If $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-3 \hat{j}-5 \hat{k}$ are the position vectors of the points $A$ and $B$ respectively, $C$ divides $AB$ in the ratio $2:3$ and $M$ is the mid-point of $AB$, then $5(\text{position vector of } C) - 2(\text{position vector of } M) =$

The magnitude of the projection of the vector $\vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ on the unit vector $\hat{i}$ is . . . . . . .

Let $A, B, C$ be three points whose position vectors are $\overrightarrow{a} = \hat{i} + 4\hat{j} + 3\hat{k}$,$\overrightarrow{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$ (where $\alpha \in R$),and $\overrightarrow{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear,then the length of the median in $\triangle ABC$ through $A$ is:

Let $|\vec{a}| = |\vec{b}| = |\vec{a} - \vec{b}| = 1$,then the angle between $\vec{a}$ and $\vec{b}$ is:

If $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$,then under what condition are the three points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ collinear?

If $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$,then $\vec{a} + \vec{b} = \dots$