Question

( )MATEMATICA VETORES S2 SEUS LINDOS

Seja V = (−7; 3). Determine um vetor paralelo a V com:

A) Módulo 1;

B) Módulo 4;

C) Módulo 6 e sentido contrário.

Answer 1

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$\Large\green{\boxed{\rm~~~\red{ A)}~\gray{\overrightarrow{\sf U_1}}~\pink{=}~\blue{ \left(\dfrac{-7}{\sqrt{58}} , \dfrac{3}{\sqrt{58}}\right) }~~~}}$

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$\Large\green{\boxed{\rm~~~\red{ B)}~\gray{\overrightarrow{\sf U_2}}~\pink{=}~\blue{ \left(\dfrac{-28}{\sqrt{58}} , \dfrac{12}{\sqrt{58}}\right) }~~~}}$

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$\Large\green{\boxed{\rm~~~\red{ C)}~\gray{\overrightarrow{\sf U_3}}~\pink{=}~\blue{ \left(\dfrac{42}{\sqrt{58}} , \dfrac{-18}{\sqrt{58}}\right) }~~~}}$

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

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☺lá, Thomas, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo. ✌

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☔ Seja $\sf \overrightarrow{\sf U}$ nosso vetor paralelo à $\sf \overrightarrow{\sf V}$. Temos portanto que

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$\LARGE\blue{\sf \overrightarrow{\sf U} = \alpha \cdot \overrightarrow{\sf V} = (-7\alpha , 3\alpha)}$

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☔ O módulo (ou norma) de um Vetor é o comprimento deste Vetor e pode ser encontrado pela fórmula da distância entre dois pontos:

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$\Large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{\sf | \overrightarrow{\sf U} | = \sqrt{x^2 + y^2}}&\\&&\\\end{array}}}}}$

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Ⓐ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~| \overrightarrow{\sf U_1} | = 1~~}}}$

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$\LARGE\blue{\sf \sqrt{(-7\alpha_1)^2 + (3\alpha_1)^2} = 1 }$

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$\LARGE\blue{\sf \sqrt{49\alpha_1^2 + 9\alpha_1^2} = 1 }$

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$\LARGE\blue{\sf \sqrt{58\alpha_1^2} = 1 }$

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$\LARGE\blue{\sf \alpha_1 \cdot \sqrt{58} = 1 }$

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$\LARGE\blue{\text{ \sf \alpha_1 = \dfrac{1}{\sqrt{58}} }}$

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☔ Com o valor de α podemos agora encontrar o valor de U1:

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$\LARGE\blue{\text{ \sf \overrightarrow{\sf U_1} = \left(\dfrac{-7}{\sqrt{58}} , \dfrac{3}{\sqrt{58}}\right) }}$

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$\Large\green{\boxed{\rm~~~\red{ A)}~\gray{\overrightarrow{\sf U_1}}~\pink{=}~\blue{ \left(\dfrac{-7}{\sqrt{58}} , \dfrac{3}{\sqrt{58}}\right) }~~~}}$ ✅

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Ⓑ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~| \overrightarrow{\sf U_2} | = 4~~}}}$

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$\LARGE\blue{\text{ \sf \alpha_2 = \dfrac{4}{\sqrt{58}} }}$

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☔ Com o valor de α podemos agora encontrar o valor de U2:

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$\LARGE\blue{\text{ \sf \overrightarrow{\sf U_2} = \left(\dfrac{-28}{\sqrt{58}} , \dfrac{12}{\sqrt{58}}\right) }}$

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$\Large\green{\boxed{\rm~~~\red{ B)}~\gray{\overrightarrow{\sf U_2}}~\pink{=}~\blue{ \left(\dfrac{-28}{\sqrt{58}} , \dfrac{12}{\sqrt{58}}\right) }~~~}}$ ✅

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Ⓒ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

$\LARGE\gray{\boxed{\sf\blue{~~| \overrightarrow{\sf U_3} | = 6~~}}}$

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$\LARGE\blue{\text{ \sf \alpha_3 = \dfrac{6}{\sqrt{58}} }}$

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☔ Como o vetor U3 terá o sentido contrário de V então o valor que teremos:

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$\LARGE\blue{\sf \overrightarrow{\sf U_3} = \alpha \cdot (-\overrightarrow{\sf V}) = (7\alpha , -3\alpha)}$

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☔ Com o valor de α podemos agora encontrar o valor de U3:

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$\LARGE\blue{\text{ \sf \overrightarrow{\sf U_3} = \left(\dfrac{42}{\sqrt{58}} , \dfrac{-18}{\sqrt{58}}\right) }}$

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$\Large\green{\boxed{\rm~~~\red{ C)}~\gray{\overrightarrow{\sf U_3}}~\pink{=}~\blue{ \left(\dfrac{42}{\sqrt{58}} , \dfrac{-18}{\sqrt{58}}\right) }~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\Large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$