Finding Triangle Area using Cross Product

the7joker7

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?

rootX

%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

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

RootX, the square root of what you have square doesn't match up with the answer you have.

RyanSchw

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]