Question

Question: In Exercise 10, 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 \(H = \left( {f:d} \right)\).

10. \(\left( {\begin{array}{*{20}{c}}1\\2\\0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\2\\{ - 1}\\{ - 3}\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\3\\2\\7\end{array}} \right),\left( {\begin{array}{*{20}{c}}3\\2\\{ - 1}\\{ - 1}\end{array}} \right)\)

Short answer

1. The normal vector is \(n = \left( {\begin{array}{*{20}{c}}{ - 2}\\3\\{ - 5}\\1\end{array}} \right)\) or a multiple
2. The linear functional f is \(f\left( x \right) = - 2{x_1} + 3{x_2} - 5{x_3} + {x_4}\) , and the real number is \(d = 4\).

Step-by-step solution

Write the given data

Let the vectors are \({v_1} = \left( {\begin{array}{*{20}{c}}1\\2\\0\\0\end{array}} \right)\), \({v_2} = \left( {\begin{array}{*{20}{c}}2\\2\\{ - 1}\\{ - 3}\end{array}} \right)\), \({v_3} = \left( {\begin{array}{*{20}{c}}1\\3\\2\\7\end{array}} \right)\)and \({v_4} = \left( {\begin{array}{*{20}{c}}3\\2\\{ - 1}\\{ - 1}\end{array}} \right)\).

Then, \({v_2} - {v_1} = \left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\\{ - 3}\end{array}} \right),{v_3} - {v_1} = \left( {\begin{array}{*{20}{c}}0\\1\\2\\7\end{array}} \right),\) and \({v_4} - {v_1} = \left( {\begin{array}{*{20}{c}}2\\0\\{ - 1}\\{ - 1}\end{array}} \right)\).

Use the cross product to compute n

(a)

\(\begin{array}{c}n = \left( {{v_2} - {v_1}} \right) \times \left( {{v_3} - {v_1}} \right) \times \left( {{v_4} - {v_1}} \right)\\ = \left| {\begin{array}{*{20}{c}}i&j&k&l\\1&0&{ - 1}&{ - 3}\\0&1&2&7\\2&0&{ - 1}&{ - 1}\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}0&{ - 1}&{ - 3}\\1&2&7\\0&{ - 1}&{ - 1}\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}1&{ - 1}&{ - 3}\\0&2&7\\2&{ - 1}&{ - 1}\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}1&0&{ - 3}\\0&1&7\\2&0&{ - 1}\end{array}} \right|k - \left| {\begin{array}{*{20}{c}}1&0&{ - 1}\\0&1&2\\2&0&{ - 1}\end{array}} \right|l\\ = - 2i + 3j - 5k + l\end{array}\)

Thus, the normal vector is \(n = \left( {\begin{array}{*{20}{c}}{ - 2}\\3\\{ - 5}\\1\end{array}} \right)\).

Find a linear functional f and a real number d

(b)

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

\(\begin{array}{c}f\left( x \right) = n \cdot x\\ = \left( {\begin{array}{*{20}{c}}{ - 2}&3&{ - 5}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right)\\f\left( x \right) = - 2{x_1} + 3{x_2} - 5{x_3} + {x_4}\end{array}\)

Note that, \({v_i}\) in \(H = \left( {f:d} \right)\) such that \(f\left( {{v_i}} \right) = d\) for \(i = 1,2,3,4\).

\(\begin{array}{c}d = f\left( {{v_1}} \right)\\ = f\left( {1,2,0,0} \right)\\ = - 2\left( 1 \right) + 3\left( 2 \right) + 0 + 0\\ = - 2 + 6\\d = 4\end{array}\)

Thus, the linear functional f is \(f\left( x \right) = - 2{x_1} + 3{x_2} - 5{x_3} + {x_4}\) , and the real number is \(d = 4\).