A triangle has vertices A (-1, 3, 4) B (3, -1, 1) and C (5, 1, 1). The area of ABC is
a) 30.1
b) 82.1
c) 9.1
d) 52.1
1
Expert's answer
2013-03-05T10:30:16-0500
Consider the vectorsAB = (3+1, -1-3, 1-4) = (4, -4, -3) AC = (5+1, 1-3, 1-4) = (6, -2, -3) Then the area of ABC is equal to S = 1/2 * AB * AC * sin BAC = 1/2 | [AB x AC] |, where [AB x AC] is a cross product of AB and AC. Let us compute vector [AB x AC] = (a,b,c). Then a = det -4 -3 -2 -3 = 12 -6 = 6
b = det -3 4 -3 6 = -18 + 12 = -6
c = det 4 -4 6 -2 = -8 + 24 = 16. Hence the length of [AB x AC] is square_root(6^2 + 6^2 + 16^2) = square_root(328) = 18.11 Therefore the area of ABC is S = 18.11/2 = 9.1. Answer c)
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