1. The problem statement, all variables and given/known data Find the area of a triangle `PQR`, where `P=(0,4,4)`, `Q=(2,-6,-5)`, and `R=(-3,-5,6)` 3. The attempt at a solution The vector PQ = (2, -10, -9) The vector PR = (-3, -9, 2) Using matrixes I set up something that looks like this... I J K 2 -10 -9 -3 -9 2 Then using the matrix methods I get. I(-20 - 81) - J(4 - 27) + K(-18 - 30) I(-101) - J(-23) + K(48) I take the square root of the squares and get. 109.4349122 Answer = 57.0832725061 What's the problem?
%was too lazy to do computations, so here's the matlab code (might help you) u = -3 -9 2 >> v v = 2 -10 -9 >> cross(v,u) ans = -101 23 -48 ans = 114.1665 >> ans/2 ans = 57.0833 why divided by 2? because axb = |a|.|b|.sin theta and area is 1/2 of that
The area of a parallelogram formed by two vectors, [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex], is [itex]|\vec{u}\times\vec{v}|[/itex]. Since a triangle is half a parallelogram, the the area of a triangle having sides [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is half that.
If you take the magnitude the negative numbers will become positive [tex] \sqrt{(-101)^2 + (23)^2 + (-48)^2}= 114.167 ~ /2 = 57.0833 [/tex]