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Q26E

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Found in: Page 437

### 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

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