In a parallelogram $ABCD$,$|\overline{AB}| = a$,$|\overline{AD}| = b$ and $|\overline{AC}| = c$,then the value of $\overline{DA} \cdot \overline{AB}$ is:

If three unit vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ satisfy $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

Find the angle $\theta$ between the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

If the position vectors of the vertices $A, B$,and $C$ of a triangle $ABC$ are $4\hat{i} + 7\hat{j} + 8\hat{k}$,$2\hat{i} + 3\hat{j} + 4\hat{k}$,and $2\hat{i} + 5\hat{j} + 7\hat{k}$ respectively,then the position vector of the point where the bisector of angle $A$ meets $BC$ is:

Let $u, v, w$ be three vectors such that $|u| = 1, |v| = 2, |w| = 3$. If the projection of $v$ along $u$ is equal to the projection of $w$ along $u$ and $v, w$ are perpendicular to each other, then $|u - v + w| = ...$

If two vectors $\vec{a}$ and $\vec{b}$ which are perpendicular to each other are such that $|\vec{a}|=8$ and $|\vec{b}|=3$, then $|\vec{a}-2\vec{b}|=$