If $2 \hat{i}+\hat{j}-\hat{k}$,$\hat{i}-3 \hat{j}+5 \hat{k}$ and $-3 \hat{i}+4 \hat{j}+4 \hat{k}$ are the position vectors of three points $A$,$B$ and $C$ respectively,then

$A$ and $B$ are two points. The position vector of $A$ is $6b - 2a$. $A$ point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,then the position vector of $B$ is given by:

If $2 \overrightarrow{a} + 3 \overrightarrow{b} - 5 \overrightarrow{c} = \overrightarrow{0}$,then the ratio in which $\overrightarrow{c}$ divides $\overrightarrow{AB}$ is

The projections of a vector on the three coordinate axes are $6, -3, 2$ respectively. The direction cosines of the vector are:

Given $\overrightarrow{p} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$,$\overrightarrow{a} = \hat{i} + \hat{j}$,$\overrightarrow{b} = \hat{j} + \hat{k}$,$\overrightarrow{c} = \hat{i} + \hat{k}$ and $\overrightarrow{p} = x \overrightarrow{a} + y \overrightarrow{b} + z \overrightarrow{c}$,then $x, y, z$ are respectively:

The vectors $a = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $b$ are collinear and $|b| = 21$,then $b =$