The system of unit vectors $i, j, k$ is

$a$ 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 $x = \ldots$

$A$ vector $\vec{r}$ of magnitude $3\sqrt{2}$ units which makes angles of $\pi/4$ and $\pi/2$ respectively with $Y$ and $Z$ axes is:

$A$ point $C$ with position vector $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}$ (where $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar vectors) divides the line segment joining $A$ and $B$ in the ratio $2:1$. If the position vector of $A$ is $\bar{a}-2 \bar{b}+3 \bar{c}$,then find the position vector of $B$.

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Reason $(R): (\vec{x} + \vec{y})^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.

The position vectors of $P$ and $Q$ are respectively $\overrightarrow{a}$ and $\overrightarrow{b}$. If $R$ is a point on the line $PQ$ such that $\overrightarrow{PR}=5 \overrightarrow{PQ}$,then the position vector of $R$ is