If $\vec{PO} + \vec{OQ} = \vec{QO} + \vec{OR}$,then

Let $OA = a, OB = b$ be two non-collinear vectors,$OP = x_1 a + y_1 b, OQ = x_2 a + y_2 b$ and $A^{\prime}O = OA, B^{\prime}O = OB$. If $x_1 = -\frac{3}{4}, x_2 = \frac{1}{3}, y_1 = \frac{7}{4}, y_2 = \frac{5}{3}$,then

Find the position vector of a point $R$ which divides the line joining the two points $P$ and $Q$ with position vectors $\overrightarrow{OP} = 2\vec{a} + \vec{b}$ and $\overrightarrow{OQ} = \vec{a} - 2\vec{b}$ respectively,in the ratio $1:2$,$(i)$ internally and (ii) externally.

If $A, B, C$ are the vertices of a triangle whose position vectors are $a, b, c$ and $G$ is the centroid of the $\Delta ABC$,then $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}$ 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 $ABCD$ is a parallelogram,$\overrightarrow{AB} = 2i + 4j - 5k$ and $\overrightarrow{AD} = i + 2j + 3k,$ then the unit vector in the direction of $\overrightarrow{BD}$ is