Finding Triangle Area using Cross Product

[the7joker7]

Homework Statement

Find the area of a triangle `PQR`, where `P=(0,4,4)`, `Q=(2,-6,-5)`, and `R=(-3,-5,6)`

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, $\vec{u}$ and $\vec{v}$, is $|\vec{u}\times\vec{v}|$. Since a triangle is half a parallelogram, the the area of a triangle having sides $\vec{u}$ and $\vec{v}$ is half that.

 

[the7joker7]

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

 

[RyanSchw]

-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

$$\sqrt{(-101)^2 + (23)^2 + (-48)^2}= 114.167 ~ /2 = 57.0833$$