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 / ii

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- Solution ii

Solution

For the triangular face, take the cross product of two edges and divide by two:

\frac{1}{2} PQ \times PR = \frac{1}{2} (- 1, 1, 0) \times (- 2, 0, 1) = \frac{1}{2} (1, 1, 2) .

The origin lies on the opposite side from this normal, since

(1, 1, 2) \cdot PO = - 3

. Thus this is already the outward vector area, with

A = \frac{6}{2}, \hat{n} = \frac{( 1 , 1 , 2 )}{6}, A \hat{n} = (\frac{1}{2}, \frac{1}{2}, 1) .

The vector areas of a closed polyhedron sum to zero, so the sum for the remaining three faces is

(- \frac{1}{2}, - \frac{1}{2}, - 1) .

Figure 1.
Tetrahedron with its highlighted face PQR and outward unit normal
. Details.

Their individual outward vector areas are

(0, 0, - 3/2)

,

(1/2, - 1, 1)

, and

(- 1, 1/2, - 1/2)

, which directly check that sum.

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