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

$\sqrt{18}$ $\sqrt{72}$ $\sqrt{33}$ $\sqrt{45}$

$\sqrt{33}$

Let D be the mid-point of BC. Then, $\vec{AD}=\frac{\vec{AB}+\vec{AC}}{2}$ $⇒\vec{AD}=\frac{(3\hat i+4\hat k)(5\hat i-2\hat j+4\hat k)}{2}=4\hat i+\hat j+4\hat k$ $⇒|\vec{AD}|=\sqrt{16+1+16}=\sqrt{33}$