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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

# Compute the quantities in Exercises 1-8 using the vectors{\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)8. $$\left\| {\mathop{\rm x}\nolimits} \right\|$$

The length of x is $$\left\| {\mathop{\rm x}\nolimits} \right\| = 7$$.

See the step by step solution

## Step 1: The length of a vector

The nonnegative scalar $$\left\| {\mathop{\rm v}\nolimits} \right\|$$ is defined as length (or norm) as shown below:

$$\left\| {\mathop{\rm v}\nolimits} \right\| = \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } = \sqrt {v_1^2 + v_2^2 + \cdots + v_n^2} ,$$ and $${\left\| {\mathop{\rm v}\nolimits} \right\|^2} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits}$$

## Step 2: Compute

$$\left\| {\mathop{\rm x}\nolimits} \right\|$$

It is given that {\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right).

Compute the length of $${\mathop{\rm x}\nolimits}$$ as shown below:

\begin{aligned}{c}\left\| {\mathop{\rm x}\nolimits} \right\| &= \sqrt {{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} } \\ &= \sqrt {{6^2} + {{\left( { - 2} \right)}^2} + {3^2}} \\ &= \sqrt {36 + 4 + 9} \\ &= \sqrt {49} \\ &= 7\end{aligned}

Thus, the length of x is $$\left\| {\mathop{\rm x}\nolimits} \right\| = 7$$.