O vetor v⃗ =(4,3,−6) em R3 pode ser escrito como combinação linear dos vetores: e⃗ 1=(1,−3,2) e e⃗ 2=(2,4,−1) ?
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Bom dia! Esta faltando um vetor!
esta assim no exercício
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Bom dia Carol!
Solução!
![Sendo~~a,b~~\in\mathbb{R}^{3}\\\\\\\
v_{1}(1,-3,2)\\\\\\\
v_{2}(2,4-1)\\\\\\\
v=(4,3,-6)\\\\\\\\\\
a(1,-3,2)+b(2,4,-1)=(4,3,-6)\\\\\\
\begin{cases}
a+2b=4\\
-3a+4b=3\\
2a-b=-6
\end{cases}\\\\\\
Fazendo!\\\\\\
-b=-6-2a.(-1)\\\\\
\boxed{b=6+2a}\\\\\\
Substituindo!\\\\\
a+2b=4\\\\\
a+2(6+2a)=4\\\\\
a+12+4a=4\\\\\\
5a=4-12\\\\\
5a=-8\\\\\
\boxed{a= -\frac{8}{5}}](https://tex.z-dn.net/?f=Sendo%7E%7Ea%2Cb%7E%7E%5Cin%5Cmathbb%7BR%7D%5E%7B3%7D%5C%5C%5C%5C%5C%5C%5C%0Av_%7B1%7D%281%2C-3%2C2%29%5C%5C%5C%5C%5C%5C%5C%0A+v_%7B2%7D%282%2C4-1%29%5C%5C%5C%5C%5C%5C%5C%0A+v%3D%284%2C3%2C-6%29%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0Aa%281%2C-3%2C2%29%2Bb%282%2C4%2C-1%29%3D%284%2C3%2C-6%29%5C%5C%5C%5C%5C%5C%0A+%5Cbegin%7Bcases%7D%0Aa%2B2b%3D4%5C%5C%0A-3a%2B4b%3D3%5C%5C%0A2a-b%3D-6%0A%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%0AFazendo%21%5C%5C%5C%5C%5C%5C%0A-b%3D-6-2a.%28-1%29%5C%5C%5C%5C%5C%0A%5Cboxed%7Bb%3D6%2B2a%7D%5C%5C%5C%5C%5C%5C%0ASubstituindo%21%5C%5C%5C%5C%5C%0Aa%2B2b%3D4%5C%5C%5C%5C%5C%0Aa%2B2%286%2B2a%29%3D4%5C%5C%5C%5C%5C%0Aa%2B12%2B4a%3D4%5C%5C%5C%5C%5C%5C%0A5a%3D4-12%5C%5C%5C%5C%5C%0A5a%3D-8%5C%5C%5C%5C%5C%0A%5Cboxed%7Ba%3D+-%5Cfrac%7B8%7D%7B5%7D%7D+%0A "Sendo~~a,b~~\in\mathbb{R}^{3}\\\\\\\
v_{1}(1,-3,2)\\\\\\\
v_{2}(2,4-1)\\\\\\\
v=(4,3,-6)\\\\\\\\\\
a(1,-3,2)+b(2,4,-1)=(4,3,-6)\\\\\\
\begin{cases}
a+2b=4\\
-3a+4b=3\\
2a-b=-6
\end{cases}\\\\\\
Fazendo!\\\\\\
-b=-6-2a.(-1)\\\\\
\boxed{b=6+2a}\\\\\\
Substituindo!\\\\\
a+2b=4\\\\\
a+2(6+2a)=4\\\\\
a+12+4a=4\\\\\\
5a=4-12\\\\\
5a=-8\\\\\
\boxed{a= -\frac{8}{5}}")
![b=6+2a\\\\\
b=6+2(- \frac{8}{5})\\\\\
b=6- \frac{16}{5}\\\\\\
b= \frac{30-16}{5}\\\\\
\boxed{b= \frac{14}{5}}](https://tex.z-dn.net/?f=b%3D6%2B2a%5C%5C%5C%5C%5C%0Ab%3D6%2B2%28-+%5Cfrac%7B8%7D%7B5%7D%29%5C%5C%5C%5C%5C%0Ab%3D6-+%5Cfrac%7B16%7D%7B5%7D%5C%5C%5C%5C%5C%5C%0Ab%3D+%5Cfrac%7B30-16%7D%7B5%7D%5C%5C%5C%5C%5C%0A%5Cboxed%7Bb%3D+%5Cfrac%7B14%7D%7B5%7D%7D++++ "b=6+2a\\\\\
b=6+2(- \frac{8}{5})\\\\\
b=6- \frac{16}{5}\\\\\\
b= \frac{30-16}{5}\\\\\
\boxed{b= \frac{14}{5}}")
Bom dia!
Bons estudos!
Solução!
Bom dia!
Bons estudos!
Muito obrigada ^
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