 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/11/b/solution

The cross product of the direction vectors is

n_{1} \times n_{2} = (1, - 1, 0)

. It is nonzero, so a common perpendicular direction exists. Projecting

a_{2} - a_{1} = (- 1, 0, 1)

s = \frac{∣ ( a _{2} - a _{1} ) \cdot ( n _{1} \times n _{2} ) ∣}{∥ n _{1} \times n _{2} ∥} = \frac{∣ - 1∣}{2} .

Thus

s = \frac{1}{2} .

The numerator is a scalar triple product; movement along either line does not change this perpendicular projection.

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