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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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 \(\left\{ {\left( \begin{array}{l}x\\y\end{array} \right):4x - 2y = 17} \right\}\).

See the step by step solution

Step by Step Solution

Step 1: Assume the vector \(n\) 

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

Step 2: Find the vector \(x\)

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:

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

Step 3: Find the required hyperplane

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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