Calculate the distance from a point to a line

Question

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: $\sqrt{20}$)

Answer 1

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.

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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!

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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.

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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!

Answer 2

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.

Answer 3

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.