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| − | __NOCACHE__
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| − | Buch S. 93 / 1 d, e
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| − | {{Lösung versteckt|1=d) <math>\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} 4 \\\ -4 \\\ 0 \end{array}\right) - \left ( \begin{array}{c} 2 \\\ -4 \\\ -5 \end{array}\right) = \left ( \begin{array}{c} 2 \\\ 0 \\\ 5 \end{array}\right)</math> <br>
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| − | <math>\vec {BA} = \vec A-\vec B = \left ( \begin{array}{c} 2 \\\ -4 \\\ -5 \end{array}\right) - \left ( \begin{array}{c} 4 \\\ -4 \\\ 0 \end{array}\right) = \left ( \begin{array}{c} -2 \\\ 0 \\\ -5 \end{array}\right)</math> <br>
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| − | e) <math>\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} -4 \\\ -8 \\\ 0 \end{array}\right) - \left ( \begin{array}{c} 0 \\\ 8 \\\ -2 \end{array}\right) = \left ( \begin{array}{c} -4 \\\ -16 \\\ 2 \end{array} \right)</math> <br>
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| − | <math>\vec {BA} = \vec A-\vec B = \left ( \begin{array}{c} 2 \\\ 3 \\\ -5 \end{array}\right) - \left ( \begin{array}{c} -4 \\\ -8 \\\ 0 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 16 \\\ -2 \end{array} \right)</math>
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| − | Es ist <math>\vec {BA} = -\vec {AB}</math> }}
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| − | Buch S. 93 / 2
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| − | {{Lösung versteckt|1=a) <math>\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math>, also <math>\vec B - \left ( \begin{array}{c} 0 \\\ 3 \\\ 6 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math> oder <math>\left ( \begin{array}{c} b_1 \\\ b_2 \\\ b_3 \end{array}\right) - \left ( \begin{array}{c} 0 \\\ 3 \\\ 6 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math> liefert <math>b_1 = 4, b_2 = 4, b_3=4</math>.
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| − | b) <math>\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math>, also <math> \left ( \begin{array}{c} 2 \\\ 3 \\\ -5 \end{array}\right) - \vec A = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math> oder <math>\left ( \begin{array}{c} 2 \\\ 3 \\\ -5 \end{array}\right) -\left ( \begin{array}{c} a_1 \\\ a_2 \\\ a_3 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math> liefert <math>a_1 = -2, b_2 = 2, b_3=-3</math>.
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| − | c) <math>\left ( \begin{array}{c} b_1 \\\ 0 \\\ 7 \end{array}\right) -\left ( \begin{array}{c} 2 \\\ a_2 \\\ a_3 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math> liefert <math>b_1 = 6, a_2=-1, a_3=9</math>
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| − | d) <math>\left ( \begin{array}{c} -a \\\ a \\\ 3a \end{array}\right) -\left ( \begin{array}{c} a \\\ -3 \\\ 2a \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)</math> liefert <br>
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| − | x<sub>2</sub>-Koordinate: a + 3 = 1, also a = -2 und <math>\vec A= \left ( \begin{array}{c} -2 \\\ -3 \\\ -6 \end{array}\right)</math> und <math>\vec B = \left ( \begin{array}{c} 2 \\\ -2 \\\ -6 \end{array}\right)</math> }}
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| − | Buch S. 93 / 3b
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| − | {{Lösung versteckt|1=<ggb_applet height="300" width="350"
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| − | filename="93-3b.ggb" />
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| − | }}
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| − | Buch S. 93 / 4
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