91Ó°ÊÓ

The vector \(p = \left( \begin{array}{l}2\\3\end{array} \right)\) , then triangle formed by the vectors \({v_1},{v_2},{v_3}\) has its closest side to \(p\) as \(\overline {{v_1}{v_3}} \) . It is shown below:

\(\begin{array}{c}\overline {{{\bf{v}}_1}{{\bf{v}}_3}} = \left( {\begin{array}{*{20}{c}}{4 - 1}\\{4 - 2}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3\\2\end{array}} \right)\end{array}\)

The vector orthogonal to another vector is shown below:

\(\begin{array}{c}\overline {{{\bf{v}}_2}{{\bf{v}}_3}} \cdot {\bf{n}} = 0\\\left( {\begin{array}{*{20}{c}}3\\2\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{{n_1}}\\{{n_2}}\end{array}} \right) = 0\\3{n_1} + 2{n_2} = 0\end{array}\)

Many orthogonal vectors are on the closest side \(\overline {{v_1}{v_2}} \). One of them is \(n = \left( {\,\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right)\).

Now the required hyperplane becomes \(f\left( {{x_1},{x_2}} \right) = - 2{x_1} + 3{x_2}\). For this hyperplane, \(f\left( p \right) = 5\) and \(f\left( {{v_1}} \right) = f\left( {{v_2}} \right) = 5\) .

So, its range of \(d\) is \(4 < d < 5\).