If $C$ is the mid-point of the line segment $AB$ and $P$ is any point outside the line $AB$,then

$A, B, C, D, E$ are five coplanar points,then $\overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC} + \overrightarrow{AE} + \overrightarrow{BE} + \overrightarrow{CE}$ is equal to

Vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are of the same length and they make equal angles with each other when taken in pairs. If $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c}$ makes an obtuse angle with the $x$-axis,find the vector $\vec{c}$.

Let $\vec{\alpha} = (\lambda - 2) \vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors where $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear is:

If a point $C$ divides the line segment joining the points with the position vectors $2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $2: 3$,then the distance of $C$ from the point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ is

The value of $k$ for which the vectors $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = -2\hat{i} + k\hat{j}$ are collinear is