Let $ABC$ be a triangle. Let $u = \overrightarrow{AB}$ and $v = \overrightarrow{AC}$. If $D$ is the midpoint of $BC$,then the length of the median of $\triangle ABD$ through the vertex $B$ is:

Let $\vec{u}, \vec{v}$ and $\vec{w}$ be vectors such that $|\vec{u}+\vec{v}+\vec{w}|=0$. If $|\vec{u}|=3$,$|\vec{v}|=4$ and $|\vec{w}|=5$,then the value of $|\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}|$ is

$A, B, C, D$ are four points in a plane with position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ respectively such that $(\overline{a}-\overline{d}) \cdot(\overline{b}-\overline{c})=(\overline{b}-\overline{d}) \cdot(\overline{c}-\overline{a})=0$. Then the point $D$ is the $\dots$ of $\triangle ABC$.

If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $\frac{\pi}{3}$,then the value of $|\vec{a}+\vec{b}|$ is

The set of all real values of $c$ such that the angle between the vectors $\vec{a} = cx \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2cx \hat{k}$ is an obtuse angle for all real $x$ is:

If $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$,then $a$ is equal to