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Finding Triangle Area using Cross Product

  • Thread starter the7joker7
  • Start date
113
0
1. Homework Statement

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?
 

Answers and Replies

351
2
%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
Science Advisor
Homework Helper
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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.
 
113
0
RootX, the square root of what you have square doesn't match up with the answer you have.
 
36
0
-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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