Question

Question: In Exercise 8, 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 \(H = \left( {f:d} \right)\).

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

Short answer

1. The normal vector is \(n = \left( {\begin{array}{*{20}{c}}4\\3\\{ - 6}\end{array}} \right)\) or a multiple
2. The linear functional is \(f\left( x \right) = 4{x_1} + 3{x_2} - 6{x_3}\) , and the real number is \(d = - 8\).

Step-by-step solution

Write the given data

Let \({v_1} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\), \({v_2} = \left( {\begin{array}{*{20}{c}}4\\{ - 2}\\3\end{array}} \right)\), and \({v_3} = \left( {\begin{array}{*{20}{c}}7\\{ - 4}\\4\end{array}} \right)\).

Then, the vectors are \({v_2} - {v_1} = \left( {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right),\) and \({v_3} - {v_1} = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\\3\end{array}} \right)\).

Use the cross product to compute n

(a)

\(\begin{array}{c}n = \left( {{v_2} - {v_1}} \right) \times \left( {{v_3} - {v_1}} \right)\\ = \left| {\begin{array}{*{20}{c}}3&6&i\\0&{ - 2}&j\\2&3&k\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}0&{ - 2}\\2&3\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}3&6\\2&3\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}3&6\\0&{ - 2}\end{array}} \right|k\\ = 4i + 3j - 6k\end{array}\)

Thus, the normal vector is \(n = \left( {\begin{array}{*{20}{c}}4\\3\\{ - 6}\end{array}} \right)\).

Find a linear functional f and a real number d

(b)

Using part (a), the linear functional f can be obtained as shown below:

\(\begin{array}{c}f\left( x \right) = n \cdot x\\ = \left[ {\begin{array}{*{20}{c}}4&3&{ - 6}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\f\left( x \right) = 4{x_1} + 3{x_2} - 6{x_3}\end{array}\)

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

\(\begin{array}{c}d = f\left( {{v_1}} \right)\\ = f\left( {1, - 2,1} \right)\\ = 4\left( 1 \right) + 3\left( { - 2} \right) - 6\left( 1 \right)\\ = 4 - 6 - 6\\d = - 8\end{array}\)

Thus, the linear function is \(f\left( x \right) = 4{x_1} + 3{x_2} - 6{x_3}\) , and the real number is \(d = - 8\).