If in the given figure $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$ and $AP : PB = m : n$,then $\overrightarrow{OP} = $

$3\overrightarrow{OD} + \overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC} = $

If the position vectors of $A$ and $B$ are $i + 3j - 7k$ and $5i - 2j + 4k$,then the direction cosine of $\overrightarrow{AB}$ along the $y$-axis is

Let $A, B, C$ be three points whose position vectors are $\overrightarrow{a} = \hat{i} + 4\hat{j} + 3\hat{k}$,$\overrightarrow{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$ (where $\alpha \in R$),and $\overrightarrow{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear,then the length of the median in $\triangle ABC$ through $A$ is:

If $O$ is the origin and the position vector of $A$ is $4\,i + 5\,j$,then a unit vector parallel to $\overrightarrow{OA}$ is:

If $a$ and $b$ are two non-collinear vectors and $xa + yb = 0$,then: