Question

Dados os vetores u = 2,-3 v= 1,-1 e w= -2,1 , determinar a) 2u - v b) v - u + 2w c) 0,5 u - 2 v - w d) 3u - 0,5 v - 0,5 w

Answer 1

Olá

$\vec{u}=(2,-3)~~~~\vec{v}=(1,-1)~~~~\vec{w}=(-2,1)$

A)  

$\displaystyle 2\vec{u}-\vec{v} \\ \\ \\ \text{Para encontrar o vetor }2\vec{u}~\text{Basta multiplicar o vetor }\vec{u}\text{ por 2} \\ \\ 2\cdot \vec{u}=2(2,-3)~~~~ \longrightarrow~~~\boxed{2\vec{u}=(4,-6)} \\ \\ \\ \text{Agora basta substrair o vetor }2\vec{u}~\text{do vetor }\vec{v} \\ \\ \\ 2\vec{u}-\vec{v}=(4,-6)-(1,-1)~=~((4-1),(-6-(-1)))~=~(3,-5) \\ \\ \boxed{\boxed{2\vec{u}-\vec{v}=(3,-5)}}$


B)

$\displaystyle \vec{v}-\vec{u}+2\vec{w}= \\ \\ \\ \text{Mesmo processo do item anterior, multiplica o vetor }\vec{w} \text{ por 2} \\ \\ 2\vec{w}=2\cdot(-2,1)~~~~~\longrightarrow~\boxed{2\vec{w}=(-4,2)} \\ \\ \\ \text{Primeiro subtrair o vetor }\vec{v}\text{ do vetor }\vec{v},~\text{e depois soma com o vetor }2\vec{w} \\ \\ \\ \vec{v}-\vec{u}=(1,-1)-(2,-3) \\ \vec{v}-\vec{u}=((1-2),(-1-(-3))) \\ \boxed{\vec{v}-\vec{u}=(-1,2)} \\ \\ \vec{v}-\vec{u}+ 2\vec{w}=(-1,2)+(-4,2)$
$\vec{v}-\vec{u}+ 2\vec{w}=(-1+(-4),2+2) \\ \\ \\\boxed{\boxed{ \vec{v}-\vec{u}+ 2\vec{w}=(-5,4)}}$


C)

Para facilitar, podemos escrever 0,5 como $\displaystyle \frac{1}{2}$

$\displaystyle \frac{1}{2} \vec{u}~-2\vec{v}-\vec{w}$

$\displaystyle \frac{1}{2}\vec{u}= \frac{1}{2}\cdot(2,-3) \\ \\ \frac{1}{2}\vec{u}=( \frac{2}{2} , \frac{-3}{2} ) \\ \\ \\ \boxed{\frac{1}{2}\vec{u}=(1,- \frac{3}{2} )} \\ \\ \\ \\ 2\vec{v}=2\cdot(1,-1) \\ \\ \boxed{2\vec{v}=(2,-2)} \\ \\ \\ \\ \frac{1}{2}\vec{u}-2\vec{v}=~(1, -\frac{3}{2} ) -(2,-2) \\ \\ \frac{1}{2}\vec{u}-2\vec{v}=~((1-2),(- \frac{3}{2}-(-2)) )$

$\displaystyle \frac{1}{2}\vec{u}-2\vec{v}=~(-1, \frac{1}{2} ) \\ \\ \\ \text{Agora temos que subtrair esse vetor do vetor }\vec{w} \\ \\ \\ \frac{1}{2}\vec{u}-2\vec{v}-\vec{w}=~ (-1, \frac{1}{2} )-(-2,1) \\ \\ \\ \frac{1}{2}\vec{u}-2\vec{v}-\vec{w}=~((-1-(-2)),( \frac{1}{2}-1 )) \\ \\ \\ \\ \boxed{\boxed{\frac{1}{2}\vec{u}-2\vec{v}-\vec{w}=~(1,- \frac{1}{2} )}}$




D)

Faremos do mesmo jeito do item anterior, no lugar de 0,5 escreveremos $\displaystyle \frac{1}{2}$

$\displaystyle 3\vec{u}- \frac{1}{2}\vec{v}-2\vec{w}$

$\displaystyle \text{Encontrando o vetor }3\vec{u} \\ \\ 3\vec{u}=3\cdot(2,-3) \\ \\ \boxed{3\vec{u}=(6,-9)} \\ \\ \\ \text{encontrando o vetor } \frac{1}{2}\vec{v} \\ \\ \frac{1}{2}\vec{v}= \frac{1}{2}\cdot(1,-1) \\ \\ \\\boxed{ \frac{1}{2}\vec{v}=( \frac{1}{2} ,- \frac{1}{2} )} \\ \\ \\ \\ \text{encontrando o vetor } \frac{1}{2}\vec{w}$
$\displaystyle \frac{1}{2}\vec{w} = \frac{1}{2} \cdot (-2,1) \\ \\ \\ \frac{1}{2}\vec{w} =(- \frac{2}{2}, \frac{1}{2} ) \\ \\ \\ \boxed{\frac{1}{2}\vec{w} =(-1, \frac{1}{2} )} \\ \\ \\ \\ \text{subtraindo o vetor }3\vec{u} ~do~\text{vetor}~\frac{1}{2}\vec{v} \\ \\ \\ 3\vec{u} -\frac{1}{2}\vec{v} =(6,-9)-( \frac{1}{2} ,- \frac{1}{2} )$

$\displaystyle 3\vec{u} -\frac{1}{2}\vec{v} =((6- \frac{1}{2}) ,(-9-(- \frac{1}{2})) ) \\ \\ \\ \boxed{3\vec{u} -\frac{1}{2}\vec{v} =( \frac{11}{2},- \frac{17}{2} )} \\ \\ \\ \\ \text{Pegando este vetor que acabamos de encontrar e subtraindo pelo vetor} \\ \frac{1}{2} \vec{w} \\ \\ \\ \\ 3\vec{u} -\frac{1}{2}\vec{v}- \frac{1}{2} \vec{w} =~( \frac{11}{2},- \frac{17}{2} )-(-1, \frac{1}{2} )$

$\displaystyle 3\vec{u} -\frac{1}{2}\vec{v}- \frac{1}{2} \vec{w} =~(( \frac{11}{2}-(-1) ), (-(\frac{17}{2}- \frac{1}{2}) ) \\ \\ \\ \boxed{\boxed{3\vec{u} -\frac{1}{2}\vec{v}- \frac{1}{2} \vec{w} =( \frac{13}{2},-9 )}}$









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