In a quadrilateral $ABCD$,$M$ and $N$ are the mid-points of the sides $AB$ and $CD$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$,then $t =$

The angle made by the position vector of the point $(5, -4, -3)$ with the positive direction of the $X$-axis is:

If in a triangle $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AC} = \vec{b}$,and $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively,then $\overrightarrow{DE}$ is equal to:

Show that the vectors $2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+6 \hat{j}-8 \hat{k}$ are collinear.

The position vectors of the points $A$ and $B$ are $\vec{a}$ and $\vec{b}$ respectively. If the position vector of the point $C$ is $\frac{\vec{a}}{2} + \frac{\vec{b}}{3}$,then:

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$,then $|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2=$