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4E

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Found in: Page 331

### 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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# In exercises 1-6, determine which sets of vectors are orthogonal.\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right], \left[ {\begin{align}0\\0\\0\end{align}} \right], \left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]

The given set is orthogonal.

See the step by step solution

## Step 1: Definition of an orthogonal set

If ${{\bf{u}}_i} \cdot {{\bf{u}}_j} = 0$ for $i \ne j$, then the set of vectors $\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}$ is said to be orthogonal.

## Step 2: Check for orthogonality of vectors

Let the given vectors be, {u_1} = \left[ {\begin{align}2\\{-5}\\{-3}\end{align}} \right], {u_2} = \left[ {\begin{align}0\\0\\0\end{align}} \right] and {u_3} = \left[ {\begin{align}4\\{ - 2}\\6\end{align}} \right].

First, find $${u_1} \cdot {u_2}$$:

\begin{align}{c}{u_1} \cdot {u_2} = \left( 2 \right)\left( 0 \right) + \left( { - 5} \right)\left( 0 \right) + \left( { - 3} \right)\left( 0 \right)\\ = 0 - 0 - 0\\ = 0\end{align}

Now, find $${u_2} \cdot {u_3}$$:

\begin{align}{c}{u_2} \cdot {u_3} = \left( 0 \right)\left( 4 \right) + \left( 0 \right)\left( { - 2} \right) + \left( 0 \right)\left( 6 \right)\\ = 0 - 0 + 0\\ = 0\end{align}

And find $${u_1} \cdot {u_3}$$:

\begin{align}{c}{u_1} \cdot {u_3} = \left( 2 \right)\left( 4 \right) + \left( { - 5} \right)\left( { - 2} \right) + \left( { - 3} \right)\left( 6 \right)\\ = 8 + 10 - 18\\ = 0\end{align}

Since all the pairs are orthogonal hence, the given set is orthogonal.

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