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I have coordinates of 3d triangle and I need to calculate its area. I know how to do it in 2D, but don't know how to calculate area in 3d. I have developed data as follows.

(119.91227722167969, 122.7717056274414, 39.3568115234375), 
(119.8951187133789, 122.7717056274414, 39.38057327270508), 
(121.11941528320312, 123.2818832397461, 38.41301345825195)
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3 Answers 3

up vote 5 down vote accepted

Say you have 3 points $\mathbf{A, B, C}$. Find the angle between $\mathbf{AB}$ and $\mathbf{AC}$ using dot product (i.e. $\mathbf{AB}\cdot\mathbf{AC}=|\mathbf{AB}||\mathbf{AC}|\cos\theta$) and then you can find the area of the triangle using $$ A=\frac{1}{2}|\mathbf{AB}||\mathbf{AC}|\sin\theta $$

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If your 3 points are A, B, C then you may use directly the (half) cross product formula : $$S=\dfrac{|\mathbf{AB}\times\mathbf{AC}|}2=\dfrac{|\mathbf{AB}||\mathbf{AC}||\sin(\theta)|}2 $$ that is (see the Wikipedia link to get the cross-product in $\mathbb{R}^3$) : $$S=\frac 12 \sqrt{(x_2\cdot y_3-x_3\cdot y_2)^2+(x_3\cdot y_1-x_1\cdot y_3)^2+(x_1\cdot y_2-x_2\cdot y_1)^2}$$ if $\mathbf{AB}=(x_1,x_2,x_3)$ and $\mathbf{AC}=(y_1,y_2,y_3)$

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If you have 3 coordinates in 3D $(x_1,y_1,z_1);(x_2,y_2,z_2);(x_3,y_3,z_3)$ make two coterminous vectors like $\vec a=(x_2-x_1)I+(y_2-y_1)j+(z_2-z_1)k$ $\vec b=(x_3-x_1)I+(y_3-y_1)j+(z_3-z_1)k$ now find vector product of $\vec a$ and $\vec b$ and area of triangle formed will be $\frac{1}{2}\vec a \times \vec b$. You are with your desired answer.

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awesome it has cleared up my douts. –  lalit arora Nov 4 '13 at 8:49
    
NOTE: This is the least-expensive formula to compute, as it does not require square root, nor trigonometric calculation. The main computation work is the cross product, which only uses multiplies and add/subtract. EDIT No I'm wrong, in 3D must take MAGNITUDE of cross-product, which requires a SQUARE-ROOT. So Raymond Manzoni's answer is about the same cost. –  ToolmakerSteve Mar 28 at 3:33

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