Vektorproblem

Kan jag ha ännu mer hjälp med mina pilar😭?

1. $AG=sAE=s(AB+BE)=s(u+1/2v)$

2. $$\begin{gathered} AG=AF+t\left(FG\right)=\frac{1}{7}AD+t\left(FB\right)=\frac{1}{7}AD+t(FD+DB)=\frac{1}{7}AD+t(FD+DA+AB) \\ =\frac{1}{7}AD+t(\frac{6}{7}AD-AD+AB)=\frac{1}{7}AD+t(-\frac{1}{7}AD+AB)=\left(\frac{1}{7}-\frac{1}{7}t\right)v+tu \end{gathered}$$

Vi sätter koordinater lika med varandra:

$$\begin{gathered} \left\{\begin{array}{l}s\cancel{u}=t\cancel{u} \\ s\frac{1}{2}\cancel{v}=\left(\frac{1}{7}-\frac{1}{7}t \right)\cancel{v}\\ \end{array}\right. \end{gathered}$$

$$\begin{gathered} s=t \\ \frac{1}{2}t=\frac{1}{7}-\frac{1}{7}t \iff \left(\frac{1}{2}+\frac{1}{7}\right)t=\frac{1}{7}\iff \frac{9}{14}t=\frac{1}{7}\iff t=\frac{1}{7}\frac{14}{9}=\frac{2}{9} \end{gathered}$$

Testning av lösningar ger:

$AG=\frac{2}{9}(u+1/2v)=\frac{2}{9}u+\frac{1}{9}v$

$\left(\frac{1}{7}-\frac{1}{7}\frac{2}{9}\right)v+\frac{2}{9}u=\left(\frac{1}{7}-\frac{2}{63}\right)v+\frac{1}{9}u=\frac{1·9}{7·9}-\frac{2}{63}v+\frac{2}{9}u=\frac{7}{63}v+\frac{2}{9}u=\frac{2}{9}u+\frac{1}{9}v$

Triangeln arean kan beräknas som en vektorprodukt, nämligen hälften av $AF \times AG$.

$$\begin{gathered} \frac{1}{2}\left|\begin{array}{cc}0& \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$7$}\right.& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\end{array}\right|=\frac{1}{2}\left(\frac{2}{63}-0\right)=\frac{1}{63}(u\times v) \\ Areor=\frac{\frac{1}{63}(u\times v)}{(u\times v)}=\frac{1}{63} \end{gathered}$$

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Why!!