Finding the Area of a Triangle (using determiants)

[Noone1982]

I know this should be really simple, but it is not working out.
Say we have three vertices (0,0,0) (4,8,3) and (3,4,5)
I need to find the area of the triangle that it makes in three space with,


$$A\mbox{re}a\; =\; \frac{1}{2}\sqrt{\det \left( A^{t}A \right)}$$

But it is not working out :(

 

[TD]

I haven't seen that formula before but have you covered the cross product yet?
If so, create two vectors by substracting two of your points twice, easiest would be to substract (0,0,0) from the other two of course, giving you the vectors (3,4,5) and (4,8,3).

Then take the cross product of these two and we know that the length of their cross product is the area of the parallellogram formed by the two vectors. The only thing that's left is dividing by two. Of course, computing the cross product can be easily done by using a determinant.

It seems as if your formula is trying to do the same thing, but the way its written now it would give 0 determinants, if I'm not mistaking.
It would depend of course on what you have to put in that matrix A...

 

[Noone1982]

I actually did try to use

$$A\mbox{re}a\; =\; \frac{1}{2}\left| A\times B \right|$$

Im getting a reallllly close answer, but not exactly what it should be. I am perplexed.

 

[TD]

Could you show that work please, together with the answer you think it should be? Normally, that should work if you're using the correct vectors

 

[amcavoy]

Your first formula will work. Note that what you wrote is equal to:

$$\frac{1}{2}\sqrt{\det{A^TA}}=\frac{1}{2}\left|\det{A}\right|$$

Where A is of the form:

$$A=\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z\end{bmatrix}$$

Where your two vectors are A=<ax,ay,az> and B=<bx,by,bz>.