Question

na figura ABCD é um trapezio. determine o vetor DE em função de a e b sabendo que: DC= a; AB= 2a; DA=b e BE=1/3BC.

Answer 1

Temos dois caminhos para sair do ponto D e chegar até o ponto E:
     
     Caminho 1:

     $\mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DA}+\overset{\longrightarrow}{AB}+\overset{\longrightarrow}{BE}}\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DA}+\overset{\longrightarrow}{AB}+\dfrac{1}{3}\,\overset{\longrightarrow}{BC}}\\\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\to}{b}+2\overset{\to}{a}+\dfrac{1}{3}\,\overset{\longrightarrow}{BC}}\\\\\\ \mathsf{\overset{\longrightarrow}{DE}-\dfrac{1}{3}\,\overset{\longrightarrow}{BC}=\overset{\to}{b}+2\overset{\to}{a}}\\\\\\ \mathsf{3\,\overset{\longrightarrow}{DE}-\overset{\longrightarrow}{BC}=3(\overset{\to}{b}+2\overset{\to}{a})\qquad\quad(i)}$


     Caminho 2:

     $\mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DC}+\overset{\longrightarrow}{CE}}\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DC}+\overset{\longrightarrow}{CB}+\overset{\longrightarrow}{BE}}\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DC}-\overset{\longrightarrow}{BC}+\overset{\longrightarrow}{BE}}\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DC}-\overset{\longrightarrow}{BC}+\dfrac{1}{3}\,\overset{\longrightarrow}{BC}}\\\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DC}+\Big(\!-1+\dfrac{1}{3}\Big)\overset{\longrightarrow}{BC}}\\\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\longrightarrow}{DC}-\dfrac{2}{3}\,\overset{\longrightarrow}{BC}}\\\\\\ \mathsf{\overset{\longrightarrow}{DE}=\overset{\to}{a}-\dfrac{2}{3}\,\overset{\longrightarrow}{BC}}\\\\\\ \mathsf{\overset{\longrightarrow}{DE}+\dfrac{2}{3}\,\overset{\longrightarrow}{BC}=\overset{\to}{a}}$

     $\mathsf{3\,\overset{\longrightarrow}{DE}+2\,\overset{\longrightarrow}{BC}=3\overset{\to}{a}\qquad\quad(ii)}$


Isole $\mathsf{\overset{\longrightarrow}{BC}}$ na equação (i) e substitua na equação (ii):

     $\mathsf{\overset{\longrightarrow}{BC}=3\,\overset{\longrightarrow}{DE}-3(\overset{\to}{b}+2\overset{\to}{a})}\\\\\\ \mathsf{3\,\overset{\longrightarrow}{DE}+2\,\big[3\,\overset{\longrightarrow}{DE}-3(\overset{\to}{b}+2\overset{\to}{a})\big]=3\overset{\to}{a}}\\\\ \mathsf{3\,\overset{\longrightarrow}{DE}+6\,\overset{\longrightarrow}{DE}-6(\overset{\to}{b}+2\overset{\to}{a})=3\overset{\to}{a}}\\\\ \mathsf{9\,\overset{\longrightarrow}{DE}-6\overset{\to}{b}-12\overset{\to}{a}=3\overset{\to}{a}}\\\\ \mathsf{9\,\overset{\longrightarrow}{DE}=3\overset{\to}{a}+6\overset{\to}{b}+12\overset{\to}{a}}\\\\ \mathsf{9\,\overset{\longrightarrow}{DE}=15\overset{\to}{a}+6\overset{\to}{b}}\\\\ \mathsf{\overset{\longrightarrow}{DE}=\dfrac{1}{9}\,(15\overset{\to}{a}+6\overset{\to}{b})}$

     $\mathsf{\overset{\longrightarrow}{DE}=\dfrac{1}{3}\,(5\overset{\to}{a}+2\overset{\to}{b})\quad\longleftarrow\quad resposta.}$


Dúvidas? Comente.


Bons estudos! :-)