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I have two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ on a line, $L$, and another point $P_0(x_0, y_0, z_0)$.

I want to find the distance between $P_0$ and $L$. Could someone help?

  • 6
    [Hmm...](http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html)2010-12-22
  • 0
    why at (9) there is only x? what about y and z?2010-12-22
  • 1
    $\mathbf{x}$ is a vector, it has all three components.2010-12-22
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    and | x_2 - x_1 | is vector magnitude or absolute value?2010-12-22
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    Vector magnitude.2010-12-22
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    It is $$\begin{eqnarray*}\left\vert \mathbf{x}_{2}-\mathbf{x}_{1}\right\vert &=&\left\vert (x_{2},y_{2},z_{2})-(x_{1},y_{1},z_{1})\right\vert\\ &=&\left\vert (x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})\right\vert \\ &=&\sqrt{\left(x_{2}-x_{1}\right) ^{2}+\left( y_{2}-y_{1}\right)^{2}+\left( z_{2}-z_{1}\right) ^{2}}\end{eqnarray*}$$2010-12-22
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    Do you know what dot and cross products of two vectors are? Otherwise the deduction of the distance formula is more difficult. Perhaps you want to look first to the 2D case. Here http://softsurfer.com/Archive/algorithm_0102/algorithm_0102.htm you find the distance formula between a point $P$ and a line $L$ that passes through the points $P_{0}$ and $P_{1}$: $d(P,L)=\dfrac{\left\vert \mathbf{v}_{L}\times \mathbf{w}\right\vert }{\left\vert \mathbf{v}_{L}\right\vert }$, with $\mathbf{v}_{L}=(P_{1}-P_{0})$, $\mathbf{w}=(P-P_{0})$.2010-12-22
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    :) thank you very much! all clear now!2010-12-23
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    now i'm studing the prove of this because i like it. i think the softsurfer.com/Archive/algorithm_0102/algorithm_0102.htm is really more clear than the profe in wolfram, even if is geometrical and not algebric. Why they use the derivate of distance to minimize it? And why they use that formula for the square distance?2011-01-11
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    "Why they use the derivate of distance to minimize it?" - because finding the optima of a function is equivalent to finding where the derivative of a function becomes zero.2011-08-25
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    I suppose somebody should be posting an answer to this question so that it is settled; who'll do the honors?2011-08-25

3 Answers 3

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I thought to furnish a picture:

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    Please cite your source of this image as it is clear that you did not create it.2014-02-18
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The shortest distance from a point to a line is always perpendicular to the given line. Here, the given line is in the direction of the vector $\langle x_2-x_1, y_2-y_1, z_2-z_1\rangle$. The plane $(x_2- x_1)(x- x_0)+ (y_2- y_1)(y- y_0)+ (z_2- z_1)(z- z_0)= 0$ has that vector as normal vector and contains the point $(x_0, y_0, z_0)$ so the shortest distance is through that plane. Determine where the given line intersects that plane. The shortest distance is the distance form that point of intersection to $(x_0, y_0, z_0)$.

1

The distance $h$ from the point $P_0=(x_0,y_0,z_0)$ to the line passing through $P_1=(x_1,y_1,z_1)$ and $P_2=(x_2,y_2,z_2)$ is given by $h=2A/r$, where $A$ is the area of a triangle defined by the three points and $r$ is the distance from $P_1$ to $P_2$. The values of $r$ and $A$ can be computed as follows:

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ and $A=\frac{1}{2}\sqrt{a_1^2+a_2^2+a_3^2},$

where

$a_1=x_0y_1+x_1y_2+x_2y_0 - (y_0x_1+y_1x_2+y_2x_0),\\ a_2=y_0z_1+y_1z_2+y_2z_0 - (z_0y_1+z_1y_2+z_2y_0),\\ a_3=x_0z_1+x_1z_2+x_2z_0 - (z_0x_1+z_1x_2+z_2x_0).$