1. 69. The Shortest Distance Between Skew Lines
Find the angle and distance between two given skew lines. (Skew lines are
non-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 the
direction of each line are given - the former by coordinates and the latter by direction
cosines.
II. Find the angle and distance between two opposite edges of a tetrahedron whose six
edges are known.
The distance between two skew lines is naturally the shortest distance between the
lines, 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
P 1 Ÿa 1 , b 1 , c 1 and P 2 Ÿa 2 , b 2 , c 2 ; d P 1 P 2 ˜a 2 a 1 , b 2 b 1 , c 2 c 1 ™ is the
direction vector from P 1 to P 2 . Let u 1 ˜l 1 , m 1 , n 1 ™ and u 2 ˜l 2 , m 2 , n 2 ™ be unit
direction vectors for the given lines; then the components of u i are the direction
cosines for the lines. Let . be the sought-for angle and k the sought-for mininum
distance between the two lines.
P2
u1 x u2
X2
l2
u2
k d
u1
l1
X1
P1
The solution to this problem becomes very simple with the introduction of the dot (or
scalar) product u 1 u 2 and the cross product u 1 _ u 2 . We have cos . uu 1 uu2
1 2
u 1 u 2 |l 1 l 2 m 1 m 2 n 1 n 2 | from which . can be found. (We can assume that . is
acute, thus the absolute values.) u 1 _ u 2 is orthogonal (perpendicular) to both lines,
so the absolute value of the (scalar) projection of d onto u 1 _ u 2 gives k.
1

2. b
θ
a
ab
Recall that the vector projection of b on a is aa
a and the scalar projection is
d u 1 _u 2 d u 1 _u 2
1
a b. Thus k u 1 _u 2
sin .
or
a
a2 a1 b2 b1 c2 c1
k det l1 m1 n1 / sin ..
l2 m2 n2
Note 1. Since skew lines are not parallel, u 1 _ u 2 F 0.
Note 2. Let X 1 Ÿx 1 , y 1 , z 1 and X 2 Ÿx 2 , y 2 , z 2 be closest points on the lines, and let
k X 1 X 2 . Let r i be the unique numbers such that X i P i r i u i . Since
X 1 X 2 X 1 P 1 P 1 P 2 P 2 X 2 , we get k r 1 u 1 d r 2 u 2 . Since k is orthogonal to
both lines, taking the dot product with u 1 and u 2 yields the system of linear equations:
u1 u1r1 u1 u2r2 u1 d 0
u1 u2r1 u2 u2r2 u2 d 0
in r 1 and r 2 . There is a solution if u 1 u 2 F o1, and this is the case since the lines
are not parallel. Then X i can be found.
Solution of II. Let the vertices of the tetrahedron be A, B, C, O, the six edges
BC, 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

3. C
a
r B
q
b
c
O
p
A
Let the angle and distance between the two opposite edges c and r be . and k
respectively.
Determination of .. First of all,
cr AB OC
AO OB OA AC
OB AC
q b,
and thus
cr cr cr qb c q q r b c b r.
However,
cr cr ccr r 2c r c 2 r 2 2cr cos .
too, 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, a 2 b 2 c 2 2bc cos BAC,
and BAC and the angle between b and c are supplemental, so
a 2 b 2 c 2 2bc cos b 2 c 2 2b c . (Similarly for dOAC.) It follows that
3

4. 2cr cos . cr cr 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 . can be found. (We can always choose . in the range 0 to 90 , since when
two lines intersect, one of the vertical angle pairs is in this range, the other being
supplemental. Note that b and q, and a and p are lengths of opposite sides of the
tetrahedron.)
Calculation of k. Let the volume of the tetrahedron be T. By No. 68, we can consider
this quantity known. Translate vector r parallel to itself so itself so that its starting
point (initial point or tail) is at A; call the translated end point (or head) Q. Then
AQ # OC.
C
a
Q
B
r
q
b
c
O
p
A
dCQA : dAOC (SSS), and thus tetrahedrons CQAB and AOCB have the same
volume T. Now consider dQAB as the base of tetrahedron CQAB and C as its apex.
The base area is 1 AQ AB sin QAB 1 rc sin .. (Remember that . is the angle
2 2
between c and r .) To find the altitude of CQAB as the (perpendicular) distance from
C to the plane QAB, note that the plane QAB can be generated by translating line OC
parallel to itself along line AB. Then line OC lies in a plane parallel to plane QAB. It
follows that the altitude from C to QAB is k, the shortest (perpendicular) distance
between lines OC and AB (and between the two planes). The volume of tetrahedron
CQAB is then 1 1 rc sin . k, and thus 6T kcr sin . or
3 2
k 6T .
cr sin .
Corollary. c_r c r sin . cr sin .. With k denoting the shortest vector between
4

5. 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 distance
between them (all thought of as vectors) equals six times the volume of the
tetrahedron.
A direct consequence of this is the famous
Theorem of Steiner. All tetrahedrons having two opposite edges of given length lying on
two fixed lines have the same volume.
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