Question

Question: In Exercise 7, let Hbe the hyperplane through the listed points. (a) Find a vector n that is normal to the hyperplane. (b) Find a linear functional f and a real number d such that \(H = \left( {f:d} \right)\).

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{4}}\\{\bf{1}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{ - {\bf{2}}}\\{\bf{5}}\end{array}} \right)\)

Short answer

1. The normal vector is \(n = \left( {\begin{array}{*{20}{c}}0\\2\\3\end{array}} \right)\) or a multiple
2. The linear function is \(f\left( x \right) = 2{x_2} + 3{x_3}\) , and the real number is \(d = 11\).

Step-by-step solution

Write the given data

Let \({v_1} = \left( {\begin{array}{*{20}{c}}1\\1\\3\end{array}} \right)\), \({v_2} = \left( {\begin{array}{*{20}{c}}2\\4\\1\end{array}} \right)\), and \({v_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\\5\end{array}} \right)\).

Then, \({v_2} - {v_1} = \left( {\begin{array}{*{20}{c}}1\\3\\{ - 2}\end{array}} \right),\) and \({v_3} - {v_1} = \left( {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\2\end{array}} \right)\).

Use the cross product to compute n

(a)

\(\begin{array}{c}n = \left( {{v_2} - {v_1}} \right) \times \left( {{v_3} - {v_1}} \right)\\ = \left| {\begin{array}{*{20}{c}}1&{ - 2}&i\\3&{ - 3}&j\\{ - 2}&2&k\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}3&{ - 3}\\{ - 2}&2\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 2}&2\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}1&{ - 2}\\3&{ - 3}\end{array}} \right|k\\ = 0i + 2j + 3k\end{array}\)

Thus, the normal vector is \(n = \left( {\begin{array}{*{20}{c}}0\\2\\3\end{array}} \right)\).

Find a linear functional f and a real number d

(b)

Using part (a) to obtain the linear functional f as shown below:

\(\begin{array}{c}f\left( x \right) = n \cdot x\\ = \left( {\begin{array}{*{20}{c}}0&2&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\f\left( x \right) = 2{x_2} + 3{x_3}\end{array}\)

Note that, \({v_i}\) in \(H = \left( {f:d} \right)\) such that, \(f\left( {{v_i}} \right) = d\) for \(i = 1,2,3\).

\(\begin{array}{c}d = f\left( {{v_1}} \right)\\ = f\left( {1,1,3} \right)\\ = 2\left( 1 \right) + 3\left( 3 \right)\\ = 2 + 9\\d = 11\end{array}\)

Thus, the linear function is \(f\left( x \right) = 2{x_2} + 3{x_3}\) , and the real number is \(d = 11\).