## Linear Algebra: A Modern Introduction

$\sqrt{17}$
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: $d=||\begin{bmatrix} -1 \\ 2 \\ \end{bmatrix}-\begin{bmatrix} 3 \\ 1 \\ \end{bmatrix}||=||\begin{bmatrix} -4 \\ 1 \\ \end{bmatrix}||=\sqrt{(-4)^2+1^2}=\sqrt{16+1}=\sqrt{17}$