Aufgabe:
Problem/Ansatz:
Text erkannt:
10:52 \&
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\( \leftarrow \quad \) mathewiederholung-Isg.pdf
\( \begin{array}{l} u=\mid \overline{2})+(2) \\ a^{2}=\frac{d^{2}}{4}+\frac{e^{2}}{4}-\frac{d e}{2} \cos \delta 1 \cdot 4 \\ 4 a^{2}=d^{2}+e^{2}-2 d e \cdot \cos \delta \\ 2 d e \cdot \cos \delta=d^{2}+e^{2}-4 a^{2} \Rightarrow \delta=\cos ^{-1}\left(\frac{d^{2}+e^{2}-4 a^{2}}{2 d e}\right) \\ \delta=72,76 \end{array} \)
Afg 10:
\( A(2 / 01-4), B(8|-2| 6), C(6 / 2 / 0), s_{t}(1|t| 1-t), t \in \mathbb{R} \)
10.1
\( |\overrightarrow{B P}|=26 \Rightarrow P=\text { ? } \)
\( \begin{array}{l} \overrightarrow{O P}=\overrightarrow{O P}-\overrightarrow{O B}=\left(\begin{array}{c} 0 \\ x \\ 0 \end{array}\right)-\left(\begin{array}{c} 8 \\ -2 \\ 6 \end{array}\right)=\left(\begin{array}{c} -8 \\ x+2 \\ -6 \end{array}\right) \\ |\overrightarrow{B P}|=26 \Leftrightarrow \sqrt{64+(x+2)^{2}+36}=26 /^{2} \\ \Rightarrow(x+2)^{2}+100=676 \Leftrightarrow(x+2)^{2}=576 / \sqrt{ } \\ \Rightarrow\left(x+2= \pm 24 \Rightarrow x_{1 / 2}= \pm 24-2\right. \\ x_{1}=24-2=22 \Rightarrow \frac{P_{1}(0 / 22 / 0)}{P_{2}(0 /-26 / 0)} \end{array} \)
10.2
\( \begin{array}{l} P \in x_{1} x_{3} \text {-Elone } \Rightarrow P\left(x_{1}|o| x_{2}\right) \\ \overrightarrow{B P} \| \vec{e}_{2} \Rightarrow \overrightarrow{B P}=\lambda \vec{e}_{2}=\left(\begin{array}{l} 0 \\ d \\ 0 \end{array}\right) \\ \Rightarrow|\overrightarrow{B P}|=\left|\lambda \vec{e}_{2}\right|=|\lambda| \end{array} \)
\( x_{1} \)
\( \overrightarrow{B P}=\overrightarrow{O P}-\overrightarrow{O B}=\left(\begin{array}{l} x_{1} \\ 0 \\ x_{2} \end{array}\right)-\left(\begin{array}{c} 8 \\ -2 \end{array}\right)=\left(\begin{array}{c} x_{1}-8 \\ x_{2}-6 \end{array}\right) \)
\( \begin{aligned} \Rightarrow\left(\begin{array}{c} x_{1}-8 \\ x_{2}-6 \end{array}\right)=\left(\begin{array}{l} 0 \\ c_{1} \\ 0 \end{array}\right) & \Rightarrow \lambda=2=d(B, p) \\ & \Rightarrow P(81016) \end{aligned} \)
10.3
\( \begin{array}{c} M=? \\ \overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O C})=\frac{1}{2}\left(\begin{array}{c} 8 \\ 2 \\ -4 \end{array}\right)=\left(\begin{array}{c} 4 \\ 1 \\ -2 \end{array}\right) \Rightarrow M(4 / 1 /-2) \end{array} \)
10.4
\( \begin{array}{c} D=? \quad \overrightarrow{A D}=\overrightarrow{B C} \\ \overrightarrow{O D}-\overrightarrow{O A}=\overrightarrow{O C}-\overrightarrow{O B} \\ \overrightarrow{O D}=\overrightarrow{O C}-\overrightarrow{O B}+\overrightarrow{O A} \\ \vec{A}=\left(\begin{array}{c} 6 \\ 2 \\ 0 \end{array}\right)-\left(\begin{array}{c} 8 \\ -2 \\ 6 \end{array}\right)+\left(\begin{array}{c} 2 \\ 0 \\ -4 \end{array}\right)=\left(\begin{array}{c} 0 \\ 4 \\ -10 \end{array}\right) \Rightarrow D(014 /-10) \end{array} \)
Hallo kann mir hier bei der markierten Aufgabe
erklären wie man auf x=2 kommt?
