Let $D, E, F$ be the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$ respectively. Then $\vec{AD} + \vec{BE} + \vec{CF} = \dots$

If $|\bar{a}|=3, |\bar{b}|=4, |\bar{a}-\bar{b}|=5$,then $|\bar{a}+\bar{b}|=$

If $r \cdot i = r \cdot j = r \cdot k$ and $|r| = 3$, then $r = $

Let $ABC$ be a triangle whose circumcentre is at $P$. If the position vectors of $A, B, C$ and $P$ are $\vec{a}, \vec{b}, \vec{c}$ and $\frac{\vec{a} + \vec{b} + \vec{c}}{4}$ respectively,then the position vector of the orthocentre of this triangle is:

Let $u$ and $v$ be two vectors. Then $|u-v|=||u|-|v||$ if and only if

If $ABCD$ is a parallelogram and the position vectors of $A, B, C$ are $i + 3j + 5k, i + j + k$ and $7i + 7j + 7k$, then the position vector of $D$ will be: