Let\({v_1} = \left[ {\begin{array}{*{20}{c}}1\\3\\{ - 6}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{7}}\\3\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{9}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{9}}\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{1.4}\\{{\bf{1}}.{\bf{5}}}\\{ - {\bf{3}}.{\bf{1}}}\end{array}} \right]\), and \({\bf{x}}\left( t \right) = {\bf{a}} + t{\bf{b}}\)for \(t \ge {\bf{0}}\).Find the point where the ray\({\bf{x}}\left( t \right)\)intersects the plane that contains the triangle with vertices\({v_1}\),\({v_{\bf{2}}}\), and\({v_{\bf{3}}}\). Is this point inside the triangle?

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:

\({{\bf{v}}_2} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}6\\0\\1\end{array}} \right]\)

And

\({{\bf{v}}_3} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}2\\6\\4\end{array}} \right]\)

Also, \({\bf{a}} - {v_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 3}\\{15}\end{array}} \right]\)

The augmented matrix is shown below:

\(\left[ {\begin{array}{*{20}{c}}6&2&{ - 1.4}&{ - 1}\\0&6&{ - 1.5}&{ - 3}\\1&4&{3.1}&{15}\end{array}} \right]\)

Row-reduce theechelon form of\(\left[ {\begin{array}{*{20}{c}}6&2&{ - 1.4}&{ - 1}\\0&6&{ - 1.5}&{ - 3}\\1&4&{3.1}&{15}\end{array}} \right]\)as shown below:

\(\left[ {\begin{array}{*{20}{c}}6&2&{ - 1.4}&{ - 1}\\0&6&{ - 1.5}&{ - 3}\\1&4&{3.1}&{15}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&4&{3.1}&{15}\\0&6&{ - 1.5}&{ - 3}\\0&{ - 22}&{ - 20}&{ - 91}\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}1&4&{3.1}&{15}\\0&6&{ - 1.5}&{ - 3}\\0&{ - 22}&{ - 20}&{ - 91}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&4&{3.1}&{15}\\0&2&{ - 0.5}&{ - 1}\\0&{ - 22}&{ - 20}&{ - 91}\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}1&4&{3.1}&{15}\\0&2&{ - 0.5}&{ - 1}\\0&{ - 22}&{ - 20}&{ - 91}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{0.6}\\0&2&0&1\\0&0&1&4\end{array}} \right]\)

Thus, the values are

\(\begin{array}{c}{c_2} = 0.6,\\{c_3} = 0.5,\\t = 4.\end{array}\)

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

\(\begin{array}{c}{\bf{x}} = {\bf{a}} + 4{\bf{b}}\\ = \left[ {\begin{array}{*{20}{c}}0\\0\\9\end{array}} \right] + 4\left[ {\begin{array}{*{20}{c}}{1.4}\\{1.5}\\{ - 3.1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5.6}\\6\\{ - 3.4}\end{array}} \right]\end{array}\)

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

\(\begin{array}{c}{\bf{x}} = \left( {1 - 0.6 - 0.5} \right){{\bf{v}}_1} + 0.6{{\bf{v}}_2} + 0.5{{\bf{v}}_3}\\ = - 0.1\left[ {\begin{array}{*{20}{c}}1\\3\\{ - 6}\end{array}} \right] + 0.6\left[ {\begin{array}{*{20}{c}}7\\3\\{ - 5}\end{array}} \right] + 0.5\left[ {\begin{array}{*{20}{c}}3\\9\\{ - 2}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5.6}\\6\\{ - 3.4}\end{array}} \right]\end{array}\)

From the above coordinate, the firstbarycentric coordinate is negative.

It shows that the intersection point is not inside the triangle.