Let the position vectors of two points $A$ and $B$ be $\vec{a}+\vec{b}+\vec{c}$ and $\vec{a}-2\vec{b}+3\vec{c}$,respectively. If the points $P$ and $Q$ divide $AB$ in the ratio $1:3$ internally and externally respectively,then $3|AB|=$

If $C$ is the mid-point of the line segment $AB$ and $P$ is any point outside the line $AB$,then

Let $ABC$ be a triangle and $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $A, B, C$ respectively. Let $D$ divide $BC$ in the ratio $3:1$ internally and $E$ divide $AD$ in the ratio $4:1$ internally. Let $BE$ meet $AC$ in $F$. If $E$ divides $BF$ in the ratio $3:2$ internally,then the position vector of $F$ is

Let $\vec{u}, \vec{v}, \vec{w}$ be vectors such that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$. If $|\vec{u}| = 3$,$|\vec{v}| = 4$,and $|\vec{w}| = 5$,then $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}$ is:

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.

$a$ 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 $x = \ldots$