Question

Exercises 7-10 use the terminology from section 8.2.

7. a. Let \(T = \left\{ {\left( {\begin{aligned}{ {20}{c}}{ - {\bf{1}}}\\{\bf{0}}\end{aligned}} \right),\,\left( {\begin{aligned}{ {20}{c}}{\bf{2}}\\{\bf{3}}\end{aligned}} \right),\,\,\left( {\begin{aligned}{ {20}{c}}{\bf{4}}\\{\bf{1}}\end{aligned}} \right)} \right\}\), and let \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{ {20}{c}}{\bf{2}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{ {20}{c}}{\bf{3}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{p}}_{\bf{3}}} = \left( {\begin{aligned}{ {20}{c}}{\bf{2}}\\{\bf{0}}\end{aligned}} \right)\), and \({{\bf{p}}_{\bf{4}}} = \left( {\begin{aligned}{ {20}{c}}{\bf{0}}\\{\bf{2}}\end{aligned}} \right)\). Find the barycentric coordinates of \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\), and \({{\bf{p}}_{\bf{4}}}\) with respect to T.

b. Use your answers in part (a) to determine whether each of \({{\bf{p}}_{\bf{1}}}\),…,\({{\bf{p}}_{\bf{4}}}\) in part (a) is inside, outside, or on the edge of conv T, a triangular region.

Short answer

a. \(\left( {\frac{1}{3},\,\frac{1}{6},\frac{1}{2}} \right)\), \(\left( {0,\frac{1}{2},\frac{1}{2}} \right)\), \(\left( {\frac{1}{2}, - \frac{1}{4},\frac{3}{4}} \right)\), and \(\left( {\frac{1}{2},\frac{3}{4}, - \frac{1}{4}} \right)\).

b. \({{\bf{p}}_1}\) will be inside conv T, \({{\bf{p}}_2}\) is on edge and, \({{\bf{p}}_3}\) and \({{\bf{p}}_4}\) is outside conv T.

Step-by-step solution

Find the augmented matrix

Using the vectors of T and \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), \({{\bf{p}}_3}\), and \({{\bf{p}}_4}\) the augmented matrix can be expressed as shown below:

\(\left( {\begin{aligned}{ {20}{c}}{\widetilde {{{\bf{v}}_1}}}&{\widetilde {{{\bf{v}}_2}}}&{\widetilde {{{\bf{v}}_3}}}&{\widetilde {{{\bf{p}}_1}}}&{\widetilde {{{\bf{p}}_2}}}&{\widetilde {{{\bf{p}}_3}}}&{\widetilde {{{\bf{p}}_4}}}\end{aligned}} \right) \sim \left( {\begin{aligned}{ {20}{c}}1&1&1&1&1&1&1\\{ - 1}&2&4&2&3&2&0\\0&3&1&1&2&0&2\end{aligned}} \right)\)

Obtain the row reduced form of the augmented matrix

\(\begin{aligned}{c}M = \left( {\begin{aligned}{ {20}{c}}1&1&1&1&1&1&1\\0&3&5&3&4&3&1\\0&3&1&1&2&0&2\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to {R_2} + {R_1}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}1&1&1&1&1&1&1\\0&1&{\frac{5}{3}}&1&{\frac{4}{3}}&1&{\frac{1}{3}}\\0&3&1&1&2&0&2\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to \frac{1}{3}{R_2}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}1&1&1&1&1&1&1\\0&1&{\frac{5}{3}}&1&{\frac{4}{3}}&1&{\frac{1}{3}}\\0&0&{ - 4}&{ - 2}&{ - 2}&{ - 3}&1\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_3} \to {R_3} - 3{R_2}} \right)\end{aligned}\)

Row reduce further,

\(\begin{aligned}{c}M = \left( {\begin{aligned}{ {20}{c}}1&1&1&1&1&1&1\\0&1&{\frac{5}{3}}&1&{\frac{4}{3}}&1&{\frac{1}{3}}\\0&0&1&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{3}{4}}&{ - \frac{1}{4}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_3} \to - \frac{1}{4}{R_3}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}1&1&1&1&1&1&1\\0&1&0&{\frac{1}{6}}&{\frac{1}{2}}&{ - \frac{1}{4}}&{\frac{3}{4}}\\0&0&1&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{3}{4}}&{ - \frac{1}{4}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to {R_2} - \frac{5}{3}{R_3}} \right)\,\\ = \left( {\begin{aligned}{ {20}{c}}1&0&0&{\frac{1}{3}}&0&{\frac{1}{2}}&{\frac{1}{2}}\\0&1&0&{\frac{1}{6}}&{\frac{1}{2}}&{ - \frac{1}{4}}&{\frac{3}{4}}\\0&0&1&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{3}{4}}&{ - \frac{1}{4}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_1} \to {R_1} - {R_2}\\{R_1} \to {R_1} - {R_3}\end{aligned} \right)\end{aligned}\)

Conclusions from the row-reduced matrix

In the row reduced matrix column 4, column 5, column 6, and column 7 represents the barycentric coordinates of \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), \({{\bf{p}}_3}\), and \({{\bf{p}}_4}\) respectively.

So, for \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), \({{\bf{p}}_3}\), and \({{\bf{p}}_4}\) the barycentric coordinate are \(\left( {\frac{1}{3},\,\frac{1}{6},\frac{1}{2}} \right)\), \(\left( {0,\frac{1}{2},\frac{1}{2}} \right)\), \(\left( {\frac{1}{2}, - \frac{1}{4},\frac{3}{4}} \right)\), and \(\left( {\frac{1}{2},\frac{3}{4}, - \frac{1}{4}} \right)\) respectively.

Find the solution for part (b)

As barycentric coordinate of \({{\bf{p}}_3}\) and \({{\bf{p}}_4}\) has one negative value, therefore \({{\bf{p}}_3}\) and \({{\bf{p}}_4}\) are outside conv T.

As all coordinates for \({{\bf{p}}_1}\) are positive, therefore, it will be inside conv T. For \({{\bf{p}}_2}\), its first coordinate is zero, therefore \({{\bf{p}}_2}\) is on the edge of conv T.