Lösung Orthogonalen Vektor finden

Version 3.1 von Daniel Stocker am 2024/02/05 17:19

$\vec u \cdot \vec v = \left(\begin{array}{c} 4 \\ 5 \\ 2\end{array}\right) \cdot \left(\begin{array}{c} \frac{2}{3} \\ a \\ 1\end{array}\right) = 4 \cdot \frac{2}{3} + 5 \cdot a + 2 \cdot (-1) \overset{!}{=} 0$

$\begin{align*} 4 \cdot \frac{2}{3} + 5 \cdot a + 2 \cdot (-1) &= 0 \\ \frac{8}{3} + 5a -2 &= 0 \\ \frac{2}{3} + 5a &= 0 \\ 5a &= -\frac{2}{3} \\ a &= -\frac{2}{15} \end{align*}$