## Linear Algebra: A Modern Introduction

$||v||=3.54$. \begin{bmatrix} \frac{3.2}{3.54} \\ \frac{-0.6}{3.54} \\ \frac{-1.4}{3.54} \\ \end{bmatrix}
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 unit vector in the direction of $v$ is $\frac{v}{||v||}$. Hence: $||v||=\sqrt{3.2^2+(-0.6)^2+(-1.4)^2}=\sqrt{10.24+0.36+1.96}=\sqrt{12.56}\approx3.54$. Thus, the unit vector in the direction of $v$ is: \begin{bmatrix} \frac{3.2}{3.54} \\ \frac{-0.6}{3.54} \\ \frac{-1.4}{3.54} \\ \end{bmatrix}