The 2 vectirs j +k and 3i-j+k represents the tow sides AB and AC respectively of a triangle ABC.the length if the median through A is

given: ABC is a triangle and D is the mid-point of BC.
let the position vector of the vertices A, B and C are $\overline{a} ,\overline{b} and \overline{c}$ .
given : $\overline{AB}=j+k and \overline{AC} =3i-j+k$
$i.e. \overline{b}-\overline{a}=j+k ....\left(1\right) and \overline{c}-\overline{a} =3i-j+k.......\left(2\right)$
adding eq(1) and eq(2): we have $\overline{b}+\overline{c}-2.\overline{a} =3i+2k$
the position vector of D is $\frac{\overline{b}+\overline{c}}{2}$  [since D is the mid-point of B and C
$$\begin{gathered} \overline{AD}=\frac{\overline{b}+\overline{c}}{2}-\overline{a}=\frac{\overline{b}+\overline{c}-2.\overline{a}}{2} \\ =\frac{3i+2k}{2} \\ \left|AD\right|=\frac{\sqrt{3^2+2^2}}{2}=\frac{\sqrt{9+4}}{2} \\ \left|AD\right|=\frac{\sqrt{13}}{2} \end{gathered}$$

hope this helps you