Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 14 / a / iii

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- Solution iii

Solution

The normal from (ii) gives the plane equation

x + y + 2 z = 3

. The distance from a point to a plane is the absolute value of the normalized plane equation at that point. At the origin this gives

d = \frac{3}{1 ^{2} + 1 ^{2} + 2 ^{2}} = \frac{3}{6} = \frac{6}{2} .

For a geometric check, the perpendicular foot is

H = (1/2, 1/2, 1)

, parallel to the normal and satisfying the plane equation. Its Euclidean norm is the same distance.

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