$ABCD$ is a parallelogram and $P$ is a point on the segment $AD$ dividing it internally in the ratio $3:1$. If the line $BP$ meets the diagonal $AC$ in $Q$, then $AQ:QC$ equals

Let the vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be such that $|\overrightarrow{a}|=2, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=4$. If the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is equal to the projection of $\overrightarrow{c}$ on $\overrightarrow{a}$ and $\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$,then the value of $|\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}|$ is

Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=|\bar{b}|$ and $|\bar{a}+2 \bar{b}|=|2 \bar{a}-\bar{b}|$. If $\bar{c}$ is a vector parallel to $\bar{a}$, then the angle between $\bar{b}$ and $\bar{c}$ is (in $^{\circ}$)

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

$ABC$ is a right-angled triangle in which $\max \{AB, BC, AC\} = BC$. If the position vectors of $B$ and $C$ are respectively $3\hat{i}-2\hat{j}+\hat{k}$ and $5\hat{i}+\hat{j}-3\hat{k}$,then find the value of $AB \cdot AC + BA \cdot BC + CA \cdot CB$.

The moment about the point $M(-2, 4, -6)$ of the force represented in magnitude and position by $\overrightarrow{AB}$,where the points $A$ and $B$ have the coordinates $(1, 2, -3)$ and $(3, -4, 2)$ respectively,is: