Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

Chapter 1 - Vectors - 1.2 Length and Angle: The Dot Product - Exercises 1.2 - Page 290: 16

Answer

$\sqrt{26.42}$

Work Step by Step

I know that for the vector $v=\begin{bmatrix} v_{1} \\ v_{2} \\ \vdots\\ v_{n} \end{bmatrix}$ $||v||=\sqrt{v_1^2+v_2^2+...+v_n^2}$ The distance of two vectors $u$ and $v$ is: $d=||u-v||$. Hence: $d=||\begin{bmatrix} 3.2 \\ -0.6 \\ -1.4 \\ \end{bmatrix}-\begin{bmatrix} 1.5 \\ 4.1 \\ -0.2\\ \end{bmatrix}||=||\begin{bmatrix} 1.7 \\ -4.7 \\ -1.2\\ \end{bmatrix}||=\sqrt{1.7^2+(-4.7)^2+(-1.2)^2}=\sqrt{2.89+22.09+1.44}=\sqrt{26.42}$
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