Es geht um Orthonormalbasen und irgendwie mache ich beim 3. Vektor immer einen Fehler, aber ich weiß nie wo. Kann vielleicht einer rübergucken?
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\( \begin{array}{l} \text { 2) } \\ \text { 2) } \vec{v}_{1}=\vec{x}=\left(\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right) \quad \vec{w}_{1}=\frac{1}{2} \cdot\left(\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right) \\ \overrightarrow{v_{2}}=\vec{y}-\frac{\vec{y}^{3} \cdot \vec{v}_{1}}{\left|\vec{v}_{n}^{2}\right|^{2}} \cdot \overrightarrow{v_{1}} \\ \vec{y} \cdot \vec{v}_{n}=4 \\ \left|\vec{v}_{1}\right|^{2}=4 \\ \vec{v}_{2}=\left(\begin{array}{c} 1 \\ 1 \\ 3 \\ -1 \end{array}\right)+1 \cdot\left(\begin{array}{c} 1 \\ -1 \\ -1 \\ -1 \end{array}\right) \\ =\left(\begin{array}{c} 2 \\ 0 \\ 4 \\ -2 \end{array}\right) \quad \overrightarrow{w_{2}}=\frac{1}{\sqrt{24}} \cdot\left(\begin{array}{c} 2 \\ 0 \\ 4 \\ -2 \end{array}\right) \\ \vec{v}_{3}=\vec{z}-\frac{\vec{z}^{2} \cdot \vec{v}_{1}^{2}}{\left|\vec{v}_{1}\right|^{2}} \cdot \vec{v}_{1}-\frac{\vec{z}^{3} \cdot \vec{v}_{2}^{3}}{\left|\vec{v}_{2}^{3}\right|^{2}} \cdot \vec{v}_{2} \\ \vec{z} \cdot \vec{v}_{1}=-11 \\ \left|\vec{v}_{1}\right|^{2}=4 \\ \vec{z} \cdot \vec{v}_{2}=2 \\ \left|\vec{v}_{2}\right|^{2}=24 \\ \vec{V}_{3}^{\prime}=\left(\begin{array}{c} -3 \\ 7 \\ 1 \\ 2 \end{array}\right)+\frac{11}{4}\left(\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right)-\frac{1}{12}\left(\begin{array}{c} 2 \\ 0 \\ 4 \\ -2 \end{array}\right) \\ =\frac{1}{4}\left(\left(\begin{array}{c} -12 \\ 28 \\ 4 \\ 8 \end{array}\right)+\left(\begin{array}{c} 11 \\ -11 \\ 11 \\ -11 \end{array}\right)-\frac{1}{3}\left(\begin{array}{c} 2 \\ 4 \\ -2 \end{array}\right)\right) \\ =\frac{1}{4}\left(\left(\begin{array}{c} -1 \\ 17 \\ 15 \\ -3 \end{array}\right)-\frac{1}{3}\left(\begin{array}{c} 2 \\ 4 \\ 4 \\ -2 \end{array}\right)\right) \\ =\frac{1}{12}\left(\begin{array}{c} -3-2 \\ 51 \\ 4 \\ 4 \\ 4 \\ 4 \\ 4 \end{array}+20\right. \end{array} \)
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\( \begin{array}{l} \text { 2) } \\ \text { 2) } \vec{v}_{1}=\vec{x}=\left(\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right) \quad \vec{w}_{1}=\frac{1}{2} \cdot\left(\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right) \\ \overrightarrow{v_{2}}=\vec{y}-\frac{\vec{y}^{3} \cdot \vec{v}_{1}}{\left|\vec{v}_{n}^{2}\right|^{2}} \cdot \overrightarrow{v_{1}} \\ \vec{y} \cdot \vec{v}_{n}=4 \\ \left|\vec{v}_{1}\right|^{2}=4 \\ \vec{v}_{2}=\left(\begin{array}{c} 1 \\ 1 \\ 3 \\ -1 \end{array}\right)+1 \cdot\left(\begin{array}{c} 1 \\ -1 \\ -1 \\ -1 \end{array}\right) \\ =\left(\begin{array}{c} 2 \\ 0 \\ 4 \\ -2 \end{array}\right) \quad \overrightarrow{w_{2}}=\frac{1}{\sqrt{24}} \cdot\left(\begin{array}{c} 2 \\ 0 \\ 4 \\ -2 \end{array}\right) \\ \vec{v}_{3}=\vec{z}-\frac{\vec{z}^{2} \cdot \vec{v}_{1}^{2}}{\left|\vec{v}_{1}\right|^{2}} \cdot \vec{v}_{1}-\frac{\vec{z}^{3} \cdot \vec{v}_{2}^{3}}{\left|\vec{v}_{2}^{3}\right|^{2}} \cdot \vec{v}_{2} \\ \vec{z} \cdot \vec{v}_{1}=-11 \\ \left|\vec{v}_{1}\right|^{2}=4 \\ \vec{z} \cdot \vec{v}_{2}=2 \\ \left|\vec{v}_{2}\right|^{2}=24 \\ \vec{V}_{3}^{\prime}=\left(\begin{array}{c} -3 \\ 7 \\ 1 \\ 2 \end{array}\right)+\frac{11}{4}\left(\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right)-\frac{1}{12}\left(\begin{array}{c} 2 \\ 0 \\ 4 \\ -2 \end{array}\right) \\ =\frac{1}{4}\left(\left(\begin{array}{c} -12 \\ 28 \\ 4 \\ 8 \end{array}\right)+\left(\begin{array}{c} 11 \\ -11 \\ 11 \\ -11 \end{array}\right)-\frac{1}{3}\left(\begin{array}{c} 2 \\ 4 \\ -2 \end{array}\right)\right) \\ =\frac{1}{4}\left(\left(\begin{array}{c} -1 \\ 17 \\ 15 \\ -3 \end{array}\right)-\frac{1}{3}\left(\begin{array}{c} 2 \\ 4 \\ 4 \\ -2 \end{array}\right)\right) \\ =\frac{1}{12}\left(\begin{array}{c} -3-2 \\ 51 \\ 4 \\ 4 \\ 4 \\ 4 \\ 4 \end{array}+20\right. \end{array} \)
