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Can you form an equation for a ''pipe'' in 3D space? It means all those points P(x,y,z) that have the same shortest distance for the same straight line l. For example what would the pipe equation be when we know that the line l is x-1=y-1=z-1 and when the distance between all points P(x,y,z) and the line l is 4?

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Do you mean a cylinder? – Hans Lundmark Jun 1 '12 at 11:46
Yeah kind of but it has no endings though, because it forms around a direct. – user30606 Jun 1 '12 at 11:56
OK, fine. And I guess "direct" means "straight line" then. – Hans Lundmark Jun 1 '12 at 12:00
Ah so that's what you call it.I just wrote what google translator told me so it may be a bit misleading – user30606 Jun 1 '12 at 12:12
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The line $x-1=y-1=z-1$ is the same as $x=y=z$ so it's all the points $(t,t,t)$, $t$ real. The plane perpendicular to this line at $(t,t,t)$ is $$x+y+z=3t$$ The points in this plane at distance 4 from $(t,t,t)$ also satisfy $$(x-t)^2+(y-t)^2+(z-t)^2=16$$ Solve the first equation for $t$ in terms of $x,y,z$, then put that in the second equation, and you should have an equation for the tube.

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For a general curve in 3-space with curvature and torsion say with arc s as parameter, find Frenet-Serret vector triad.

For convenience, make them unit vectors, then multiply by r = 4 the desired radius. Then rotate the normal or binomial vector N or B between 0 to 2 pi in Normal plane through a parameter theta.We have the 3 (x,y,z) coordinates as functions of arc s and theta.

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