Find point between two skew lines I have two skew lines.
and I was wondering how I could find a point, C when given the distance AC and CB (AC = CB).
Thank you very much in advance.

 A: Knowing the distance from $C$ to one of your lines will restrict the locus of $C$ to a cylinder around that line, with the distance as the radius. If you have the same distance to a second line, you have a second cylinder of equal radius. So you are investigating the curve you get from intersecting two cylinders, and trying to find a point (any point?) on that curve.
You could use a parametric form for one cylinder, but turn the other into a quadratic equation instead. Then you could use the quadratic equation to express one parameter in terms of the other, thus obtaining a one-parameter description of the curve from which you could pick any point.
A: I'm going to assume that you're talking about the three-dimensional case here...
Let $\vec u = u_x i + u_y j+u_zl$
Let $\vec v = v_x i + v_y j+v_zk$
Let $A$ have coordinates $(x_A,y_A, z_A)$
Let $B$ have coordinates $(x_B,y_B, z_A)$
Let $C$ have coordinates $(x,y,z)$
Then $\vec {AC}=(x-x_A)i+(y-y_A)j+(z-z_A)k$
and $\vec {BC}=(x-x_B)i+(y-y_B)j+(z-z_B)k$
$\vec {AC} \perp \vec u \Rightarrow \vec {AC} . \vec u=0 \Rightarrow (x-x_A)u_x+(y-y_A)u_y+(z-z_A)u_z=0$
$\vec {BC} \perp \vec v \Rightarrow \vec {BC} . \vec v=0 \Rightarrow (x-x_B)v_x+(y-y_B)v_y+(z-z_B)v_z=0$
So far this gives two linear equations for three unknowns.
The third equation comes from knowing that $AC=BC$
So $(x-x_A)^2+(y-y_A)^2+(z-z_A)^2=(x-x_B)^2+(y-y_B)^2+(z-z_B)^2$
$-2xx_A+x_A^2-2yy_A+y_A^2-2zz_A+z_A^2=-2xx_B+x_B^2-2yy_B+y_B^2-2zz_B+z_B^2$
These three linear equations are:
$$u_xx+u_yy+u_zz=x_Au_x+y_Au_y+z_Au_z$$
$$v_xx+v_yy+v_zz=x_Bv_x+y_Bv_y+z_Bv_z$$
$$2(x_B-x_A)x+2(y_B-y_A)y+2(z_B-z_A)z=x_B^2-x_A^2+y_B^2-y_A^2+z_B^2-z_A^2$$
Pretty straightforward from there...
A: Let $d$ be the required distance from the lines. Choose convenient fixed points $A_1$, $A_2$ on the first line and $B_1$, $B_2$ on the second. Using the formula for the distance from a point to a line, we have the system of equations $$\begin{align}{\|(A_2-A_1)\times(A_1-C)\|^2\over\|A_2-A_1\|^2}&=d^2 \\ {\|(B_2-B_1)\times(B_1-C)\|^2\over\|B_2-B_1\|^2}&=d^2. \end{align}$$ From here, proceed with your favorite way of solving simultaneous second-degree polynomial equations. There will be zero, one, or an infinite number of solutions.
A: I am going to try to explain why there are possibly many solutions to this problem.
The point C can be thought of as the centre of a sphere that is resting on your two lines. Points A and B are the points of contact between the sphere and the lines.
If you had two parallel lines instead of two skew lines, then there are three possible scenarios:
1) The distance between the lines is greater than the diameter of the sphere.
In this case the sphere would not be able to be in contact with both lines simultaneously - no solution.
2) The distance between the lines is equal to the diameter of the sphere.
In this case the solution set for C would be the line exactly midway between the two lines and parallel to both. Infinitely many solutions.
3) The distance between the lines is less than the diameter of the sphere.
In this case you can imagine the sphere rolling along the top or the bottom of a set of rails (which are the lines). There are an infinite number of solutions above and below the lines.
In the case you describe the lines are skew lines. You can think of this as approximating two rails that are far apart from each other, grow closer and then grow further apart from each other. There is a shortest distance between the two lines and this can be found easily.
Imagine again the sphere and where it might be in contact with the two lines.
There are three scenarios again:
1) The shortest distance between the lines is greater than the diameter of the sphere.
As before, the sphere would not be able to be in contact with both lines simultaneously - no solution.
2) The shortest distance between the lines is equal to the diameter of the sphere.
In this case there is only one solution for C. Find the points of closest approach on the two lines and the midpoint will be C. There is exactly one solution.
3) The distance between the lines is less than the diameter of the sphere.
Again you can imagine the sphere sitting on the rails at the points of closest approach. It can then roll along the rails. As they grow further apart it begins to sink lower through the rails until you reach the point at which the distance apart is equal to the diameter. At this point the sphere sits halfway down between the rails and can go no further in the same direction without falling away from the lines. But you can move in the opposite direction, this time on the underside of the rails. The sphere will pass the points of closest approach and continue in a loop, rising through the rails until it reaches a point beyond which it again can not go. At this point it returns along the top of the rails.
There are infinitely many solutions, but they lie in a loop.
A: Fist of all the question is a bit ambiguous.
If the question is
Given two skew lines, is there a point C in space equidistant from them?
The answer is yes.
The midpoint of the shortest segment joining these lines is equidistant from them. The existence (and construction) of such a shortest segment joining two skew lines could be found in any solid geometry book.
If the question is
Given two skew lines, is there a point C in space whose distance from both lines is equal to a given value, say 2 units?
The answer is, it depends.
If the distance between the given lines is less than 4 units, there are infinitely many such points C. The reason is that, the cylinders whose axes are the given lines and whose radii are 2 units each, will contain in their interiors the midpoint of the shortest segment joining the given lines, and hence will certainly intersect at infinitely many points, each of which will be at a distance of 2 units from both lines, and hence could serve as point C. 
If the distance between the lines is 4 units, the midpoint of the shortest segment joining the lines is the only point in space at a distance of 2 units from the given lines.
If the distance between the lines exceeds 4 units, then such a point C does not exist.
