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Concept 03 / 06Three Dimensional Geometry
Skew Lines, Shortest Distance and Intersection of Two Lines
Classifying two lines as intersecting, parallel or skew, the shortest distance formula using the scalar triple product, distance between parallel lines, the condition for coplanar lines, finding the point of intersection, and the equation of the common perpendicular.
Figure 1
Two lines in space can be related in three ways: they intersect, they are parallel, or they are skew (neither parallel nor intersecting). Skew lines do not exist in the plane; they are a genuinely three-dimensional phenomenon, like two roads at different levels crossing on a flyover.
Let the lines be r=a1+λb1 and r=a2+μb2.
Shortest distance between skew lines. There is a unique segment perpendicular to both lines (the common perpendicular), shown in the figure, and its length is the shortest distance. Its direction is b1×b2. The shortest distance is the projection of a2−a1 (joining a point of one line to a point of the other) onto this direction:
d=∣b1×b2∣∣(a2−a1)⋅(b1×b2)∣.
The numerator is a scalar triple product ∣[a2−a1b1b2]∣, which is a 3×3 determinant in Cartesian form:
d=∣b1×b2∣x2−x1a1a2y2−y1b1b2z2−z1c1c2.
Coplanar (intersecting) lines. Two non-parallel lines intersect iff the shortest distance is zero, i.e.
[a2−a1b1b2]=0.
This is also the condition for the lines to be coplanar, since the three vectors then lie in one plane. The plane containing both lines has normal b1×b2 and passes through a1.
Parallel lines (b1∥b2). The formula above fails because b1×b2=0. The distance between parallel lines is the distance from any point of one to the other: d=∣b∣∣(a2−a1)×b∣.
Finding the point of intersection. Write the general points of both lines, (x1+a1λ,…) and (x2+a2μ,…), and equate coordinates. Solve two of the three equations for λ and μ, then check the third. If it is satisfied, the lines intersect at the corresponding point; if not, they are skew (assuming they are not parallel).
Equation of the common perpendicular. Take general points P(λ) on the first line and Q(μ) on the second, and impose PQ⋅b1=0 and PQ⋅b2=0. Solving gives the feet P and Q; the common perpendicular is the line PQ, and PQ is the shortest distance. This method also confirms the formula.
Key Takeaways:
- Intersecting, parallel or skew: check parallelism of b1,b2, then the triple product.
- Skew lines: d=∣b1×b2∣∣(a2−a1)⋅(b1×b2)∣.
- Coplanar / intersecting (non-parallel): [a2−a1b1b2]=0.
- Parallel lines: d=∣b∣∣(a2−a1)×b∣.
- Intersection point: equate general points, solve two equations, verify the third.
- Common perpendicular: PQ⊥b1 and PQ⊥b2.
- Pitfall: applying the skew-line formula to parallel lines (division by zero).
Brain-Teaser 1: Find the shortest distance between r=(i^+2j^+k^)+λ(i^−j^+k^) and r=(2i^−j^−k^)+μ(2i^+j^+2k^).
Solution: a2−a1=(1,−3,−2). b1×b2=(1,−1,1)×(2,1,2)=((−1)(2)−(1)(1),(1)(2)−(1)(2),(1)(1)−(−1)(2))=(−3,0,3), with magnitude 32. The dot product is −3+0−6=−9. So d=329=23. The lines are skew, since d=0 and the directions are not parallel.
Brain-Teaser 2: Show that the lines 2x−1=3y−2=4z−3 and 5x−4=2y−1=z intersect, and find the point.
Solution: General points: (1+2λ,2+3λ,3+4λ) and (4+5μ,1+2μ,μ). Equate x and y: 2λ−5μ=3 and 3λ−2μ=−1. Solving: from the first, λ=23+5μ; substitute: 29+15μ−2μ=−1, so 9+11μ=−2, μ=−1, λ=−1. Check z: 3+4(−1)=−1 and μ=−1. It matches, so the lines intersect at (−1,−1,−1). The third equation is the real test; the first two can always be solved.
Brain-Teaser 3: Find the distance between the parallel lines r=(i^+j^)+λ(2i^−j^+k^) and r=(2i^+j^−k^)+μ(4i^−2j^+2k^).
Solution: The directions are parallel ((4,−2,2)=2(2,−1,1)). With b=(2,−1,1) and a2−a1=(1,0,−1): (1,0,−1)×(2,−1,1)=(0⋅1−(−1)(−1),(−1)(2)−1⋅1,1⋅(−1)−0⋅2)=(−1,−3,−1), with magnitude 11. So d=611=611. Recognising parallelism first is what selects the right formula.
