Question: 25. Let \(p = \left( \begin{array}{l}1\\1\end{array} \right)\). Find a hyperplane \(\left( {f:d} \right)\) that strictly separates \(B\left( {0,3} \right)\) and \(B\left( {p,1} \right)\). (Hint: After finding \(f\), show that the point \(v = \left( {1 - .75} \right)0 + .75p\) is neither in \(B\left( {0,3} \right)\) nor in \(B\left( {p,1} \right)\)).

Short Answer

Expert verified

The hyperplane is \(f\left( {x:d} \right) = f\left( {x:n \cdot x = 12.75} \right)\).

Step by step solution

01

Find the hyper plane with normal vector \(p\)

Consider the length of the line \(L\) joining thecentercentreof \(B\left( {0,3} \right)\) and \(B\left( {p,1} \right)\) is \(\sqrt {{4^2} + {1^2}} = 4.12\).

This distance is greater than the overall radii of the balls. So, thedisk areisuntouched.

This does not require the hyperplane to be normaltoto \(L\). So, a hyperplane whose normal vector is p would work. The hyperplane \(H\) with normal vector \(p\) is \(f\left( x \right) = p \cdot x\).

02

Find a point \(q\) on the line \(L\)

As the radii of the balls are 3 and 1, the point \(d\) should be \(\frac{3}{4}\) of the distance from the centers of the ball. So, obtain a point \(q\) at the distance ratio of \(\frac{3}{4}\) on line \(L\) as shown below:

\(\begin{array}{c}q = \left( {.25} \right)0 + \left( {0.75} \right)p\\ = \left( {3,\,\,.75} \right)\end{array}\)

03

Evaluate \(d\)as \(d = f\left( q \right)\)

The real number d is shown below:

\(\begin{array}{c}d = f\left( q \right)\\ = p \cdot q\\ = 4\left( 3 \right) + 1\left( {.75} \right)\\ = 12.75\end{array}\)

04

Find the hyperplane \(f\left( {x:d} \right)\)

The required hyper plane is \(f\left( {x:n \cdot x = 12.75} \right)\).

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