 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/1/b/i/solution

The three lines have the convenient parametric equations

L_{1} : (s, 0, 1), L_{2} : (1, s, 0), L_{3} : (0, 1, s), s \in R .

For

L_{1}

, substitute

a = i

b = j

j + i \land y = (0, 1 - y_{3}, y_{2})

. Cyclically permuting the coordinates gives all three planes:

Π_{1} : Π_{2} : Π_{3} : (1 - y_{3}) r_{2} + y_{2} r_{3} = y_{2}, y_{3} r_{1} + (1 - y_{1}) r_{3} = y_{3}, (1 - y_{2}) r_{1} + y_{1} r_{2} = y_{1} .

Each normal vector is nonzero because

y

is not on the corresponding line. Thus these are genuine planes, and each is unique.

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