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Finding Triangle Area using Cross Product

by the7joker7
Tags: cross, product, triangle
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the7joker7
#1
Apr5-08, 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?
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rootX
#2
Apr5-08, 11:30 PM
rootX's Avatar
P: 1,294
%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
HallsofIvy
#3
Apr6-08, 06:55 AM
Math
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Thanks
PF Gold
P: 39,345
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.

the7joker7
#4
Apr6-08, 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.
RyanSchw
#5
Apr6-08, 06:24 PM
RyanSchw's Avatar
P: 37
-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
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]


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