Answer
$||v||=\sqrt{6}$.
\begin{bmatrix}
\frac{1}{\sqrt{6}} \\
\frac{\sqrt{2}}{\sqrt{6}} \\
\frac{\sqrt{3}}{\sqrt{6}} \\
\frac{0}{\sqrt{6}} \\
\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: $||v||=\sqrt{1^2+\sqrt2^2+\sqrt3^2+0^2}=\sqrt{1+2+3+0}=\sqrt{6}$.
Thus, the unit vector in the direction of $v$ is: \begin{bmatrix}
\frac{1}{\sqrt{6}} \\
\frac{\sqrt{2}}{\sqrt{6}} \\
\frac{\sqrt{3}}{\sqrt{6}} \\
\frac{0}{\sqrt{6}} \\
\end{bmatrix}
