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 $AB$?

If the vector $\bar{i}-7 \bar{j}+2 \bar{k}$ is along the internal bisector of the angle between the vectors $\bar{a}$ and $-2 \bar{i}-\bar{j}+2 \bar{k}$ and the unit vector along $\bar{a}$ is $x \bar{i}+y \bar{j}+z \bar{k}$ then $x=$

If $D, E, F$ are the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$ respectively,find the position vector of the centroid of $\Delta DEF$,given that the position vectors of $A, B, C$ are $i + j, j + k, k + i$ respectively.

Answer the following as true or false.
Two collinear vectors are always equal in magnitude.

Let the three sides of a triangle $ABC$ be represented by the vectors $\vec{AB} = 2\hat{i}-\hat{j}+\hat{k}$,$\vec{BC} = 3\hat{i}-4\hat{j}-4\hat{k}$,and $\vec{CA} = \hat{i}-3\hat{j}-5\hat{k}$. Let $G$ be the centroid of the triangle $ABC$. Then $6(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2)$ is equal to

The vector which is parallel to the resultant vector of $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} - \hat{k}$ and having a magnitude of $5$ units is . . . . . . .