point inside a triangle

Question

I wanted to ask you if anyone knew how to check if a point was inside a given triangle, on a reference system in space. I am aware that speaking of 2d systems I can obtain this point with the following procedures: To determine if a given point v lies within a given triangle, consider a single vertex, denoted v0, v1 and v2 are the vectors from the other two vertices v0. Expressing the vector from v0 to v in terms of v1 and v2 then gives

v = v0 + av1 + bv2

where a, b are constant. Solve for a, b

a= (det (v v2) - det (v0 v2)) / (det (v1 v2))
b= - (det (v v1) - det (v0 v1)) / (det (v1 v2))

So the point v is inside the triangle if a, b> 0 e a + b <1. I was wondering if there was a similar procedure or not.

After MBo's advice I wrote the following code:

    def pointLocatedOnTheTriangularFace(self, point):
    vertList = list(self.vertices)
    system = []
    for i in range(4):
        p = vertList[1].coords-vertList[0].coords
        q = vertList[2].coords-vertList[0].coords
        n = np.cross(p,q)
        for i in range(3):
            system.append([p[i],q[i],n[i]])
        try:
            solution = list(np.linalg.solve(system,np.array(point-vertList[0].coords)))
            if solution[0] > 0 and solution[1] > 0 and solution[0]+solution[1]<1 and solution[2]==0:
                return True
        except np.linalg.LinAlgError:
            print("The system having the floor {0} {1} {2} with the point {3} does not admit solutions.\n".format(vertList[0].coords,vertList[1].coords,vertList[2].coords,point))
        v = vertList.pop(0)
        vertList.append(v)
    return False

I am working with tetrahedra, and the triangles that interest me are the faces of the tetrahedron. I state that it does not work but I can not understand why. Can anyone tell me what am I doing wrong? I changed the code in the following way and it seems to work.

def pointLocatedOnTheTriangularFace(self, point):
    vertList = list(self.vertices)
    system = []
    for i in range(4):
        r = point-vertList[0].coords
        p = vertList[1].coords-vertList[0].coords
        q = vertList[2].coords-vertList[0].coords
        n = np.cross(p,q)
        if np.inner(r,n) == 0:
            system = [[p[0],q[0]],[p[1],q[1]]]
            try:
                solution = list(np.linalg.solve(system,np.array([r[0],r[1]])))
                if solution[0] > 0 and solution[1] > 0 and solution[0]+solution[1]<1:
                    return True
            except np.linalg.LinAlgError:
                print("The system having the floor {0} {1} {2} with the point {3} does not admit solutions.\n".format(vertList[0].coords,vertList[1].coords,vertList[2].coords,point))
        v = vertList.pop(0)
        vertList.append(v)
    return False

With the following tests:

def test_PointPlacedOnTheTriangularFace(self):
    tr1Vertices = [vertex([0,4.32978e-17,0.5],1),vertex([-0.433013,0.25,-0.5],2),vertex([-4.32978e-17,-0.5,-0.5],3),vertex([0.433013,0.25,-0.5],4)]
    tr1= tetrahedron(tr1Vertices)
    point = vertex([0,0,-0.5],1)
    self.assertTrue(tr1.pointLocatedOnTheTriangularFace(point.coords), msg="In this test the point is inside the face")

def test_PointNotPlacedLocatedOnTheTriangularFace(self):
    tr1Vertices = [vertex([0,4.32978e-17,0.5],1),vertex([-0.433013,0.25,-0.5],2),vertex([-4.32978e-17,-0.5,-0.5],3),vertex([0.433013,0.25,-0.5],4)]
    tr1= tetrahedron(tr1Vertices)
    point = vertex([1,0,-0.5],1)
    self.assertFalse(tr1.pointLocatedOnTheTriangularFace(point.coords), msg="In this test the point is outside the face")

If anyone has any advice for me I will surely treasure it. Thanks.

Answer 1

Described 2D approach is essentially decomposition of vector r = v-v0 by basis vectors p=v1-v0 and q=v2-v0.

In 3D you can decompose vector r by vectors p, q and n = p x q (where x denotes vector product operation)

If resulting coefficients a,b,c fulfill limits a, b > 0, a + b < 1, c=0, then point v lies in triangle plane inside it.

For decomposition solve this linear system for unknowns a,b,c:

rx = a * px + b * qx + c * nx
ry = a * py + b * qy + c * ny
rz = a * pz + b * qz + c * nz

Alternative approach - check that dot product r.dot.n is zero - in this case point lies in the plane, coefficient с is zero, and you can solve simplified system for a and b choosing a pair of equations and excluding the third summand (same method as in 2D)