«K12/K13» 98. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 98. Hausaufgabe

0.0.1.1 ↑ Geometrie-Buch 208, Aufgabe 1

Berechne die Beträge von

a)

\left|\left(\!\begin{smallmatrix}3\\4\\12\end{smallmatrix}\!\right)\right| = \sqrt{3^2 + 4^2 + 12^2} = 13;$\left|\left(\begin{array}{c}3\\ 4\\ 12\end{array}\right)\right|=\sqrt{3^2+4^2+12^2}=13;$

b)

\left|\left(\!\begin{smallmatrix}4\\-12\\-3\end{smallmatrix}\!\right)\right| = \left|\left(\!\begin{smallmatrix}3\\4\\12\end{smallmatrix}\!\right)\right| = 13;$\left|\left(\begin{array}{c}4\\ -12\\ -3\end{array}\right)\right|=\left|\left(\begin{array}{c}3\\ 4\\ 12\end{array}\right)\right|=13;$

c)

\left|\left(\!\begin{smallmatrix}12\\-15\\16\end{smallmatrix}\!\right)\right| = \sqrt{12^2 + 15^2 + 16^2} = 25;$\left|\left(\begin{array}{c}12\\ -15\\ 16\end{array}\right)\right|=\sqrt{12^2+15^2+16^2}=25;$

0.0.1.2 ↑ Geometrie-Buch 208, Aufgabe 3

Berechne die Einheitsvektoren in Richtung

d)

\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)\!;$\frac{\left(\begin{array}{c}0\\ -2\\ 0\end{array}\right)}{\left|\left(\begin{array}{c}0\\ -2\\ 0\end{array}\right)\right|}=\frac{\left(\begin{array}{c}0\\ -2\\ 0\end{array}\right)}{2}=\left(\begin{array}{c}0\\ -1\\ 0\end{array}\right);$

e)

\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)\!;$\frac{\left(\begin{array}{c}-1\\ -1\\ -1\end{array}\right)}{\left|\left(\begin{array}{c}-1\\ -1\\ -1\end{array}\right)\right|}=\frac{\left(\begin{array}{c}-1\\ -1\\ -1\end{array}\right)}{\sqrt{3}}=\left(\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{array}\right);$

f)

\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)\!;$\frac{\left(\begin{array}{c}8\\ -1\\ 4\end{array}\right)}{\left|\left(\begin{array}{c}8\\ -1\\ 4\end{array}\right)\right|}=\frac{\left(\begin{array}{c}8\\ -1\\ 4\end{array}\right)}{9}=\left(\begin{array}{c}\frac{8}{9}\\ -\frac{1}{9}\\ \frac{4}{9}\end{array}\right);$

g)

\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)\!;$\frac{13\left(\begin{array}{c}8\\ -1\\ 4\end{array}\right)}{\left|13\left(\begin{array}{c}8\\ -1\\ 4\end{array}\right)\right|}=\frac{\left(\begin{array}{c}8\\ -1\\ 4\end{array}\right)}{\left|\left(\begin{array}{c}8\\ -1\\ 4\end{array}\right)\right|}=\left(\begin{array}{c}\frac{8}{9}\\ -\frac{1}{9}\\ \frac{4}{9}\end{array}\right);$

h)

\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)\!;$\frac{\left(\begin{array}{c}1\\ 1\\ 2,3\end{array}\right)}{\left|\left(\begin{array}{c}1\\ 1\\ 2,3\end{array}\right)\right|}=\frac{\left(\begin{array}{c}1\\ 1\\ 2,3\end{array}\right)}{2,7}=\left(\begin{array}{c}\frac{1}{2,7}\\ \frac{1}{2,7}\\ \frac{2,3}{2,7}\end{array}\right);$

i)

\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)\!;$\frac{\left(\begin{array}{c}-1\\ -1\\ \frac{1}{12}\end{array}\right)}{\left|\left(\begin{array}{c}-\\ -1\\ \frac{1}{12}\end{array}\right)\right|}=\frac{\left(\begin{array}{c}-1\\ -1\\ \frac{1}{12}\end{array}\right)}{\frac{17}{12}}=\left(\begin{array}{c}-\frac{12}{17}\\ -\frac{12}{17}\\ \frac{1}{17}\end{array}\right);$

j)

\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)\!;$\frac{9\left(\begin{array}{c}\frac{7}{5}\\ \frac{1}{2}\\ \frac{1}{5}\end{array}\right)}{\left|9\left(\begin{array}{c}\frac{7}{5}\\ \frac{1}{2}\\ \frac{1}{5}\end{array}\right)\right|}=\frac{\left(\begin{array}{c}\frac{7}{5}\\ \frac{1}{2}\\ \frac{1}{5}\end{array}\right)}{\frac{3}{2}}=\left(\begin{array}{c}\frac{14}{15}\\ \frac{1}{3}\\ \frac{2}{15}\end{array}\right);$

k)

\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)\!;$\frac{\frac{1}{3}\left(\begin{array}{c}-1\\ -1\\ \frac{7}{4}\end{array}\right)}{\left|\frac{1}{3}\left(\begin{array}{c}-1\\ -1\\ \frac{7}{4}\end{array}\right)\right|}=\frac{\frac{1}{3}\left(\begin{array}{c}-1\\ -1\\ \frac{7}{4}\end{array}\right)}{\frac{9}{4}}=\left(\begin{array}{c}-\frac{4}{9}\\ -\frac{4}{9}\\ \frac{7}{9}\end{array}\right);$

l)

\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)\!;$\frac{\left(\begin{array}{c}-3a\\ 2,4a\\ 3,2a\end{array}\right)}{\left|\left(\begin{array}{c}-3a\\ 2,4a\\ 3,2a\end{array}\right)\right|}=\frac{\left(\begin{array}{c}-3a\\ 2,4a\\ 3,2a\end{array}\right)}{5\left|a\right|}=\pm \left(\begin{array}{c}-\frac{3}{5}\\ \frac{2,4}{5}\\ \frac{3,2}{5}\end{array}\right);$

0.0.1.3 ↑ Geometrie-Buch 208, Aufgabe 4

Berechne a$a$.

a)

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

⇔ \left|a\right| = 2;$\left|a\right|=2;$

b)

\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;$\left|\left(\begin{array}{c}a\\ 2a\\ a-1\end{array}\right)\right|=\sqrt{a^2+4a^2+a^2-2a+1}=\sqrt{6a^2-2a+1}\stackrel{!}{=}7;$

⇔ a_1 = -\frac{8}{3}; \quad a_2 = 3;$a_1=-\frac{8}{3};a_2=3;$

"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..."