Question

Geometria Analitica
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Answer 1

Ver figura em anexo.


$P(1,\,1,\,4)\in r\\\\ A(1,\,1,\,1)\in\mathbb{R}^3$


Um vetor diretor de $r$ é $\overrightarrow{\mathbf{v}}=(1,\,-1,\,0).$


$\bullet\;\;$ Encontrando o vetor $\overrightarrow{PA}:$

$\overrightarrow{PA}=A-P\\\\ \overrightarrow{PA}=(1,\,1,\,1)-(1,\,1,\,4)\\\\ \overrightarrow{PA}=(1-1,\,1-1,\,1-4)\\\\ \overrightarrow{PA}=(0,\,0,\,-3)~~\Rightarrow~~\|\overrightarrow{PA}\|=3$

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Queremos encontrar os pontos $X$ da reta $r\,,$ de forma que a distância de $X$ até $A$ seja igual a $\sqrt{11}:$

$\|\overrightarrow{XA}\|=\sqrt{11}$

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Como $X$ é um ponto da reta $r\,,$ devemos ter

$X=P+\lambda \overrightarrow{\mathbf{v}}\\\\ X-P=\lambda \overrightarrow{\mathbf{v}}\\\\ \overrightarrow{PX}=\lambda \overrightarrow{\mathbf{v}}\\\\ \overrightarrow{PX}=\lambda \,(1,\,-1,\,0)\\\\ \overrightarrow{PX}=\,(\lambda,\,-\lambda,\,0)~~~~\text{para algum }\lambda \in \mathbb{R}~~~~~~\mathbf{(i)}$

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Por soma de vetores, temos

$\overrightarrow{PX}+\overrightarrow{XA}=\overrightarrow{PA}\\\\ \overrightarrow{XA}=\overrightarrow{PA}-\overrightarrow{PX}\\\\ \overrightarrow{XA}=(0,\,0,\,-3)-\,(\lambda,\,-\lambda,\,0)\\\\ \overrightarrow{XA}=(0-\lambda,\,0+\lambda,\,-3-0)\\\\ \overrightarrow{XA}=(-\lambda,\,\lambda,\,-3)$


Calculando o módulo do vetor $\overrightarrow{XA}\,,$ devemos ter

$\|\overrightarrow{XA}\|=\sqrt{11}\\\\ \|\overrightarrow{XA}\|^2=11\\\\ \|(-\lambda,\,\lambda,\,-3)\|^2=11\\\\ (-\lambda)^2+(\lambda)^2+(-3)^2=11\\\\ \lambda^2+\lambda^2+9=11\\\\ 2\lambda^2=11-9\\\\ 2\lambda^2=2\\\\ \lambda^2=1\\\\ \lambda=\pm \sqrt{1}\\\\ \lambda=\pm 1\\\\ \boxed{\begin{array}{c} \lambda_1=-1~~\text{ e }~~\lambda_2=1 \end{array}}$

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Ora, encontramos dois valores para $\lambda\,,$ de modo que $\|\overrightarrow{XA}\|=\sqrt{11}.$ Então existe duas possibilidades para o ponto $X$ procurado:

$\bullet\;\;$ Para $\lambda=-1:$

$X=P+\lambda \overrightarrow{\mathbf{v}}\\\\ X=(1,\,1,\,4)+(-1)\cdot (1,\,-1,\,0)\\\\ X=(1,\,1,\,4)+(-1,\,1,\,0)\\\\ X=(1-1,\,1+1,\,4+0)\\\\ X=(0,\,2,\,4)\\\\\\ \therefore~~\boxed{\begin{array}{c}X_1=(0,\,2,\,4) \end{array}}$


$\bullet\;\;$ Para $\lambda=1:$

$X=P+\lambda \overrightarrow{\mathbf{v}}\\\\ X=(1,\,1,\,4)+1\cdot (1,\,-1,\,0)\\\\ X=(1,\,1,\,4)+(1,\,-1,\,0)\\\\ X=(1+1,\,1-1,\,4+0)\\\\ X=(2,\,0,\,4)\\\\\\ \therefore~~\boxed{\begin{array}{c}X_2=(2,\,0,\,4) \end{array}}$

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Existem dois pontos da reta $r\,,$ cuja distância até $A$ é $\sqrt{11}.$

Answer 2

Resposta:

Explicação passo-a-passo: