# how to calculate area of 3D triangle?

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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Say you have 3 points 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 $\mathbf{AC}=(y_1,y_2,y_3)$
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.