69. The Shortest Distance Between Skew Lines

Find the angle and distance between two given skew lines. (Skew lines arenon-parallel non-intersecting lines.)

This important problem is usually encountered in one of the following forms:

I. Find the angle and distance between two skew lines when a point on each line and thedirection of each line are given - the former by coordinates and the latter by directioncosines.

II. Find the angle and distance between two opposite edges of a tetrahedron whose sixedges are known.

The distance between two skew lines is naturally the shortest distance between thelines, i.e., the length of a perpendicular to both lines.

Solution of I. In the usual rectangular xyz-coordinate system, let the two points be

P1 � �a1,b1,c1  and P2 � �a2, b2, c2 ; d � P1P2 � �a2 " a1,b2 " b1,c2 " c1 � is thedirection vector from P1 to P2. Let u1 � �l1,m1,n1 � and u2 � �l2,m2, n2 � be unitdirection vectors for the given lines; then the components of ui are the directioncosines for the lines. Let F be the sought-for angle and k the sought-for mininumdistance between the two lines.

l2

l1

k

u1 x u2

u1

u2

d

X2

X1

P2

P1

The solution to this problem becomes very simple with the introduction of the dot (or

scalar) product u1 � u2 and the cross product u1 � u2. We have cosF � u1�u2

u1 u2

�u1 � u2 � |l1l2 � m1m2 � n1n2 | from which F can be found. (We can assume that F is

acute, thus the absolute values.) u1 � u2 is orthogonal (perpendicular) to both lines,

so the absolute value of the (scalar) projection of d onto u1 � u2 gives k.

1

θ

a

b

Recall that the vector projection of b on a is a�b

a�aa and the scalar projection is

1

aa � b . Thus k � d �u1�u2

u1�u2

� d�u1�u2

sinF or

k � det

a2 " a1 b2 " b1 c2 " c1

l1 m1 n1

l2 m2 n2

/ sinF.

Note 1. Since skew lines are not parallel, u1 � u2 p 0.

Note 2. Let X1 � �x1,y1, z1  and X2 � �x2,y2, z2  be closest points on the lines, and let

k � X1X2. Let ri be the unique numbers such that X i � Pi � riui. Since

X1X2 � X1P1 � P1P2 � P2X2, we get k � "r1u1 � d � r2u2. Since k is orthogonal toboth lines, taking the dot product with u1 and u2 yields the system of linear equations:

u1 � u1r1 " u1 � u2r2 " u1 � d � 0

u1 � u2r1 " u2 � u2r2 " u2 � d � 0

in r1 and r2. There is a solution if u1 � u2 p o1, and this is the case since the linesare not parallel. Then X i can be found.

Solution of II. Let the vertices of the tetrahedron be A, B,C,O, the six edgesBC,CA,AB, OA,OB,OC have lengths a,b, c,p,q, r respectively, and the vectors

BC,CA,AB, OA,OB,OC be a , b , c , p , q , r respectively.

2

a

b

c

p

r

q

C

A

B

O

Let the angle and distance between the two opposite edges c and r be F and k

respectively.

Determination of F. First of all,

c � r � AB � OC

� AO � OB � OA � AC

� OB � AC

� q " b ,

and thus

c � r � c � r � c � r � q " b � c � q � q � r " b � c " b � r .

However,

c � r � c � r � c � c � r � r � 2c � r � c2 � r2 � 2crcosFtoo, and

2c � q � c2 � q2 " p2, 2q � r � q2 � r2 " a2,

2b � c � a2 " b2 " c2, 2b � r � p2 " b2 " r2.

Note that when the law of cosines is used with dABC, a2 � b2 � c2 " 2bc cos0BAC,

and 0BAC and the angle 2 between b and c are supplemental, so

a2 � b2 � c2 � 2bc cos2 � b2 � c2 � 2b � c . (Similarly for dOAC.) It follows that

3

2crcosF � c � r � c � r " c2 " r2

� c � q � q � r " b � c " b � r " c2 " r2

� a2 � b2 � c2 " p2 � q2 � r2 " c2 " r2

� b2 � q2 " a2 " p2

and F can be found. (We can always choose F in the range 0( to 90(, since whentwo lines intersect, one of the vertical angle pairs is in this range, the other beingsupplemental. Note that b and q, and a and p are lengths of opposite sides of thetetrahedron.)

Calculation of k. Let the volume of the tetrahedron be T. By No. 68, we can considerthis quantity known. Translate vector r parallel to itself so itself so that its startingpoint (initial point or tail) is at A; call the translated end point (or head) Q. ThenAQ 5 OC.

a

b

c

p

r

q

Q

C

A

B

O

dCQA X dAOC (SSS), and thus tetrahedrons CQAB and AOCB have the samevolume T. Now consider dQAB as the base of tetrahedron CQAB and C as its apex.The base area is 1

2AQ � AB � sin0QAB � 1

2rc sinF. (Remember that F is the angle

between c and r .) To find the altitude of CQAB as the (perpendicular) distance fromC to the plane QAB, note that the plane QAB can be generated by translating line OCparallel to itself along line AB. Then line OC lies in a plane parallel to plane QAB. Itfollows that the altitude from C to QAB is k, the shortest (perpendicular) distancebetween lines OC and AB (and between the two planes). The volume of tetrahedronCQAB is then 1

3� 1

2rc sinF � k, and thus 6T � kcr sinF or

k � 6Tcr sinF .

Corollary. c � r � c r sinF � cr sinF. With k denoting the shortest vector between

4

lines OC and AB, we have 6T � k � c � r . This is sometimes expressed as the

Theorem. The mixed product of the two opposite sides of a tetrahedron and the distancebetween them (all thought of as vectors) equals six times the volume of thetetrahedron.

A direct consequence of this is the famous

Theorem of Steiner. All tetrahedrons having two opposite edges of given length lying ontwo fixed lines have the same volume.

5