 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/4/solution

The traversal fixes the orientation of the vector area as half the cross product of the edges

A B

and

A C

S_{A BC A} = \frac{1}{2} (1, 1, 0)^{T} \times (1, 1, 1)^{T} = \frac{1}{2} (1, - 1, 0)^{T} .

1/ 2

is the area of the triangle, and its direction follows the right-hand rule.

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