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:

$2 \hat{i}-3 \hat{j}+\hat{k}$ and $\hat{i}+2 \hat{j}-3 \hat{k}$ are the position vectors of two points $A$ and $B$ respectively and $C$ divides $AB$ in the ratio $3:2$. If $3 \hat{i}-\hat{j}+2 \hat{k}$ is the position vector of a point $D$,then the unit vector in the direction of $\overrightarrow{CD}$ is

Given $p = 2a - 3b$,$q = a - 2b + c$,and $r = -3a + b + 2c$,where $a, b,$ and $c$ are non-zero,non-coplanar vectors,then the vector $-2a + 3b - c$ is equal to:

$A$ line segment $PQ$ has the length $63$ and direction ratios $(3, -2, 6)$. If this line makes an obtuse angle with the $X$-axis,then the components of the vector $\vec{PQ}$ 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:

The vector in the direction of the sum of the vectors $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$ is