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!
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$\begingroup$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 $\tfrac12\sqrt{29} \approx 2.69258$.
$\endgroup$ 1 $\begingroup$So solution is
$\text{Area} = 2.693$
$\text{Sides}: a = 3,\ b = 5,\ c = 2.449$
Using $3D$ triangle calculator:
Answer by Ross Millikan is the best algorithm to solve this.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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
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