Question

Sabendo que a reta R: { x = 1+2K y=k z= 3-k } forma um ângulo de 60 com a reta determinada pelos pontos A( 3,1,-2) B(4,0,m). Determine o valor de M.

Answer 1

A reta $\mathtt{r}$ é descrita por suas equações paramétricas:

$\mathtt{r:~}\left\{ \!\begin{array}{l} \mathtt{x=1+2k}\\ \mathtt{y=k}\\ \mathtt{z=3-k} \end{array} \right.\quad\quad\quad\texttt{com }\mathtt{k\in\mathbb{R}}$


Podemos obter as coordenadas de um vetor diretor para $\mathtt{r}$ apenas olhando para os coeficientes do parâmetro $\mathtt{k:}$

$\overrightarrow{\mathtt{v}}\mathtt{=(2,\,1,\,-1)}.$

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Encontrando o vetor determinado pelos pontos $\mathtt{A(3,\,1,\,-2)}$ e $\mathtt{B(4,\,0,\,m)}:$

$\overrightarrow{\mathtt{w}}=\overrightarrow{\mathtt{AB}}\\\\ \overrightarrow{\mathtt{w}}\mathtt{=B-A}\\\\ \overrightarrow{\mathtt{w}}\mathtt{=(4,\,0,\,m)-(3,\,1,\,-2)}\\\\ \overrightarrow{\mathtt{w}}\mathtt{=(4-3,\,0-1,\,m-(-2))}\\\\ \overrightarrow{\mathtt{w}}\mathtt{=(1,\,-1,\,m+2)}$

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Queremos encontrar $\mathtt{m}$ de forma que o ângulo $\theta$ entre os vetores $\overrightarrow{\mathtt{v}}$ e $\overrightarrow{\mathtt{w}}$ seja igual a 60°.


Pela definição de produto escalar, temos

$\overrightarrow{\mathtt{v}}\cdot\overrightarrow{\mathtt{w}}=\|\overrightarrow{\mathtt{v}}\|\cdot\|\overrightarrow{\mathtt{w}}\|\cdot \mathtt{cos\,\theta}\quad\quad\mathtt{(i)}$


Encontrando as normas dos vetores:

$\|\overrightarrow{\mathtt{v}}\|=\mathtt{\|(2,\,1,\,-1)\|}\\\\ =\mathtt{\sqrt{2^2+1^2+(-1)^2}}\\\\ =\mathtt{\sqrt{4+1+1}}\\\\ \therefore~~\|\overrightarrow{\mathtt{v}}\|=\mathtt{\sqrt{6}}$


$\|\overrightarrow{\mathtt{w}}\|=\mathtt{\|(1,\,-1,\,m+2)\|}\\\\ =\mathtt{\sqrt{1^2+(-1)^2+(m+2)^2}}\\\\ =\mathtt{\sqrt{1+1+m^2+4m+4}}\\\\ \therefore~~\|\overrightarrow{\mathtt{w}}\|=\mathtt{\sqrt{m^2+4m+6}}$

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Encontrando o produto escalar:

$\overrightarrow{\mathtt{v}}\cdot \overrightarrow{\mathtt{w}}=\mathtt{(2,\,1,\,-1)\cdot(1,\,-1,\,m+2)}\\\\ \mathtt{=2\cdot 1+1\cdot (-1)+(-1)\cdot (m+2)}\\\\ \mathtt{=2-1-m-2}\\\\ \therefore~~\overrightarrow{\mathtt{v}}\cdot \overrightarrow{\mathtt{w}}=\mathtt{-1-m}$

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Substituindo em $\mathtt{(i)},$ devemos ter

$\mathtt{-1-m=\sqrt{6}\cdot \sqrt{m^2+4m+6}\cdot cos\,60^\circ}\\\\\\ \mathtt{-1-m=\sqrt{6(m^2+4m+6)}\cdot \dfrac{1}{2}}\\\\\\ \mathtt{-2-2m=\sqrt{6(m^2+4m+6)}}$


Elevando os dois lados ao quadrado, temos

$\mathtt{(-2-2m)^2=\left(\sqrt{6(m^2+4m+6)}\right)^2}\\\\ \mathtt{4+8m+4m^2=6(m^2+4m+6)}\\\\ \mathtt{4+8m+4m^2=6m^2+24m+36}\\\\ \mathtt{0=6m^2+24m+36-4-8m-4m^2}\\\\ \mathtt{0=2m^2+16m+32}\\\\ \mathtt{m^2+8m+16=0}\\\\ \mathtt{(m+4)^2=0}\\\\ \mathtt{m+4=0}\\\\ \mathtt{m=-4}$


ATENÇÃO: Acabamos de resolver uma equação irracional, portanto temos que testar o valor encontrado na equação inicial:

•   Testando $\mathtt{m=-4}:$

$\mathtt{\sqrt{6}\cdot \sqrt{(-4)^2+4(-4)+6}\cdot\dfrac{1}{2}}\\\\ \mathtt{=\sqrt{6}\cdot \sqrt{16-16+6}\cdot \dfrac{1}{2}}\\\\ \mathtt{=\sqrt{6}\cdot \sqrt{6}\cdot \dfrac{1}{2}}\\\\ \mathtt{=6\cdot \dfrac{1}{2}}\\\\ \mathtt{=3}\\\\ \mathtt{=-1+4}\\\\ \mathtt{=-1-(-4)\quad\quad\checkmark}$


Logo, o valor procurado é

$\boxed{\begin{array}{c}{\mathtt{m=-4}} \end{array}}$


Dúvidas? Comente.


Bons estudos! :-)