If $\overline{a} = m \overline{b} + n \overline{c}$,where $\overline{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$,$\overline{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$,and $\overline{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$,then $m + n =$

If $a$ is collinear with $b = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$ and $a \cdot b = 27$,then $|a| =$

The vectors $\overrightarrow{AB} = 3 \hat{i} + 4 \hat{k}$ and $\overrightarrow{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $ABC$. The length of the median through $A$ is

If $2\vec{a} - 3\vec{b}$,$\vec{b}$,and $\vec{a} - \vec{b}$ are the position vectors of three points $A$,$B$,and $C$ respectively,then they are:

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 5$ and $|\vec{a} + \vec{b} - \vec{c}|^2 + |\vec{b} + \vec{c} - \vec{a}|^2 + |\vec{c} + \vec{a} - \vec{b}|^2 = 50$. Then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = $

If $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ of triangle $ABC$ respectively,then $\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} =$