In the above figure,$P$ divides $AC$ in the ratio $3:4$ and $Q$ divides $BC$ in the ratio $4:3$. Then $M$ divides $AQ$ in the ratio:

The vector projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)$ and $D \equiv(7,0,10)$,is

If a particle is acted upon by forces of magnitude $6$ and $7$ units in the directions of $-\hat{i} - 2\hat{j} + 2\hat{k}$ and $2\hat{i} - 3\hat{j} - 6\hat{k}$ respectively,and it undergoes a displacement from point $P(2, -1, -3)$ to $Q(5, -1, 1)$,then the work done by the forces is .......... units.

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

If $\vec{i}+\vec{j}-\vec{k}, -\vec{i}+2\vec{j}+\vec{k}, \vec{j}+2\vec{k}, 2\vec{i}-\vec{j}+2\vec{k}$ are the position vectors of four points $A, B, C, D$ respectively,then the shortest distance between the lines $AB$ and $CD$ is

The angle between two diagonals of a cube is: