Question

2) Dados os vetores u= ( -4, -1/3,4) e v= (1,0, -3), calcular u+v, 5u, 2u-3v e u-1/2v

Answer 1

No exercício, temos expressões envolvendo soma e subtração entre vetores e produto de vetor por escalar.

Sejam dois vetores u=(x₁, y₁, z₁) e v=(x₂ ,y₂ ,z₂), a soma/subtração entre vetores se dá somando/subtraindo as coordenadas de mesma posição nos vetores (x₁ com x₂, y₁ com y₂...).

Já o produto por escalar se dá multiplicando cada coordenada do vetor pelo escalar.

$\vec{u}~+~\vec{v}~=~\left(-4~,\,-\dfrac{1}{3}~,~4\right)~+~(1~,~0~,\,-3)\\\\\\\vec{u}~+~\vec{v}~=~\left(-4+1~,\,-\dfrac{1}{3}+0~,~4+(-3)\right)\\\\\\\boxed{\vec{u}~+~\vec{v}~=~\left(-3~,\,-\dfrac{1}{3}~,~1\right)}\\\\\\\\5\vec{u}~=~5\cdot\left(-4~,\,-\dfrac{1}{3}~,~4\right)\\\\\\5\vec{u}~=~\left(5\cdot(-4)~,~5\cdot\left(-\dfrac{1}{3}\right)~,~5\cdot4\right)\\\\\\\boxed{5\vec{u}~=~\left(-20~,\,-\dfrac{5}{3}~,~20\right)}$

$2\vec{u}~-~3\vec{v}~=~2\cdot\left(-4~,\,-\dfrac{1}{3}~,~4\right)~-~3\cdot(1~,~0~,\,-3)\\\\\\2\vec{u}~-~3\vec{v}~=~\left(2\cdot(-4)~,~2\cdot \left(-\dfrac{1}{3}\right)~,~2\cdot 4\right)~-~\left(3\cdot1~,~3\cdot 0~,~3\cdot (-3)\right)\\\\\\2\vec{u}~-~3\vec{v}~=~\left(-8~,\,-\dfrac{2}{3}~,~8\right)~-~\left(3~,~0~,\,-9\right)\\\\\\2\vec{u}~-~3\vec{v}~=~\left(-8-3~,\,-\dfrac{2}{3}-0~,~8-(-9)\right)\\\\\\\boxed{2\vec{u}~-~3\vec{v}~=~\left(-11~,\,-\dfrac{2}{3}~,~17\right)}$

$\vec{u}~-~\dfrac{1}{2}\vec{v}~=~\left(-4~,\,-\dfrac{1}{3}~,~4\right)~-~\dfrac{1}{2}\cdot(1~,~0~,\,-3)\\\\\\\vec{u}~-~\dfrac{1}{2}\vec{v}~=~\left(-4~,\,-\dfrac{1}{3}~,~ 4\right)~-~\left(\dfrac{1}{2}\cdot1~,~\dfrac{1}{2}\cdot 0~,~\dfrac{1}{2}\cdot (-3)\right)\\\\\\\vec{u}~-~\dfrac{1}{2}\vec{v}~=~\left(-4~,\,-\dfrac{1}{3}~,~ 4\right)~-~\left(\dfrac{1}{2}~,~0~,\,-\dfrac{3}{2}\right)\\\\\\\vec{u}~-~\dfrac{1}{2}\vec{v}~=~\left(-4-\dfrac{1}{2}~,\,-\dfrac{1}{3}-0~,~4-\left(-\dfrac{3}{2}\right)\right)$

$\boxed{\vec{u}~-~\dfrac{1}{2}\vec{v}~=~\left(-\dfrac{9}{2}~,\,-\dfrac{1}{3}~,\,-\dfrac{11}{2}\right)}\\\\\\\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$