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Por favor, alguém me ajude com essa questão de Geometria:

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Please, can someone help me with this geometry question?

Given the point $A(3,4,-2)$ and the line $$r:\left\{\begin{array}{l} x = 1 + t \\ y = 2 - t \\ z = 4 + 2t \end{array} \right.$$ compute the distance from $A$ to $r$. (Answer: $\sqrt{20}$)

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Bem-vindo à Matemática Stack Exchange. Informamos que as perguntas deverão ser em Inglês. – John Jul 6 '14 at 22:28
Non-English posts are acceptable. Hopefully someone who knows Portuguese will translate it to English. – Ayman Hourieh Jul 6 '14 at 22:34
Although I already answered it, I'll translate, so everyone else will get it too. – Ivo Terek Jul 6 '14 at 22:37
@IvoTerek Thank you! – Ayman Hourieh Jul 6 '14 at 22:40

Pra achar a distância de um ponto à uma reta faça o seguinte: Pegue um ponto qualquer da reta, digamos, $P$. Então calcule o vetor $\vec{PA} = A - P$. Depois, calcule a projeção deste vetor sobre a reta $r$, usando a fórmula: $$\mathrm{proj}_v \vec{PA} = \frac{\vec{PA} \cdot v}{v \cdot v}~v$$ onde $v$ é o vetor diretor da reta dada. Assim, o vetor $w = \vec{PA} - \mathrm{proj}_v \vec{PA}$ é perpendicular à reta. Seja $s$ outra reta, dada por $$s: X = A + tw, \quad t \in \Bbb R$$ Ache a interseção entre as retas $r$ e $s$, chamemos esse ponto de $B$. A distância entre $A$ e $B$ é a distância procurada. Siga os passos com calma que você deve conseguir.

Neste site, é melhor sempre postar as dúvidas em inglês. Você deu sorte que eu sei português e apareci aqui, mas nem sempre vai aparecer alguém que saiba!

In English: To find the distance from the point to the line, do the following: take any point in the line, let's call it $P$. Then find the vector $\vec{PA} = A - P$. After, compute the projection of this vector on the line $r$, using the formula $$\mathrm{proj}_v \vec{PA} = \frac{\vec{PA} \cdot v}{v \cdot v}~v$$ where $v$ is the direction of the given line. This way, the vector $w = \vec{PA} - \mathrm{proj}_v \vec{PA}$ is orthogonal to the line. Let $s$ be another line, given by $$s: X = A + tw, \quad t \in \Bbb R$$ Find the intersection between the lines $r$ and $s$, and let's call this point $B$. The distance between $A$ and $B$ is the distance you seek. Follow the steps calmly and you should manage it.

In this site, it is better to always ask your questions in english. You were lucky that I know portuguese and I showed up here, but not always will show up someone who do!

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One can only love Portuguese. =) – Pedro Tamaroff Jul 6 '14 at 22:45
The more languages you know.. (; – Ivo Terek Jul 6 '14 at 22:47
@PedroTamaroff The similarity with Spanish is just too lovely! I don't even know Portuguese, but I could read both posts without problems :D – chubakueno Jul 7 '14 at 0:54
Until and unless you provide a translation into English, this is Not An Answer. – tchrist Jul 7 '14 at 1:37
You do realize that I answered in portuguese on purpose, right? My main concern is that OP, who is obviously brazilian (and so am I) gets the idea. And I don't remember seeing written anywhere that answers must be in english. In fact, on the comments directly on the question, there is link to meta that states that questions in other languages are, up to a certain point, acceptable. And OP just contacted me via facebook, and it happens that he doesn't know english whatsoever, neither how to accepts answers, etc. My guess is that he won't come back here anymore. – Ivo Terek Jul 7 '14 at 1:43

Consider the vector $\vec{PA}=(-2,-2,-6)$ and the vector that gives the direction of the line $\vec{v}=(1,-1,2).$ These two vectors form a parallelogram and the height of this parallelogram is the distance between the point and the line (since distance is realized in the direction perperdicular to the line through $A$). Now, to get the height of the parallelogram we divide its area by the basis, that is

$$\frac{|\vec{PA}\times \vec{v}|}{|\vec{v}|}.$$

You compute

$$\vec{PA}\times \vec{v}=\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k}\\ -2 & -2 & - 6\\ 1 & -1 & 2\end{array}\right|,$$ get its module, divide by the module of $\vec{v}$ and you have the desired solution.

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Let $P(t)$ is a point of the line, $\vec v$ a direction of a line. Take $\vec{AP}$ and request it be normal to $\vec v$: $$\vec{AP} \cdot \vec v = 0.$$ Solve it for $t$, substitute the $t$ to $P(t)$ and calculate $AP = |\vec{AP}|.$ Done.

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