Question

Sejam os pontos m(1,-2,-2) e p(0,-1,2) determine um vetor v colinear a PM e tal que |v|=raiz3
resposta; v=(+-1/rais6, +-1/rais6, +-4rais6)

Answer 1

Olá

M = (1,-2,-2)        P = (0,-1,2)

$\vec{PM}=M-P=(1-0,~-2-(-1),~,-2-2) \\ \\ \vec{PM}=(1, -1, -4) \\ \\ \text{Se PM e V sao colineares, entao podemos escrever um como} \\ \text{combinacao linear do outro} \\ \\ \vec{v}= \beta ~\vec{ PM} \\ \\ \vec{v}= \beta (1, -1,-4) \\ \\ \vec{v}=( \beta ,- \beta ,-4 \beta ) \\ \\ \text{Calcula o modulo} \\ \\ |\vec{v}|= \sqrt{ (\beta) ^2+(- \beta )^2+(-4 \beta )^2} \\ \\ |\vec{v}| } = \sqrt{ \beta ^2+ \beta ^2+16 \beta ^2} \\ \\ \sqrt{3}= \sqrt{18 \beta ^2} \\ \\ \text{Eleva ao quadrado dos dois lados}$
$(\sqrt{3})^2= (\sqrt{18 \beta ^2})^2 \\ \\ 3=18 \beta ^2 \\ \\ \beta ^2= \frac{3}{18} ~~~~~ ~~~~~~~~~~~~~ ~~~~~ ~~\text{Simplifica por 3}\\ \\ \beta ^2= \frac{1}{6} \\ \\ \boxed{ \boxed{\beta ~=+ \sqrt{ \frac{1}{6} } }} ~~~~ ~~~~ ~~~~ ~~~~ou~~ ~~ ~~~~~ ~~~~~ \boxed{\boxed{ \beta =- \sqrt{ \frac{1}{6} } }} \\ \\ \\ \text{Entao} \\ \\\boxed{\boxed{ \vec{v}=+ \frac{1}{ \sqrt{6} } \cdot(1,-1,-4) }} \\ \\ \\ ou \\ \\ \\ \boxed{\boxed{\vec{v}=- \frac{1}{ \sqrt{6} } \cdot(1,-1,-4)}}$