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Q26E

Expert-verifiedFound in: Page 437

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Question:** **26. Let \({\rm{q}} = \left( \begin{array}{l}2\\3\end{array} \right)\), \({\rm{p}} = \left( \begin{array}{l}6\\1\end{array} \right)\). Find a hyperplane \(\left( {f:d} \right)\) that strictly separates \(B\left( {{\rm{q}},3} \right)\) and \(B\left( {{\rm{p}},1} \right)\).**

The equation of the hyperplane is \(\left\{ {\left( \begin{array}{l}x\\y\end{array} \right):4x - 2y = 17} \right\}\).

The line segment joins \(p\) and \(q\) is perpendicular to the separating hyperplane. Thus, \(n = \left( {p - q} \right)\) is equal to \(\left( \begin{array}{l}4\\ - 2\end{array} \right)\).

The distance between \(p\) and \(q\) is \(\sqrt {{4^2} + {{\left( { - 2} \right)}^2}} = \sqrt {20} \), which is greater than the overall radii of the balls.

The centre of the larger ball is \(q\). The point situated at the \(\frac{3}{4}\) of the distance between p and q is 3 units far from q and 1 unit far from p.

Thus, the corresponding point is shown below:

\(\begin{array}{c}x = .75p + .25q\\ = .75\left( \begin{array}{l}6\\1\end{array} \right) + .25\left( \begin{array}{l}2\\3\end{array} \right)\\ = \left( \begin{array}{c}5.0\\1.5\end{array} \right)\end{array}\)

The dot product n.x is shown below:

\begin{gathered} nx = 4 \cdot 5 - 2 \cdot \left( {1.5} \right) \\ = 20 - 3 \\ = 17 \\ \end{gathered}

So, the required hyperplane is \(\left\{ {\left( \begin{array}{l}x\\y\end{array} \right):4x - 2y = 17} \right\}\).

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