If $D, E, F$ are the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$,then $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}$ is:

$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. Also,it is given that $\frac{OA}{CB}=2$,$\frac{OD}{AB}=\frac{1}{3}$. If $\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$,then $\vec{AD}+\vec{OC}+\vec{DC}=$

$a$ and $b$ are non-collinear vectors. If $c=(x-2)a+b$ and $d=(2x+1)a-b$ are collinear vectors,then the value of $x = \ldots$

If the position vectors of $A, B, C,$ and $D$ are $2i + j,$ $i - 3j,$ $3i + 2j,$ and $i + \lambda j$ respectively and $\overrightarrow{AB} \parallel \overrightarrow{CD},$ then the value of $\lambda$ is:

If $\vec{a}, \vec{b}, \vec{c}$ are $3$ vectors such that $|\vec{a}|=5, |\vec{b}|=8, |\vec{c}|=11$ and $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

Let $u, v$ and $w$ be three vectors in $R^3$. Then,any vector $z \in R^3$ can be written as $z = au + bv + cw$ for some scalars $a, b$ and $c$ if and only if: