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Answer to Question #25561 in Vector Calculus for jon

Question #25561
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
Expert's answer
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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