Länge des Vektors berechnen

Berechne die Länge des Vektors.

$\mathrm{\mid }\vec{a}\mathrm{\mid }=\sqrt{3^2+0^2+4^2}LE=\sqrt{25}LE=5LE|\backslash vec\; a|=\backslash sqrt\{3^2+0^2+4^2\}\; LE=\backslash sqrt\{25\}\; LE=5\; LE$

Vektor normieren

Teile den Vektor durch seine Länge

${\vec{a}}^0=\frac{\vec{a}}{\mathrm{\mid }\vec{a}\mathrm{\mid }}=\frac{1}{5}\cdot \left(\begin{array}{c}3\\ 0\\ 4\end{array}\right)=\left(\begin{array}{c}\mathrm{0,6}\\ 0\\ \mathrm{0,8}\end{array}\right)\backslash vec\; a^0=\backslash frac\; \{\backslash vec\; a\}\; \{|\backslash vec\; a|\}=\backslash frac\; 1\; 5\; \backslash cdot\; \backslash begin\{pmatrix\}3\backslash \backslash 0\backslash \backslash 4\backslash end\{pmatrix\}\; =\; \backslash begin\{pmatrix\}0\{,\}6\backslash \backslash 0\backslash \backslash 0\{,\}8\backslash end\{pmatrix\}$

Probe

Falls du deine Lösung überprüfen möchtest, bestimme die Länge des normierten Vektors.

$\mathrm{\mid }{\vec{a}}^0\mathrm{\mid }=\sqrt{{\mathrm{0,6}}^2+0^2+{\mathrm{0,8}}^2}LE=\sqrt{1}LE=1LE|\backslash vec\; a\; ^0|=\backslash sqrt\{0\{,\}6^2+0^2+0\{,\}8^2\}LE=\backslash sqrt\{1\}LE=1LE$

Länge des Vektors berechnen

Berechne die Länge des Vektors.

$\mathrm{\mid }\vec{a}\mathrm{\mid }=\sqrt{(-2{)}^2+1^2+2^2}LE=\sqrt{9}LE=3LE|\backslash vec\; a|=\backslash sqrt\{(-2)^2+1^2+2^2\}\; LE=\backslash sqrt\{9\}\; LE=3\; LE$

Vektor normieren

Teile den Vektor durch seine Länge

${\displaystyle {\vec{a}}^0=\frac{\vec{a}}{\mathrm{\mid }\vec{a}\mathrm{\mid }}=\frac{1}{3}\cdot \left(\begin{array}{c}-2\\ 1\\ 2\end{array}\right)=\left(\begin{array}{c}-\frac{2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right)}\backslash displaystyle\{\backslash vec\; a^0=\backslash frac\; \{\backslash vec\; a\}\; \{|\backslash vec\; a|\}=\backslash frac\; 1\; 3\; \backslash cdot\; \backslash begin\{pmatrix\}-2\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\; =\; \backslash begin\{pmatrix\}-\backslash frac\; 2\; 3\backslash \backslash \backslash frac\; 1\; 3\backslash \backslash \backslash frac\; 2\; 3\backslash end\{pmatrix\}\}$

Probe

Falls du deine Lösung überprüfen möchtest, bestimme die Länge des normierten Vektors.

$\mathrm{\mid }{\vec{a}}^0\mathrm{\mid }=\sqrt{{(-\frac{2}{3})}^2+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2}LE=\sqrt{1}LE=1LE|\backslash vec\; a\; ^0|=\backslash sqrt\{\backslash left(\; -\backslash frac\; 2\; 3\backslash right\; )^2+\backslash left\; (\; \backslash frac\; 1\; 3\; \backslash right)^2+\backslash left\; (\; \backslash frac\; 1\; 3\; \backslash right)^2\}LE=\backslash sqrt\{1\}LE=1LE$