Question

Question: Let \({{\bf{a}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{1}}}\\{\bf{5}}\end{array}} \right)\), \({{\bf{a}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{a}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{6}}\\{\bf{0}}\end{array}} \right)\), \({{\bf{b}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{ - {\bf{2}}}\end{array}} \right)\),\({{\bf{b}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{2}}\\{\bf{1}}\end{array}} \right)\) and \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\), and let \(A = \left\{ {{{\bf{a}}_{\bf{1}}},{{\bf{a}}_{\bf{2}}},{{\bf{a}}_{\bf{3}}}} \right\}\) and \(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\). Find a hyperplane H with normal n that separates A and B. Is there a hyperplane parallel to H that strictly separates A and B?

Short answer

\(\left( {{H_0}:f} \right) = 3{x_1} + {x_2} - 2{x_3}\)

Hyperplane H does not separate A and B.

Step-by-step solution

Solve the equation \(n{\bf{v}} = {\bf{0}}\)

Consider \({\bf{v}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right)\).

Let \({H_0}\) be the origin that contains the normal vector, and then it can be written as shown below:

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}3&1&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = 0\\3{x_1} + {x_2} - 2{x_3} = 0\end{array}\)

So, it can be written as \(\left( {{H_0}:f} \right) = 3{x_1} + {x_2} - 2{x_3}\).

Find the value of d

As hyperplaneis passing through 6 nonzero vectors. So, the value of d is shown below:

\(d = 6\)

Find the dot products of \({\bf{n}}\) with given vectors

The dot product of \(n\) and \({{\bf{a}}_1}\) is:

\(\begin{array}{c}f\left( {{{\bf{a}}_1}} \right) = n \cdot {{\bf{a}}_1}\\ = 3\left( 2 \right) + \left( { - 1} \right) - 2\left( 5 \right)\\ = - 5\end{array}\)

The dot product of \(n\) and \({{\bf{a}}_2}\) is:

\(\begin{array}{c}f\left( {{{\bf{a}}_2}} \right) = n \cdot {{\bf{a}}_2}\\ = 3\left( 3 \right) + \left( 1 \right) - 2\left( 3 \right)\\ = 4\end{array}\)

The dot product of \(n\) and \({{\bf{a}}_3}\) is:

\(\begin{array}{c}f\left( {{{\bf{a}}_3}} \right) = n \cdot {{\bf{a}}_3}\\ = 3\left( { - 1} \right) + \left( 6 \right) - 2\left( 0 \right)\\ = 3\end{array}\)

The dot product of \(n\) and \({{\bf{b}}_1}\) is:

\(\begin{array}{c}f\left( {{{\bf{b}}_1}} \right) = n \cdot {{\bf{b}}_1}\\ = 3\left( 0 \right) + \left( 5 \right) - 2\left( { - 1} \right)\\ = 7\end{array}\)

The dot product of \(n\) and \({{\bf{b}}_2}\) is:

\(\begin{array}{c}f\left( {{{\bf{b}}_2}} \right) = n \cdot {{\bf{b}}_2}\\ = 3\left( 1 \right) + \left( { - 3} \right) - 2\left( { - 2} \right)\\ = 4\end{array}\)

The dot product of \(n\) and \({{\bf{b}}_3}\) is:

\(\begin{array}{c}f\left( {{{\bf{b}}_3}} \right) = n \cdot {{\bf{b}}_3}\\ = 3\left( 2 \right) + \left( 2 \right) - 2\left( 1 \right)\\ = 6\end{array}\)

It can be observed from the above equations \(f\left( A \right) < 4\) and \(f\left( B \right) > 4\), the parallel hyperplane \(\left( {f:4} \right)\) strictly separates A and B, then by theorem 13 H does not separate A and B.