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I think the area for the parallelogram would be the length of the cross product.
|u x v|

Therefore, the area of the triangle would be the half of }u x v|

Thanks!

Could you write it out? I'm a little confused, I guess.

what have you some up with in the cross product so far?

I have (20, 0, 0) but I feel like I've done something wrong...
I'm not very good with vectors.

thats the "formal" confusion for it:
me, i just do this
x 4 5 x = 1
y 0 -1 y = 5
z 1 0 z = -4

How does what you just did work?

<1,5,-4> ^2 = 1 + 25 + 16 = sqrt(42)
sqrt(42)/2 = area triangle

thank you!!

there are other ways to do it if you cant cross that well :)

but the cross bypasses any trig you might not know

Ok. I think this works pretty well for me:P