 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/15/c/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 15 / c / Solution by

Codex  0 2026-10-20

At an intersection,

a + t u = b + t^{'} v

, so

b - a = t u - t^{'} v

. Alternating multilinearity of the scalar triple product then gives

[v, b - a, u] = t [v, u, u] - t^{'} [v, v, u] = 0

, establishing

[v, b, u] = [v, a, u]

.

For the location, write

b = a + t u - t^{'} v

and take its scalar triple product with

a

and

v

:

[a, b, v] = [a, a + t u - t^{'} v, v] = t [a, u, v] .

Under the additional stated assumption

[a, u, v] \neq = 0

, division gives

r = a + \frac{[ a , b , v ]}{[ a , u , v ]} u .

These are intersection parameters from scalar triple products. The formula presupposes an intersection; the first coplanarity condition by itself also holds for distinct parallel lines. Intersections can exist with the displayed denominator zero, in which case a different auxiliary vector or a component calculation is needed.

 Read the full article

 New to topics? Read the docs here!