If the vectors  $\overline{AB}=3\stackrel{\wedge }{i}+4\stackrel{\wedge }{k}and \overline{AC}=5\stackrel{\wedge }{i}-2\stackrel{\wedge }{j}+4\stackrel{\wedge }{k}$  are the sides of a triangle ABC, then    the length of the median through A is 

We have,
 $\overline{AB}+\overline{BC}+\overline{CA}=0\Rightarrow \overline{BC}=\overline{AC}-\overline{AB}$
Let M be mid point of BC
Now,  $\overline{BM}=\frac{\overline{AC}-\overline{AB}}{2}(\because \overline{BM}=\frac{\overline{BC}}{2})$

Also, we have

$$\begin{gathered} \Rightarrow \overline{AB}+\frac{\overline{AC}-\overline{AB}}{2}=\overline{AM} \\ \Rightarrow \frac{\overline{AB}+\overline{AC}}{2}=4\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+4\stackrel{\wedge }{k} \\ \Rightarrow \overline{AM}=\sqrt{33} \end{gathered}$$