 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-1/5c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Assume the direction vectors are nonzero. The cross product equation is equivalent to

(x - a) \times u = 0,

x - a

is parallel to

u

and the locus is the line

x = a + λ u

For nonparallel directions,

u_{1} \times u_{2}

is normal to both lines. Projecting their displacement onto the corresponding unit vector gives the distance between skew lines

d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( u _{1} \times u _{2} ) ∣}{∣ u _{1} \times u _{2} ∣} .

If the directions are parallel, the corresponding formula is

d = \frac{∣ ( a _{2} - a _{1} ) \times u _{1} ∣}{∣ u _{1} ∣} .

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