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Analytical Geometry 11

Figure 1.12: The distance between two lines in 3D.

two parameters s and t that yield these q

1

and q

2

:

q

1

= l

1

(s) = p

1

+ su ,

q

2

= l

2

(t) = p

2

+ tv .

The vector q

1

− q

2

is perpendicular to u and v:

h(p

1

− p

2

) + su − tv, ui = 0 ,

h(p

1

− p

2

) + su − tv, vi = 0 ,

leading to:

hp

1

− p

2

, ui + skuk

2

− thv, ui = 0 ,

hp

1

− p

2

, vi − tkvk

2

+ shu, vi = 0 .

Hence, we have two equations and two unknowns. Therefore we

solve for s and t:

˜s =

βhw, vi− kvk

2

hw, ui

kuk

2

kvk

2

− β

2

,

˜

t =

kuk

2

hw, vi− βhw, ui

kuk

2

kvk

2

− β

2

,

where β = hv, ui and w = p

1

− p

2

. Finally, we have

dist(l

1

, l

2

) = kl

1

(˜s) −l

2

(

˜

t )k.

Cross product.

1

We end this chapter with a quick review of

the notion of a