How do you find the area of a triangle in a 3D graph? How do you find the area of a triangle in a 3 dimensional graph? Is it any different than a regular 2d graph? How would you solve it, if these were your three points? A(1,-4,-2), B(3,-3,-3), C(5,-1,-2)? Thanks for any help!
 A: If you know what a vector cross product is, just take half of the magnitude of the cross product of the vectors $\overrightarrow{AB} = B-A$ and $\overrightarrow{AC} = C-A$
This is because
\begin{align}
\text{Area} &= \tfrac12 \times \text{base} \times \text{height}    \\
            &= \tfrac12 \|B - A\| \cdot \|C - A\|\, \sin \angle BAC  \\ 
            &= \tfrac12\|(B-A) \times (C-A)   \|
\end{align}
In your example $\;B-A =(2, 1, -1)$ and $\;C-A = (4, 3, 0)$ so 
$(B-A) \times (C-A) = (3, -4, 2)$ and the triangle area is $\tfrac12\sqrt{29} \approx 2.69258$.
A: So solution is 
$\text{Area} = 2.693$
$\text{Sides}: a = 3,\ b = 5,\ c = 2.449$
Using $3D$ triangle calculator: 
http://www.triangle-calculator.com/?what=vc&a=1&a1=-4&a2=-2&b=3&b1=-3&b2=-3&c=5&c1=-1&c2=-2&submit=Solve&3d=1
Answer by Ross Millikan is the best algorithm to solve this.
A: Heron's formula gives you the area of a triangle from the length of the sides.  Let the sides be $a,b,c$ and $s=\frac{a+b+c}2$ then the area $A=\sqrt{s(s-a)(s-b)(s-c)}$.  Works in any dimension.
A: Let $M$ be the matrix with column vectors $C-A$ and $C -B$.  Then $|\det(M^tM)|$ is the area of the parallelogram spanned by the column vectors of $M$.
Works in any dimension.
Michael
