Repeat Exercise 25 with\({v_1} = \left[ {\begin{array}{*{20}{c}}1\\{\bf{2}}\\{ - {\bf{4}}}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{8}}\\{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{{\bf{10}}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{8}}\end{array}} \right]\), and \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{.{\bf{9}}}\\{{\bf{2}}.{\bf{0}}}\\{ - {\bf{3}}.{\bf{7}}}\end{array}} \right]\).

Recall that the typical point in the plane can be written as

\(x = \left( {1 - {c_2} - {c_3}} \right){v_1} + {c_2}{v_2} + {c_3}{v_3}\).

For the intersection of the plane with the ray, use the equation\({\bf{x}} = {\bf{a}} + t{\bf{b}}\).

It becomes as shown:

\(\begin{array}{c}\left( {1 - {c_2} - {c_3}} \right){v_1} + {c_2}{v_2} + {c_3}{v_3} = {\bf{a}} + t{\bf{b}}\\{c_2}\left( {{v_2} - {v_1}} \right) + {c_3}\left( {{v_3} - {v_1}} \right) + t\left( { - {\bf{b}}} \right) = {\bf{a}} - {v_1}\end{array}\)

Write the equation\({c_2}\left( {{v_2} - {v_1}} \right) + {c_3}\left( {{v_3} - {v_1}} \right) + t\left( { - {\bf{b}}} \right) = {\bf{a}} - {v_1}\)in the matrix form as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{v_2} - {v_1}}&{{v_3} - {v_1}}&{ - {\bf{b}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{c_2}}\\{{c_3}}\\t\end{array}} \right] = {\bf{a}} - {v_1}\)

Obtain the vectors\({v_2} - {v_1}\)and \({v_3} - {v_1}\) as shown below:

\({{\bf{v}}_2} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}7\\0\\{ - 1}\end{array}} \right]\)

And

\({{\bf{v}}_3} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}2\\8\\2\end{array}} \right]\)

Also, \({\bf{a}} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\\{12}\end{array}} \right]\).

The augmented matrix is shown below:

\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 0.9}&{ - 1}\\0&8&{ - 2}&{ - 2}\\{ - 1}&2&{3.7}&{12}\end{array}} \right]\)

Row-reduce theechelon form of\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 0.9}&{ - 1}\\0&8&{ - 2}&{ - 2}\\{ - 1}&2&{3.7}&{12}\end{array}} \right]\)as shown below:

\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 0.9}&{ - 1}\\0&8&{ - 2}&{ - 2}\\{ - 1}&2&{3.7}&{12}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3.7}&{ - 12}\\0&8&{ - 2}&{ - 2}\\0&{16}&{25}&{83}\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3.7}&{ - 12}\\0&8&{ - 2}&{ - 2}\\0&{16}&{25}&{83}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3.7}&{ - 12}\\0&4&{ - 1}&{ - 1}\\0&0&{29}&{87}\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3.7}&{ - 12}\\0&4&{ - 1}&{ - 1}\\0&0&{29}&{87}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 2}&0&{ - 0.9}\\0&4&0&2\\0&0&1&3\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&0&{ - 0.9}\\0&4&0&2\\0&0&1&3\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{0.1}\\0&1&0&{0.5}\\0&0&1&3\end{array}} \right]\)

Thus, the values are

\(\begin{array}{c}{c_2} = 0.1,\\{c_3} = 0.5,\\t = 3.\end{array}\)

Obtain the intersection point using the equation\({\bf{x}} = {\bf{a}} + t{\bf{b}}\).

\(\begin{array}{c}{\bf{x}} = {\bf{a}} + 3{\bf{b}}\\ = \left[ {\begin{array}{*{20}{c}}0\\0\\8\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}{.9}\\{2.0}\\{ - 3.7}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{2.7}\\6\\{ - 3.1}\end{array}} \right]\end{array}\)

The equation\(x = \left( {1 - {c_2} - {c_3}} \right){v_1} + {c_2}{v_2} + {c_3}{v_3}\)becomes as shown below:

\(\begin{array}{c}{\bf{x}} = \left( {1 - 0.1 - 0.5} \right){{\bf{v}}_1} + 0.1{{\bf{v}}_2} + 0.5{{\bf{v}}_3}\\ = 0.4\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 4}\end{array}} \right] + 0.1\left[ {\begin{array}{*{20}{c}}8\\2\\5\end{array}} \right] + 0.5\left[ {\begin{array}{*{20}{c}}3\\{10}\\{ - 2}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{2.7}\\6\\{ - 3.1}\end{array}} \right]\end{array}\)

From the above coordinate, all thebarycentric coordinates are positive.

It shows that the intersection point lies inside the triangle.