Simulation · Maths · Class 12

Skew gap

From the lesson Skew lines and the shortest distance in Three Dimensional Geometry. Change the values and watch what happens.

The idea behind it

NCERT §11.5–11.5.1

Two lines in space either meet, are parallel, or are skew: neither parallel nor meeting, so no plane holds both. In a box-shaped room, a diagonal of the ceiling and a diagonal of a side wall can be skew.
The shortest distance between two lines is the length of the shortest segment joining a point of one to a point of the other. It is 0 for meeting lines.
For skew lines the shortest segment is perpendicular to both lines, so it runs along b₁ × b₂.
For r = a₁ + λb₁ and r = a₂ + μb₂ the shortest distance is d = |(b₁ × b₂)·(a₂ − a₁)| / |b₁ × b₂|: the projection of any joining vector a₂ − a₁ on the common perpendicular.
With coordinates, the top is a determinant: stack the joining vector a₂ − a₁ above the two ratio triples and take its absolute value. The bottom is |b₁ × b₂|, the length of the cross product of the ratio triples.
Worked: r = î + ĵ + λ(2î − ĵ + k̂) and r = 2î + ĵ − k̂ + μ(3î − 5ĵ + 2k̂). Here a₂ − a₁ = î − k̂ and b₁ × b₂ = 3î − ĵ − 7k̂ of length √59, so d = |3 + 7|/√59 = 10/√59.
If the numerator is 0 (and the lines are not parallel), the lines meet: that determinant being 0 is the test for two lines to intersect.

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