 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/11/a/solution

The direction vector is

b - a = (1, 2, - 2)

. The vector equation and symmetric Cartesian equation of the line are

r = (1, 2, 4) + t (1, 2, - 2), x - 1 = \frac{y - 2}{2} = \frac{z - 4}{- 2} .

The

(x, y)

-plane has

z = 0

, which gives

t = 2

. Thus its intersection with the line has position

r = (3, 6, 0) .

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