• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! 6E

Expert-verified 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 # In exercises 1-6, determine which sets of vectors are orthogonal.$$\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]$$, $$\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]$$, $$\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]$$

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}{ 5}\\{-4}\\{ 0}\\3\end{align}} \right], {u_2} = \left[ {\begin{align}{ - 4}\\1\\{ - 3}\\8\end{align}} \right] and {u_3} = \left[ {\begin{align}3\\3\\5\\{ - 1}\end{align}} \right].

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

\begin{align}{c}{u_1} \cdot {u_2} = \left( 5 \right)\left( { - 4} \right) + \left( { - 4} \right)\left( 1 \right) + \left( 0 \right)\left( { - 3} \right) + \left( 3 \right)\left( 8 \right)\\ = - 20 - 4 + 0 + 24\\ = 0\end{align}

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

\begin{align}{c}{u_2} \cdot {u_3} = \left( { - 4} \right)\left( 3 \right) + \left( 1 \right)\left( 3 \right) + \left( { - 3} \right)\left( 5 \right) + \left( 8 \right)\left( { - 1} \right)\\ = - 12 + 3 - 15 - 8\\ = - 32\end{align}

Since $${u_2} \cdot {u_3} \ne 0$$, hence, the given set is not orthogonal. ### Want to see more solutions like these? 