Finding the area of a triangle on the space given three points

**Ulysses** · June 5th 2010, 02:36 PM

Hi there, I must solve the next problem. The sentence says:
Find the area of a triangle wich vertex are the points: $V_1(1,5,-2)$ $V_2(0,0,0)$ and $V_3(3,5,1)$ .

What I did was define the vectors: $\vec{v_1}=(1,5,-2)$ y $\vec{v_3}=(3,5,1)$ , considering the point $v_2$ is the origin, these vectors are $\vec{v_2v_1}$ y $\vec{v_2v_3}$ .

So I took the vectors $\vec{v_1}$ and $\vec{v_3}$ and then what I did was getting the module of the vectorial product, which is the area of the parallelogram conformed by these two vectors, and then I've divide it on two:

$A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}$

$v_1\wedge{v_3}=(15,-7,-10)\Rightarrow{||(15,-7,-10)||=\sqrt[ ]{347}}$

Then: $A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}=\displ aystyle\frac{\sqrt[ ]{347}}{2}$

Is it right?

Bye, and thanks for commenting.

**skeeter** · June 5th 2010, 03:18 PM

Originally Posted by Ulysses

Hi there, I must solve the next problem. The sentence says:
Find the area of a triangle wich vertex are the points: $V_1(1,5,-2)$ $V_2(0,0,0)$ and $V_3(3,5,1)$ .

What I did was define the vectors: $\vec{v_1}=(1,5,-2)$ y $\vec{v_3}=(3,5,1)$ , considering the point $v_2$ is the origin, these vectors are $\vec{v_2v_1}$ y $\vec{v_2v_3}$ .

So I took the vectors $\vec{v_1}$ and $\vec{v_3}$ and then what I did was getting the module of the vectorial product, which is the area of the parallelogram conformed by these two vectors, and then I've divide it on two:

$A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}$

$v_1\wedge{v_3}=(15,-7,-10)\Rightarrow{||(15,-7,-10)||=\sqrt[ ]{347}}$

Then: $A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}=\displ aystyle\frac{\sqrt[ ]{347}}{2}$

Is it right?

Bye, and thanks for commenting.

$\frac{\sqrt{374}}{2}$

**Ulysses** · June 5th 2010, 03:22 PM

Thanks, didn't note that :P

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