Answer
$||v||=4.41$.
\begin{bmatrix}
\frac{1.12}{4.41} \\
\frac{-3.25}{4.41} \\
\frac{2.07}{4.41} \\
\frac{-1.83}{4.41} \\
\end{bmatrix}
Work Step by Step
I know that for the vector $v=\begin{bmatrix}
v_{1} \\
v_{2} \\
\vdots\\
v_{n}
\end{bmatrix}$
The unit vector in the direction of $v$ is $\frac{v}{||v||}$.
Hence: $||v||=\sqrt{1.12^2+(-3.25)^2+2.07^2+(-1.83)^2}=\sqrt{1.2544+10.5625+4.2849+3.3489}=\sqrt{19.4507}\approx4.41$.
Thus, the unit vector in the direction of $v$ is: \begin{bmatrix}
\frac{1.12}{4.41} \\
\frac{-3.25}{4.41} \\
\frac{2.07}{4.41} \\
\frac{-1.83}{4.41} \\
\end{bmatrix}
