 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/d/i/solution

For a point

p = r_{2} (t)

, the magnitude of

(p - a_{1}) \times n_{1}

is the area of a parallelogram. Dividing by its base length

∥ n_{1} ∥

p - a_{1} = (t - 1, t, t + 1), (p - a_{1}) \times n_{1} = (- (t + 1), t + 1, - 1) .

Hence

d (t) = \frac{2 ( t + 1 ) ^{2} + 1}{2} = (t + 1)^{2} + \frac{1}{2} .

This use of the cross product removes any component parallel to the first line automatically.

 New to topics? Read the docs here!