Let \({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{{}}{ - {\bf{1}}}\\{ - {\bf{3}}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{{}}{\bf{0}}\\{ - {\bf{3}}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{ - {\bf{1}}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right)\), \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{ - {\bf{1}}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{{}}{\bf{0}}\\{ - {\bf{2}}}\\{\bf{2}}\end{aligned}} \right)\),

and \(S = \left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}},\,{{\bf{v}}_{\bf{3}}},\,{{\bf{v}}_{\bf{4}}}} \right\}\). Determine whether \({{\bf{p}}_{\bf{1}}}\) and \({{\bf{p}}_{\bf{2}}}\) are in conv S?

Since the points \(\left\{ {{{\bf{v}}_1},\,{{\bf{v}}_2},\,{{\bf{v}}_3},\,{{\bf{v}}_4}} \right\}\) are not orthogonal, the translated points can be calculated as shown below:

\(\begin{aligned}{}{{\bf{v}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}1\\0\\{ - 3}\end{aligned}} \right)\\{{\bf{v}}_3} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}2\\2\\0\end{aligned}} \right)\\{{\bf{v}}_4} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}2\\4\\{ - 6}\end{aligned}} \right)\\{{\bf{p}}_1} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}2\\2\\{ - 2}\end{aligned}} \right)\\{{\bf{p}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{{}}1\\1\\{ - 2}\end{aligned}} \right)\end{aligned}\)

The system of the equation can be written as shown below:

\(\begin{aligned}{l}{c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + {c_4}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_1} - {{\bf{v}}_1}\\{c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + {c_4}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_2} - {{\bf{v}}_1}\end{aligned}\)

The augmented matrix can be written as shown below:

\(\left( {\left. {\begin{aligned}{{}}{{{\bf{v}}_2} - {{\bf{v}}_1}}&{{{\bf{v}}_3} - {{\bf{v}}_1}}&{{{\bf{v}}_4} - {{\bf{v}}_1}}\end{aligned}} \right|\,{{\bf{p}}_1} - {{\bf{v}}_1}} \right) = \left( {\begin{aligned}{{}}1&2&2&2\\0&2&4&2\\{ - 3}&0&{ - 6}&{ - 2}\end{aligned}} \right)\)

The augmented matrix is shown below:

\(M = \left( {\begin{aligned}{{}}1&2&2&2\\0&2&4&2\\{ - 3}&0&{ - 6}&{ - 2}\end{aligned}} \right)\)

Apply row operations:

\(\begin{aligned}{}M &= \left( {\begin{aligned}{{}}1&2&2&2\\0&2&4&2\\0&6&0&4\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to {R_3} + 3{R_1}} \right\}\\ &= \left( {\begin{aligned}{{}}1&2&2&2\\0&1&2&1\\0&0&{ - 12}&{ - 2}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{aligned}{}{R_3} \to {R_3} - 3{R_2}\\{R_2} \to \frac{1}{2}{R_2}\end{aligned} \right\}\\ &= \left( {\begin{aligned}{{}}1&2&2&2\\0&1&2&1\\0&0&1&{\frac{1}{6}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to - \frac{1}{{12}}{R_3}} \right\}\end{aligned}\)

So, by row reduced form, the values of scalars are:

\({c_4} = \frac{1}{6}\)

And,

\(\begin{aligned}{}{c_3} + 2{c_4} &= 1\\{c_3} + 2\left( {\frac{1}{6}} \right) &= 1\\{c_3} &= \frac{2}{3}\end{aligned}\)

And,

\(\begin{aligned}{}{c_2} + 2{c_3} + 2{c_4} &= 2\\{c_2} + 2\left( {\frac{2}{3}} \right) + 2\left( {\frac{1}{6}} \right) &= 2\\{c_2} &= \frac{1}{3}\end{aligned}\)

So, the value of \({{\bf{p}}_1}\) can be calculated as shown below:

\(\begin{aligned}{}\frac{1}{3}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + \frac{2}{3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + \frac{1}{6}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) &= {{\bf{p}}_1} - {{\bf{v}}_1}\\{{\bf{p}}_1} &= - \frac{1}{6}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{2}{3}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\end{aligned}\)

As all the coefficients are not positive, therefore \({{\bf{p}}_1} \notin {\rm{conv}}\,\,S\).

For the equation \({c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + {c_4}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_2} - {{\bf{v}}_1}\) augmented matrix:

\(\left( {\left. {\begin{aligned}{{}}{{{\bf{v}}_2} - {{\bf{v}}_1}}&{{{\bf{v}}_3} - {{\bf{v}}_1}}&{{{\bf{v}}_4} - {{\bf{v}}_1}}\end{aligned}} \right|\,\,{{\bf{p}}_2} - {{\bf{v}}_1}} \right) = \left( {\begin{aligned}{{}}1&2&2&1\\0&2&4&1\\{ - 3}&0&{ - 6}&{ - 2}\end{aligned}} \right)\)

Apply row operations:

\(\begin{aligned}{}M = \left( {\begin{aligned}{{}}1&2&2&1\\0&2&4&1\\0&6&0&1\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to {R_3} + 3{R_1}} \right\}\\ = \left( {\begin{aligned}{{}}1&2&2&1\\0&1&2&{\frac{1}{2}}\\0&0&{ - 12}&{ - 2}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{aligned}{l}{R_3} \to {R_3} - 3{R_2}\\{R_2} \to \frac{1}{2}{R_2}\end{aligned} \right\}\\ = \left( {\begin{aligned}{{}}1&2&2&1\\0&1&2&{\frac{1}{2}}\\0&0&1&{\frac{1}{6}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to \frac{1}{{12}}{R_3}} \right\}\end{aligned}\)

So, by row reduced form, the values of scalars are shown below:

\({c_4} = \frac{1}{6}\)

And,

\(\begin{aligned}{}{c_3} + 2{c_4} = 1\\{c_3} + 2\left( {\frac{1}{6}} \right) = \frac{1}{2}\\{c_3} = \frac{1}{6}\end{aligned}\)

And,

\(\begin{aligned}{}{c_2} + 2{c_3} + 2{c_4} = \frac{1}{3}\\{c_2} + 2\left( {\frac{1}{6}} \right) + 2\left( {\frac{1}{6}} \right) = 1\\{c_2} = \frac{1}{3}\end{aligned}\)

So, the value of \({{\bf{p}}_1}\) can be calculated as,

\(\begin{aligned}{}\frac{1}{3}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + \frac{1}{6}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + \frac{1}{6}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right) = {{\bf{p}}_2} - {{\bf{v}}_1}\\{{\bf{p}}_2} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\end{aligned}\)

As all the coefficients are positive, so\(\left( {\frac{1}{3} + \frac{1}{3} + \frac{1}{6} + \frac{1}{6} = 1} \right)\). Therefore,\({{\bf{p}}_2} \in {\rm{conv}}\,\,S\).

Therefore, \({{\bf{p}}_1} \notin {\rm{conv}}\,\,S\) and \({{\bf{p}}_2} \in {\rm{conv}}\,\,S\).