\section{Aufgabe 3}
Gegeben sind: $\vec{a}=\left(3,-1,2\right)\text{, }\vec{b}=\left(1,2,4\right)\text{, }\vec{c}=\left(1,1,1\right)$.

Berechnen Sie:
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\item[a.]$\vec{a}+\vec{b}\text{, }\vec{a}-\vec{b}\text{, }4\cdot\vec{a}\text{, }-\frac{1}{4}\cdot\vec{b}\text{, }-5\cdot\vec{c}$,
\item[b.]$\left|\vec{a}\right|\text{, }\left|\vec{b}\right|\text{, }\left|\vec{c}\right|$,
\item[c.]$\alpha\left(\vec{a},\vec{b}\right)$,
\item[d.]$\vec{a}\times\vec{b}$ und den Flächeninhalt des von $\vec{a}$ und $\vec{b}$ aufgespannten Parallelogramms,
\item[e.]$\left[\vec{a},\vec{b},\vec{c}\right]$ und das Volumen des von $\vec{a},\vec{b}$ und $\vec{c}$ aufgespannten Spats.
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\subsection{Lösung}
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\item[a.]$\underline{\underline{\vec{a}+\vec{b}}}=\left(3,-1,2\right)+\left(1,2,4\right)=\left(3+1,-1+2,2+4\right)=\underline{\underline{\left(4,1,6\right)}}$ \\
$\underline{\underline{\vec{a}-\vec{b}}}=\left(3,-1,2\right)-\left(1,2,4\right)=\left(3-1,-1-2,2-4\right)=\underline{\underline{\left(2,-3,-2\right)}}$ \\
$\underline{\underline {4 \cdot \vec a}}  = 4 \cdot \left( {3. - 1,2} \right) = \left( {4 \cdot 3,4 \cdot \left( { - 1} \right),4 \cdot 2} \right) = \underline{\underline {\left( {12, - 4,8} \right)}} $\\
$ \underline{\underline { - \frac{1}{4} \cdot \vec b}}  =  - \frac{1}{4} \cdot \left( {1,2,4} \right) = \left( {\left( { - \frac{1}{4}} \right) \cdot 1,\left( { - \frac{1}{4}} \right) \cdot 2,\left( { - \frac{1}{4}} \right) \cdot 4} \right) = \underline{\underline {\left( { - \frac{1}{4}, - \frac{1}{2}, - 1} \right)}} $ \\
$ \underline{\underline { - 5 \cdot \vec c}}  =  - 5 \cdot \left( {1,1,1} \right) = \underline{\underline {\left( { - 5, - 5, - 5} \right)}}  $
\item[b.]$ \underline{\underline {\left| {\vec a} \right|}}  = \left| {\left( {3, - 1,4} \right)} \right| = \sqrt {3^2  + \left( { - 1} \right)^2  + 2^2 }  = \sqrt {9 + 1 + 4}  = \underline{\underline {\sqrt {14} }}  \cong 3.741657387 \\
 \underline{\underline {\left| {\vec b} \right|}}  = \left| {\left( {1,2,4} \right)} \right| = \sqrt {1^2  + 2^2  + 4^2 }  = \sqrt {1 + 4 + 16}  = \underline{\underline {\sqrt {21} }}  \cong 4.582575695 \\
 \underline{\underline {\left| {\vec c} \right|}}  = \left| {\left( {1,1,1} \right)} \right| = \sqrt {1^2  + 1^2  + 1^2 }  = \underline{\underline {\sqrt 3 }}  \cong 1.732050808 $
 \item[c.] Sei $ \alpha = \alpha \left( {\vec a,\vec b} \right)$, dann gilt $0 \le \alpha  \le \pi$ und
 $\cos \left( \alpha  \right) = \frac{{\left\langle {\vec a,\vec b} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}}\mathop  = \limits_{vgl.b.} \frac{{\left\langle {\left( {3, - 1,2} \right),\left( {1,2,4} \right)} \right\rangle }}{{\sqrt {14}  \cdot \sqrt {21} }} = \frac{{3 \cdot 1 + \left( { - 1} \right) \cdot 2 + 2 \cdot 4}}{{7 \cdot \sqrt 2  \cdot \sqrt 3 }} = \frac{{3 - 2 + 8}}{{7 \cdot \sqrt 2  \cdot \sqrt 3 }} = \frac{9}{{7 \cdot \sqrt 2  \cdot \sqrt 3 }} $\\
 Der Taschenrechner liefert : $\underline{\underline {\alpha  \cong 1.018209678}}\entspricht 58.33911721^ \circ  $
\item[d.]$ \underline{\underline{\vec a \times \vec b }}= \left( {3, - 1,2} \right) \times \left( {1,2,4} \right) = \left( {\left( { - 1} \right) \cdot 4 - 2 \cdot 2,2 \cdot 1 - 4 \cdot 3,3 \cdot 2 - 1 \cdot \left( { - 1} \right)} \right) = \left( { - 4 - 4,2 - 12,6 + 1} \right) = \underline{\underline {\left( { - 8, - 10,7} \right)}}$  \\
    Der gesuchte Flächeninhalt $F$ ist $F = \left| {\vec a \times \vec b} \right| = \sqrt {\left( { - 8} \right)^2  + \left( { - 10} \right)^2  + 7^2 }  = \sqrt {64 + 100 + 49}  = \underline{\underline {\sqrt {213} }}$  $ \left( { \cong 14.59451952} \right)$
\item[e.]$ \left[ {\vec a,\vec b,\vec c} \right] = \left\langle {\vec a \times \vec b,\vec c} \right\rangle \mathop  = \limits_{vgl.d.} \left\langle {\left( { - 8, - 10,7} \right),\left( {1,1,1} \right)} \right\rangle  =  - 8 - 10 + 7 = \underline{\underline { - 11}} $ \\
    Das gesuchte Volumen ist $\underline{\underline V}  = \left| {\left[ {\vec a,\vec b,\vec c} \right]} \right| = \left| { - 11} \right| = \underline{\underline {11}}$
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