M11 Aufgabe zu Vektoren

Buch S. 93 / 1 d, e

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d) $\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} 4 \\\ -4 \\\ 0 \end{array}\right) - \left ( \begin{array}{c} 2 \\\ -4 \\\ -5 \end{array}\right) = \left ( \begin{array}{c} 2 \\\ 0 \\\ 5 \end{array}\right)$
$\vec {BA} = \vec A-\vec B = \left ( \begin{array}{c} 2 \\\ -4 \\\ -5 \end{array}\right) - \left ( \begin{array}{c} 4 \\\ -4 \\\ 0 \end{array}\right) = \left ( \begin{array}{c} -2 \\\ 0 \\\ -5 \end{array}\right)$
e) $\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} -4 \\\ -8 \\\ 0 \end{array}\right) - \left ( \begin{array}{c} 0 \\\ 8 \\\ -2 \end{array}\right) = \left ( \begin{array}{c} -4 \\\ -16 \\\ 2 \end{array} \right)$
$\vec {BA} = \vec A-\vec B = \left ( \begin{array}{c} 2 \\\ 3 \\\ -5 \end{array}\right) - \left ( \begin{array}{c} -4 \\\ -8 \\\ 0 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 16 \\\ -2 \end{array} \right)$

Es ist $\vec {BA} = -\vec {AB}$

Buch S. 93 / 2

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a) $\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$, also $\vec B - \left ( \begin{array}{c} 0 \\\ 3 \\\ 6 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$ oder $\left ( \begin{array}{c} b_1 \\\ b_2 \\\ b_3 \end{array}\right) - \left ( \begin{array}{c} 0 \\\ 3 \\\ 6 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$ liefert $b_1 = 4, b_2 = 4, b_3=4$.

b) $\vec {AB} = \vec B-\vec A = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$, also $\left ( \begin{array}{c} 2 \\\ 3 \\\ -5 \end{array}\right) - \vec A = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$ oder $\left ( \begin{array}{c} 2 \\\ 3 \\\ -5 \end{array}\right) -\left ( \begin{array}{c} a_1 \\\ a_2 \\\ a_3 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$ liefert $a_1 = -2, b_2 = 2, b_3=-3$.

c) $\left ( \begin{array}{c} b_1 \\\ 0 \\\ 7 \end{array}\right) -\left ( \begin{array}{c} 2 \\\ a_2 \\\ a_3 \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$ liefert $b_1 = 6, a_2=-1, a_3=9$

d) $\left ( \begin{array}{c} -a \\\ a \\\ 3a \end{array}\right) -\left ( \begin{array}{c} a \\\ -3 \\\ 2a \end{array}\right) = \left ( \begin{array}{c} 4 \\\ 1 \\\ -2 \end{array}\right)$ liefert

x₂-Koordinate: a + 3 = 1, also a = -2 und $\vec A= \left ( \begin{array}{c} -2 \\\ -3 \\\ -6 \end{array}\right)$ und $\vec B = \left ( \begin{array}{c} 2 \\\ -2 \\\ -6 \end{array}\right)$

Buch S. 93 / 3b

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Buch S. 93 / 4