latex
\begin{multicols}{2}

=over

=item d)

M<\displaystyle\frac{\left(\!\begin{smallmatrix}0\\-2\\0\end{smallmatrix}\!\right)}{\left|\left(\!\begin{smallmatrix}0\\-2\\0\end{smallmatrix}\!\right)\right|}
= \frac{\left(\!\begin{smallmatrix}0\\-2\\0\end{smallmatrix}\!\right)}{2} =
\left(\!\begin{smallmatrix}0\\-1\\0\end{smallmatrix}\!\right)\!;>

=item e)

M<\displaystyle\frac{\left(\!\begin{smallmatrix}-1\\-1\\-1\end{smallmatrix}\!\right)}{\left|\left(\!\begin{smallmatrix}-1\\-1\\-1\end{smallmatrix}\!\right)\right|}
=
\frac{\left(\!\begin{smallmatrix}-1\\-1\\-1\end{smallmatrix}\!\right)}{\sqrt{3}} =
\left(\!\begin{smallmatrix}\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{smallmatrix}\!\right)\!;>

=item f)

M<\displaystyle\frac{\left(\!\begin{smallmatrix}8\\-1\\4\end{smallmatrix}\!\right)}{\left|\left(\!\begin{smallmatrix}8\\-1\\4\end{smallmatrix}\!\right)\right|}
=
\frac{\left(\!\begin{smallmatrix}8\\-1\\4\end{smallmatrix}\!\right)}{9} =
\left(\!\begin{smallmatrix}\frac{8}{9}\\-\frac{1}{9}\\\frac{4}{9}\end{smallmatrix}\!\right)\!;>

=item g)

M<\displaystyle\frac{13
\left(\!\begin{smallmatrix}8\\-1\\4\end{smallmatrix}\!\right)}{\left|13
\left(\!\begin{smallmatrix}8\\-1\\4\end{smallmatrix}\!\right)\right|} =
\frac{\left(\!\begin{smallmatrix}8\\-1\\4\end{smallmatrix}\!\right)}{\left|\left(\!\begin{smallmatrix}8\\-1\\4\end{smallmatrix}\!\right)\right|}
=
\left(\!\begin{smallmatrix}\frac{8}{9}\\-\frac{1}{9}\\\frac{4}{9}\end{smallmatrix}\!\right)\!;>

=item h)

M<\displaystyle\frac{\left(\!\begin{smallmatrix}1\\1\\2{,}3\end{smallmatrix}\!\right)}{\left|\left(\!\begin{smallmatrix}1\\1\\2{,}3\end{smallmatrix}\!\right)\right|}
=
\frac{\left(\!\begin{smallmatrix}1\\1\\2{,}3\end{smallmatrix}\!\right)}{2{,}7} =
\left(\!\begin{smallmatrix}\frac{1}{2{,}7}\\\frac{1}{2{,}7}\\\frac{2{,}3}{2{,}7}\end{smallmatrix}\!\right)\!;>

=item i)

M<\displaystyle\frac{\left(\!\begin{smallmatrix}-1\\-1\\\frac{1}{12}\end{smallmatrix}\!\right)}{\left|\left(\!\begin{smallmatrix}-\\-1\\\frac{1}{12}\end{smallmatrix}\!\right)\right|}
=
\frac{\left(\!\begin{smallmatrix}-1\\-1\\\frac{1}{12}\end{smallmatrix}\!\right)}{\frac{17}{12}}
=
\left(\!\begin{smallmatrix}-\frac{12}{17}\\-\frac{12}{17}\\\frac{1}{17}\end{smallmatrix}\!\right)\!;>

=item j)

M<\displaystyle\frac{9
\left(\!\begin{smallmatrix}\frac{7}{5}\\\frac{1}{2}\\\frac{1}{5}\end{smallmatrix}\!\right)}{\left|9
\left(\!\begin{smallmatrix}\frac{7}{5}\\\frac{1}{2}\\\frac{1}{5}\end{smallmatrix}\!\right)\right|}
=
\frac{\left(\!\begin{smallmatrix}\frac{7}{5}\\\frac{1}{2}\\\frac{1}{5}\end{smallmatrix}\!\right)}{\frac{3}{2}}
=
\left(\!\begin{smallmatrix}\frac{14}{15}\\\frac{1}{3}\\\frac{2}{15}\end{smallmatrix}\!\right)\!;>

=item k)

M<\displaystyle\frac{\frac{1}{3}
\left(\!\begin{smallmatrix}-1\\-1\\\frac{7}{4}\end{smallmatrix}\!\right)}{\left|\frac{1}{3}
\left(\!\begin{smallmatrix}-1\\-1\\\frac{7}{4}\end{smallmatrix}\!\right)\right|}
=
\frac{\frac{1}{3}\left(\!\begin{smallmatrix}-1\\-1\\\frac{7}{4}\end{smallmatrix}\!\right)}{\frac{9}{4}}
=
\left(\!\begin{smallmatrix}-\frac{4}{9}\\-\frac{4}{9}\\\frac{7}{9}\end{smallmatrix}\!\right)\!;>

=item l)

M<\displaystyle\frac{
\left(\!\begin{smallmatrix}-3a\\2{,}4a\\3{,}2a\end{smallmatrix}\!\right)}{\left|
\left(\!\begin{smallmatrix}-3a\\2{,}4a\\3{,}2a\end{smallmatrix}\!\right)\right|}
=
\frac{\left(\!\begin{smallmatrix}-3a\\2{,}4a\\3{,}2a\end{smallmatrix}\!\right)}{5
\left|a\right|}
=
\pm \left(\!\begin{smallmatrix}-\frac{3}{5}\\\frac{2{,}4}{5}\\\frac{3{,}2}{5}\end{smallmatrix}\!\right)\!;>

=back

=for latex
\end{multicols}

=head3 Geometrie-Buch 208, Aufgabe 4

Berechne M<a>.

=over

=item a)

M<\left|\left(\!\begin{smallmatrix}3a\\-6a\\2a\end{smallmatrix}\!\right)\right|
= \sqrt{9a^2 + 36a^2 + 4a^2} = 7 \left|a\right| \stackrel{!}{=} 14;>

⇔ M<\left|a\right| = 2;>

=item b)

M<\left|\left(\!\begin{smallmatrix}a\\2a\\a-1\end{smallmatrix}\!\right)\right|
= \sqrt{a^2 + 4a^2 + a^2 - 2a + 1} = \sqrt{6a^2 - 2a + 1} \stackrel{!}{=} 7;>

⇔ M<a_1 = -\frac{8}{3}; \quad a_2 = 3;>

=back

"z.B. [will länger ausholen]... ah ne; die, die in Physik sind, wissen's eh,
und die, die's nicht sind, wissen's halt nicht..."