Question

Repeat Exercise 9 for the points \({{\bf{q}}_{\bf{1}}}\),….\({{\bf{q}}_{\bf{5}}}\) whose barycentric coordinates with respect to S are given by \(\left( {\frac{{\bf{1}}}{{\bf{8}}},\frac{{\bf{1}}}{{\bf{4}}},\frac{{\bf{1}}}{{\bf{8}}},\frac{{\bf{1}}}{{\bf{2}}}} \right)\), \(\left( {\frac{{\bf{3}}}{{\bf{4}}}, - \frac{{\bf{1}}}{{\bf{4}}},{\bf{0}},\frac{{\bf{1}}}{{\bf{2}}}} \right)\),\(\left( {{\bf{0}},\frac{{\bf{3}}}{{\bf{4}}},\frac{{\bf{1}}}{{\bf{4}}},{\bf{0}}} \right)\),\(\left( {{\bf{0}}, - {\bf{2}},{\bf{0}},{\bf{3}}} \right)\), and \(\left( {\frac{{\bf{1}}}{{\bf{3}}},\frac{{\bf{1}}}{{\bf{3}}},\frac{{\bf{1}}}{{\bf{3}}},{\bf{0}}} \right)\), respectively.

Short answer

\({{\bf{q}}_1}\) is inside the tetrahedron \({\rm{conv}}\,S\), \({{\bf{q}}_2}\) is outside \({\rm{conv}}\,S\), \({{\bf{q}}_3}\) is the edge between \({{\bf{v}}_2}\) and \({{\bf{v}}_3}\). \({{\bf{q}}_4}\) is outside the tetrahedron and \({{\bf{q}}_5}\) is on the face with vertices \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\).

Step-by-step solution

Check for the first coordinate

The barycentric coordinateof the point \({{\bf{q}}_1}\) are all positive, so the point \({{\bf{q}}_1}\) is inside the tetrahedron convex S.

Check for the second coordinate

The barycentric coordinate of the point \({{\bf{q}}_2}\)is not positive, so the point \({{\bf{q}}_2}\) is outside the tetrahedron convexS.

Check for the third coordinate

The first and fourth barycentric coordinate of the point \({{\bf{q}}_3}\)is zero, representing the edge between \({{\bf{v}}_2}\) and \({{\bf{v}}_3}\).

Check for the fourth coordinate

The barycentric coordinate of the point \({{\bf{q}}_4}\)is not positive, so the point \({{\bf{q}}_4}\) is outside the tetrahedron convex S.

Check for the fifth coordinate

The point \({{\bf{q}}_5}\) is a convex combination of vectors \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\), hence \({{\bf{q}}_5}\) lies on face containing the vertices \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\).