Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors such that $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c} = \frac{3}{2}$. Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is:

Let the unit vectors $a$ and $b$ be perpendicular and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c = \alpha a + \beta b + \gamma (a \times b)$,then

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$,then $|\vec{a} \times \vec{c}|$ is equal to

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{14}$ and $\vec{a} \cdot \vec{b}=-7$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=$

$M$ and $N$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$ respectively,then $\overline{AM} + \overline{AN} =$

Let $a = \sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}$, where $x \in R$. If the pairs of vectors $(a, \hat{i})$, $(a, \hat{j})$, and $(a, \hat{k})$ are adjacent sides of $3$ distinct parallelograms and $A$ is the sum of the squares of the areas of these parallelograms, then $A$ lies in the interval