If the vectors $\vec{AB} = 3\hat i +4\hat j + 4\hat k$ and $\vec{AC}=5\hat i-2\hat j + 4\hat k$ are the sides of a ΔABC, then length of the median through A is

$\sqrt{33}$

Let D be the mid-point of BC. Then,

$\vec{AD}=\frac{\vec{AB}+\vec{AC}}{2}=4\hat i+\hat j+4\hat k$

$⇒|\vec{AD}|=\sqrt{16+1+16}=\sqrt{33}$

Hence, required length = $\sqrt{33}$ units.