# Chapter 1 - Vectors - 1.2 Length and Angle: The Dot Product - Exercises 1.2 - Page 290: 15

$\sqrt{6}$

#### 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: $d=||\begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix}-\begin{bmatrix} 2 \\ 3 \\ 1\\ \end{bmatrix}||=||\begin{bmatrix} -1 \\ -1 \\ 2\\ \end{bmatrix}||=\sqrt{(-1)^2+(-1)^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.