If the vectors $\vec{a}=2 \hat{i}+p \hat{j}+4 \hat{k}$ and $\vec{b}=6 \hat{i}-9 \hat{j}+q \hat{k}$ are collinear,then the values of $p$ and $q$ are:

The vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$. The length of the median through $A$ is:

$\hat{i}-2 \hat{j}+\hat{k}$, $2 \hat{i}+\hat{j}-\hat{k}$, and $\hat{i}-\hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$, and $C$ of a triangle $ABC$ respectively. If $D$ and $E$ are the midpoints of $BC$ and $CA$ respectively, then the unit vector along $\overrightarrow{DE}$ is

If the position vectors of the points $A$ and $B$ are $\vec{a} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} - 3\hat{k}$,then what will be the position vector of the midpoint of $AB$?

The position vector of point $C$ relative to $B$ is $(\hat{i} + \hat{j})$ and the position vector of $B$ relative to $A$ is $(\hat{i} - \hat{j})$. The position vector of $C$ relative to $A$ is:

Let $p = (x + 4y)\vec{a} + (2x + y + 1)\vec{b}$ and $q = (y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}$,where $\vec{a}$ and $\vec{b}$ are non-collinear vectors. If $3p = 2q$,then the values of $x$ and $y$ are: