-- Final Practice Solutions
- Express the point (0,0.5,0) in barycentric coordinates with respect to the triangle with vertices (−1,0,0), (0,1,0), (0,0,1)
if given point p = (0, .5, 0) & triangle vertices p1 = (-1, 0, 0), p2 = (0, 1, 0), p3 = (0, 0, 1) we can calculate
a1 = s(d22d31 - d12d32)
= 1/3(21.5 - 11) = 1/3(3-1) = 1/3(2)
= '''2/3'''
a2 = s(d11d32 - d12d31)
= 1/3(21 - 11.5) = 1/3(2-1.5) = 1/3(.5)
= '''1/6'''
a3 = (1 - a1 - a2)
= 1 - 2/3 - 1/6
= '''1/6'''
where
s = 1/(d11d22 - d12d12)
= 1/(22 - 11) = 1/(4-1)
= '''1/3'''
where
dij = ei · ej
where
e1 = p2 - p1 = '''(1, 1, 0)'''
e2 = p3 - p1 = '''(1, 0, 1)'''
e3 = p - p1 = '''(1, .5. 0)'''
then plug in a1, a2, a3 to calculate the barycentric coordinates:
pb = a1p1 + a2p2 + a3*p3
= 2/3(-1, 0, 0) + 1/6(0, 1, 0) + 1/6(0, 0, 1)
= 2/3, 0, 0) + (0, 1/6, 0) + (0, 0, 1/6)
= '''(-2/3, 1/6, 1/6)'''
(
Edited: 2019-05-19)
9. Express the point (0,0.5,0) in barycentric coordinates with respect to the triangle with vertices (−1,0,0), (0,1,0), (0,0,1)
if given point p = (0, .5, 0) & triangle vertices p1 = (-1, 0, 0), p2 = (0, 1, 0), p3 = (0, 0, 1) we can calculate
a1 = s(d22*d31 - d12*d32)
= 1/3(2*1.5 - 1*1) = 1/3(3-1) = 1/3(2)
= '''2/3'''
a2 = s(d11*d32 - d12*d31)
= 1/3(2*1 - 1*1.5) = 1/3(2-1.5) = 1/3(.5)
= '''1/6'''
a3 = (1 - a1 - a2)
= 1 - 2/3 - 1/6
= '''1/6'''
where
s = 1/(d11*d22 - d12*d12)
= 1/(2*2 - 1*1) = 1/(4-1)
= '''1/3'''
where
dij = ei · ej
where
e1 = p2 - p1 = '''(1, 1, 0)'''
e2 = p3 - p1 = '''(1, 0, 1)'''
e3 = p - p1 = '''(1, .5. 0)'''
then plug in a1, a2, a3 to calculate the barycentric coordinates:
pb = a1*p1 + a2*p2 + a3*p3
= 2/3(-1, 0, 0) + 1/6(0, 1, 0) + 1/6(0, 0, 1)
= 2/3, 0, 0) + (0, 1/6, 0) + (0, 0, 1/6)
= '''(-2/3, 1/6, 1/6)'''