Analytic Geometry (Coordinate Geometry) in Three-dimensional Space
Point, Line and Plane - orthogonal projections, distances, perpendicularity of line and plane
Distance between parallel lines example

Distance between parallel lines
 Distance between two parallel lines we calculate as the distance between intersections of the lines and a plane orthogonal to the given lines. The direction vector of the plane orthogonal to the given lines is collinear or coincides with their direction vectors that is N = s = ai + b j + ck By plugging any of two points, which are included in the equations of the given lines, into the equation of the plane, determined is the parameter D, as well as determined is one of the intersection points.
Example:   Find the distance between given parallel lines,
Solution:  The direction vector of a plane orthogonal to the parallel lines is collinear with the direction vectors of these lines, so  N = s = 2i - 9 j - 2k.
Let the plane passes through the point A´2(-5, -3, 6)  of the second line, then
A´2(-5, -3, 6)  =>  Ax + By + Cz + D = 0  or  2 · (-5) - 9 · (-3) + (-2) · 6 + D = 0,    D = -5
thus, the equation of the plane 2x - 9y - 2z - 5 = 0, besides the point A´2(-5, -3, 6) is the intersection of the second line and the plane.
The intersection of the first line and the plane we calculate as follows,
The distance between the intersection points 1 and 2 is at the same time the distance between given lines, thus
Distance between two skew lines
Through one of a given skew lines lay a plane parallel to another line and calculate the distance between any point of that line and the plane.
 The direction vector of planes, which are parallel to both lines, is coincident with the vector product of direction vectors of given lines, so we can write N = s1 ´ s2 A plane to lie on the one of the given lines, we should plug the point, that is included into the equation of the line, into the equation of the plane and so determined is the parameter D of the plane.

Example:   Find the distance between given skew lines,
Solution:  Through the line l1 lay a plane parallel to the line l2. The direction vector of the plane,
The point included in the equation of the line l1 must satisfy the equation of the plane, so
P1(7, -10, -5)  =>   Ax + By + Cz + D = 0,   20 · 7 +  (-4) · (-10) + (-22) · (-5) + D = 0,
D = -290  thus, the equation of the plane  20x - 4y - 22z - 290 = 0.
The distance between the point T2(-6, -1, 2), of the line l2, and the plane which is parallel to it
As was already mentioned, the sign of the square root is taken to be opposite to the sign of the parameter D
Therefore, the negative value of the result only informs us that the point is positioned at the same side as the origin of the coordinate system.
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