Answer
$||v||=\sqrt{13}$.
\begin{bmatrix}
\frac{3}{\sqrt{13}} \\
\frac{-2}{\sqrt{13}} \\
\end{bmatrix}
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 unit vector in the direction of $v$ is $\frac{v}{||v||}$.
Hence: $||v||=\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}$.
Thus, the unit vector in the direction of $v$ is: \begin{bmatrix}
\frac{3}{\sqrt{13}} \\
\frac{-2}{\sqrt{13}} \\
\end{bmatrix}
