
#1
Apr508, 10:09 PM

P: 115

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? 



#2
Apr508, 11:30 PM

P: 1,295

%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 



#3
Apr608, 06:55 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

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.




#4
Apr608, 05:26 PM

P: 115

Finding Triangle Area using Cross Product
RootX, the square root of what you have square doesn't match up with the answer you have.



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