Finding the area of a triangle in two-space.

I know how to find the area of a triangle in three-space, say ABC: First I would find $\displaystyle \vec{AB}$ and then $\displaystyle \vec{AC}$ and then use the formula for the area of a parallelogram:

$\displaystyle A = {\vec{AB} \times \vec{AC}$

Finding the magnitude of that and dividing it by two (since the area of a triangle is half that of the area of a parallelogram), I would find the area of the triangle ABC.

But, how would I find the area of a triangle that does not have a Z component (is in two-space)? For example: Find the area of the triangle with vertices of (1, -2), (-1, 3), (2, 4). Since this is in two-space, I can't take the cross product of any two of these vectors to find the area. According to my book, the area of this triangle is $\displaystyle \frac{17}{2}$ and I have no idea how they computed that. Any help? I would appreciate learning a systematic approach to solving the areas of a triangle in two-space. Thanks in advance.

Re: Finding the area of a triangle in two-space.

Why can't you add z=0 to all of them?

Re: Finding the area of a triangle in two-space.

Quote:

Originally Posted by

**TKHunny** Why can't you add z=0 to all of them?

I did just that:

A(1, -2, 0), B(-1, 3, 0), and C(2, 4, 0)

$\displaystyle \vec{AB} = (-1, 3, 0) - (1, -2, 0) = (-2, 5, 0)$

$\displaystyle \vec{AC} = (2, 4, 0) - (1, -2, 0) = (1, 6, 0)$

The cross product of these two direction vectors:

$\displaystyle \vec{AB} \times \vec{AC} = (0, 0, -18) $

Now to find the magnitude of the normal:

$\displaystyle |\vec{AB} \times \vec{AC}| = \sqrt{(0)^{2}+(0)^{2}+(-18)^{2}$

$\displaystyle |\vec{AB} \times \vec{AC}| = 18$

Since this is only the area of the parallelogram and the area of the triangle is half of that:

$\displaystyle A = \frac{18}{2} = 9$ square units.

But in my textbook, the area of the triangle is $\displaystyle \frac{17}{2}$ or 8.5. So, did I compute the area incorrectly?

Re: Finding the area of a triangle in two-space.

Check the calculations. How did you get -18?

-2*6 - 1*5 = ?

Re: Finding the area of a triangle in two-space.

Quote:

Originally Posted by

**sa-ri-ga-ma** Check the calculations. How did you get -18?

-2*6 - 1*5 = ?

Minor mistake of mine, thanks for catching it. I usually use the dot product to confirm I computed the cross product correctly, but since z = 0 for all z, I foolishly assumed it was correct even though I was off.

Many thanks to TKHunny for giving me an invaluable tool to calculating problems like these.