If $|\vec{a}|=1, |\vec{b}|=2, |\vec{a}-\vec{b}|^2+|\vec{a}+2\vec{b}|^2=20$,then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is:

The position vectors of the vertices of $\triangle ABC$ are $4\hat{i} - 2\hat{j}$,$\hat{i} + 4\hat{j} - 3\hat{k}$,and $-\hat{i} + 5\hat{j} + \hat{k}$ respectively. Then,$m \angle ABC = $

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}+\bar{c}|=1$ and $\bar{b}$ is perpendicular to $\bar{c}$. If $\bar{a}$ makes angles $\alpha$ and $\beta$ with $\bar{b}$ and $\bar{c}$ respectively,then the value of $\cos \alpha+\cos \beta$ is:

If $4\hat{i}+\hat{j}-\hat{k}$ and $3\hat{i}+m\hat{j}+2\hat{k}$ are at a right angle,then $m = $

Let $S$ be the set of all $a \in \mathbb{R}$ for which the angle between the vectors $\vec{u} = a(\log_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{v} = (\log_{e} b) \hat{i} + 2 \hat{j} + 2a(\log_{e} b) \hat{k}$ is acute,where $b > 1$. Then $S$ is equal to:

$ABCD$ 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