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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

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

The given set is an orthogonal set.

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 are, {u_1} = \left[ {\begin{align}{*{20}{c}}1\\{ - 2}\\1\end{align}} \right], {u_2} = \left[ {\begin{align}{*{20}{c}}0\\1\\2\end{align}} \right] and {u_3} = \left[ {\begin{align}{*{20}{c}}{ - 5}\\{ - 2}\\1\end{align}} \right].

First, find :

\begin{align}{c}{u_1} \cdot {u_2} = \left( 1 \right)\left( 0 \right) + \left( { - 2} \right)\left( 1 \right) + \left( 1 \right)\left( 2 \right)\\ = 0 - 2 + 2\\ = 0\end{align}

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

\begin{align}{c}{u_2} \cdot {u_3} = \left( 0 \right)\left( { - 5} \right) + \left( 1 \right)\left( { - 2} \right) + \left( 2 \right)\left( 1 \right)\\ = 0 - 2 + 2\\ = 0\end{align}

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

\begin{align}{c}{u_1} \cdot {u_3} = \left( 1 \right)\left( { - 5} \right) + \left( { - 2} \right)\left( { - 2} \right) + \left( 1 \right)\left( 1 \right)\\ = - 5 + 4 + 1\\ = 0\end{align}

Since all the pairs are orthogonal. Hence, the given set is an orthogonal set.