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# The shortest line between two lines in 3D

Two lines in 3 dimensions generally don’t intersect at a point, they may be parallel (no intersections) or they may be coincident (infinite intersections) but most often only their projection onto a plane intersect.. When they don’t exactly intersect at a point they can be connected by a line segment, the shortest line segment is unique and is often considered to be their intersection in 3D.

The following will show how to compute this shortest line segment that joins two lines in 3D, it will as a biproduct identify parrallel lines. In what follows a line will be defined by two points lying on it, a point on line “a” defined by points P1 and P2 has an equation

Pa = P1 + mua (P2
P1)

similarly a point on a second line “b” defined by points
P4 and P4 will be written as

Pb = P3 + mub
(P4 – P3)

The values of mua and mub range from negative to positive infinity. The line segments between P1
P2 and P3 P4 have their corresponding mu
between 0 and 1.

There are two approaches to finding the shortest line segment between lines “a” and “b”. The first is to write down the length of the line segment joining the two lines and then find the minimum. That is, minimize the following

|| Pb – Pa ||2

Substituting the equations of the lines gives

|| P1 – P3 + mua (P2
P1) – mub (P4 – P3) ||2

The above can then be expanded out in the (x,y,z) components. There are conditions to be met at the minimum, the derivative with respect to mua and mub must be zero. Note: it is easy to convince
oneself that the above function only has one minima and no other minima or maxima. These two equations can then be solved for mua and mub, the actual intersection points found by substituting the values of mu into the original equations of the line.

An alternative approach but one that gives the exact same equations is to realize that the shortest line segment between the two lines will be perpendicular to the two lines. This allows us to write two equations for the dot product as

(Pa – Pb) dot (P2 – P1) = 0

(Pa – Pb) dot (P4 – P3) = 0

Expanding these given the equation of the lines

( P1 – P3 + mua (P2
P1) – mub (P4 – P3) ) dot
(P2 – P1) = 0

( P1 – P3 + mua (P2
P1) – mub (P4 – P3) ) dot
(P4 – P3) = 0

Expanding these in terms of the coordinates (x,y,z) is a nightmare but the

result is as follows

d1321 + mua d2121 – mub
d4321 = 0

d1343 + mua d4321 – mub
d4343 = 0

where

dmnop = (xm – xn)(xo
xp) + (ym – yn)(yo – yp)
+ (zm – zn)(zo – zp)

Note that dmnop = dopmn

Finally, solving for mua gives

mua = ( d1343 d4321 – d1321
d4343 ) / ( d2121 d4343 – d4321
d4321 )

and backsubstituting gives mub

mub = ( d1343 + mua d4321 )
/ d4343