Ensino Superior ⇒ Vetores: Combinação Linear Tópico resolvido

[aprendiz123]

Dados os pontos [tex3]A[/tex3], [tex3]B[/tex3] e [tex3]C[/tex3] quaisquer, construir o vetor [tex3]\bar{x} = D-A[/tex3], tal que [tex3](D-B) + \frac{2}{3}(A-D) + \frac{5}{3}(C-B)=0[/tex3]

Resposta

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[tex3]x= 8(B-A)-5(C-A)[/tex3]

[Karl Weierstrass]

Dados os pontos [tex3]A,\, B[/tex3] e [tex3]C[/tex3] quaisquer, construir o vetor [tex3]\vec{x} = D\,-\,A,[/tex3] tal que [tex3](D\,-\,B)\, +\, \frac{2}{3} (A\,-\,D)\, +\, \frac{5}{3} (C\,-\,B)\,=\,0.[/tex3]

[tex3]\hspace{70pt}(D\,-\,B)\, +\, \frac{2}{3} (A\,-\,D)\, +\, \frac{5}{3} (C\,-\,B)\,=\,0\,\Longrightarrow\, D\,-\,B\, +\, \frac{2}{3} A\,-\, \frac{2}{3} D\, +\, \frac{5}{3} C\,-\, \frac{5}{3} B\,=\,0[/tex3]

[tex3]\Longrightarrow\, \frac{1}{3} D\, +\, \frac{2}{3} A\, -\,A\,+\,A\,+\, \frac{5}{3} C\,-\, \frac{8}{3} B\,=\,0[/tex3]

[tex3]\Longrightarrow\, \frac{1}{3} D\, -\, \frac{1}{3} A\,\,=\, \frac{8}{3} B\,-\, \frac{5}{3} C\,-\,A[/tex3]

[tex3]\Longrightarrow\,D\, -\, A\,=\,8B\,-\, 5C\,-\,3A[/tex3]

[tex3]\Longrightarrow\,D\, -\, A\,=\,8B\,-\, 5C\,-\,3A\,+\,5A\,-\,5A[/tex3]

[tex3]\Longrightarrow\,\vec{x}\,=\,8(B\,-\,A)\,- \,5(C\,-\,A).[/tex3]

[aprendiz123]

essa do A-A é difícil em.... vlw cara