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Distance between triangle incenter and vertices Ask Question Asked 2 years ago Modified 2 years ago
I would approach the issue from a completely different direction. Consider a triangle in 3D with vertices at $\vec {v}_0$, $\vec {v}_1$, and $\vec {v}_2$. It has a directed normal $\vec {n}$, $$\vec {n} = \left (\vec {v}_1 - \vec {v}_0\right)\times\left (\vec {v}_2 - \vec {v}_0\right) \tag {1}\label {1}$$ If we look along $\vec {n}$ in one direction, the vertices are clockwise; in the opposite ...
geometry - Orientation of a triangle's vertices in 3D space: Clockwise ...
To get a formula where the vertices can be anywhere, just subtract the coordinates of the third vertex from the coordinates of the other two (translating the triangle) and then use the above formula.
Anyone know of an online tool available for making graphs (as in graph theory - consisting of edges and vertices)? I have about 36 vertices and even more edges that I wish to draw. (why do I have so
Think about the vertices of the polygon as potential candidates for vertices of the triangle. Using that, you get (n choose 3) as the number of possible triangles that can be formed by the vertices of a regular polygon of n sides.