Finding a Skew Line in R3 at Distance d

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hello

Homework Statement


given a line in R3 and a distance d how do i find a line that skew with that line in that distance between them?

Homework Equations



The Attempt at a Solution


i tried to create a plane with the given line but it doesn't seem to work for me. cause too many assumptions have to be made
and the way i did it was really complicated in my opinion and not the right way to go

if you can help
 
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in what form is the equation of line given?

basically, find any line perpendicular to the given line(let perpendicular be p and given line be l) and take a point at a distance d from l, on p. Then write the general equation of any line thorough this point..

How easy or tough it is to do this depends on the the form of the given equation...
 
I think there are infinite solutions.

Draw a cylinder with radius d across the line .
and you will see many lines going in different directions ... and touching the cylinder surface (tangent to the surface .. )_
 
The crucial point is what rootX said: there are an infinite number of skew lines a given distance from a given line. Since you are asked to find only one of them you can make a number of simplifying assumptions:

1) Choose any point you like on the given line.

2) Find a point the given distance from that point in a convenient direction- for example, you might assume that only z changes; that is, if your point on the line is (x0, y0, z0) choose (x0, y0, z0+ d) where d is the given distance.

3) Make your skew line, through that point, perpendicular to both the given line and the line connecting the two points in (2).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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