# 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

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)

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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