$A$ vector of magnitude $\sqrt{2}$ units along the internal bisector of the angle between the vectors $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ is

If the position vectors of two points $A$ and $B$ are $\vec{a} + 3\vec{b}$ and $\vec{a} - 2\vec{b}$ respectively,find the position vector of the point that divides $AB$ in the ratio $2:5$.

In the given figure,identify which of the vectors are coinitial.

If $M_1, M_2, M_3$ and $M_4$ are respectively the magnitudes of the vectors $\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = -3\hat{i} - 4\hat{j} - 4\hat{k}$,$\vec{a}_3 = -\hat{i} + \hat{j} - \hat{k}$,and $\vec{a}_4 = -\hat{i} + 3\hat{j} + \hat{k}$,then the correct order of $M_1, M_2, M_3$ and $M_4$ is:

The position vectors of two points $A$ and $B$ are $\bar{i}+2\bar{j}+3\bar{k}$ and $7\bar{i}-\bar{k}$ respectively. The point $P$ with position vector $-2\bar{i}+3\bar{j}+5\bar{k}$ is on the line $AB$. If the point $Q$ is the harmonic conjugate of $P$ with respect to $A$ and $B$, then the sum of the scalar components of the position vector of $Q$ is

The points with position vectors $60\,i + 3\,j$,$40\,i - 8\,j$,and $a\,i - 52\,j$ are collinear,if $a = $